Quay lại đáp án

	2	bớt 87 ký tự trong đáp án		Sửa 02-03-16 11:25 PM Confusion 851 1 7
Sử dụng bất đẳng thức AM-GM,ta có:12(a2+b2+c2)+6abc+48−30(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">6VT=6(a2+b2+c2)+6abc+30−18(a+b+c)12(a2+b2+c2)+6abc+48−30(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">=6(a2+b2+c2)+3(2abc+1)+27−3.2.3(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥3(a2+b2+c2)+9a2b2c2−−−−−√3+27−3.((a+b+c)2+9)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥3(a2+b2+c2)+27a+b+c−6(ab+bc+ca)≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)Mặt khác sử dụng bất đẳng thức Schur,9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)Do đó7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">3(a2+b2+c2)+27a+b+c−6(ab+bc+ca)7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\geq 3(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−6(ab+bc+ca)a2=0≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0Bất đẳng thức được chứng minh. Sử dụng bất đẳng thức AM-GM,ta có:12(a2+b2+c2)+6abc+48−30(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">12(a2+b2+c2)+6abc+48−30(a+b+c)12(a2+b2+c2)+6abc+48−30(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥12(a2+b2+c2)+9a2b2c2−−−−−√3+45−5.((a+b+c)2+9)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)Mặt khác sử dụng bất đẳng thức Schur,9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)Do đó7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0Bất đẳng thức được chứng minh! Sử dụng bất đẳng thức AM-GM,ta có:12(a2+b2+c2)+6abc+48−30(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">6VT=6(a2+b2+c2)+6abc+30−18(a+b+c)12(a2+b2+c2)+6abc+48−30(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">=6(a2+b2+c2)+3(2abc+1)+27−3.2.3(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥3(a2+b2+c2)+9a2b2c2−−−−−√3+27−3.((a+b+c)2+9)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥3(a2+b2+c2)+27a+b+c−6(ab+bc+ca)≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)Mặt khác sử dụng bất đẳng thức Schur,9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)Do đó7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">3(a2+b2+c2)+27a+b+c−6(ab+bc+ca)7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\geq 3(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−6(ab+bc+ca)a2=0≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0Bất đẳng thức được chứng minh.

	1	tạo mới		Trả lời 02-03-16 11:02 PM Confusion 851 1 7
Sử dụng bất đẳng thức AM-GM,ta có:12(a2+b2+c2)+6abc+48−30(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">12(a2+b2+c2)+6abc+48−30(a+b+c)12(a2+b2+c2)+6abc+48−30(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)=12(a2+b2+c2)+3(2abc+1)+45−5.2.3(a+b+c)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥12(a2+b2+c2)+9a2b2c2−−−−−√3+45−5.((a+b+c)2+9)≥12(a2+b2+c2)+9a2b2c23+45−5.((a+b+c)2+9)\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}\displaystyle{=7(a^2+b^2+c^2)+\dfrac{9abc}{\sqrt[3]{abc}-10(ab+bc+ca)}≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)Mặt khác sử dụng bất đẳng thức Schur,9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)9a+b+c≥4(ab+bc+ca)−(a+b+c)2=2(ab+bc+ca)−(a2+b2+c2)Do đó7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)7(a2+b2+c2)+27a+b+c−10(ab+bc+ca)≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0" role="presentation" style="display: inline; font-size: 14px; word-spacing: 0px; color: rgb(40, 40, 40); font-family: helvetica, arial, sans-serif; position: relative; background-color: rgb(255, 255, 255);">≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0≥7(a2+b2+c2)+6(ab+bc+ca)−3(a2+b2+c2)−10(ab+bc+ca)=4(a2+b2+c2−ab−bc−ca)≥0Bất đẳng thức được chứng minh!

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Ngothikhuyen886: cho đoạn mạch như hình vẽ, dây nối A kể có điện trở k đáng kể, V rất lớn, 2 đầu đoạn mạch nối với hiệu điện thế U=2V / a, chỉnh biến trở để vôn kế chỉ 4A . Khi đó cường độ dòng điện qua A kế 5A. Tính điện trở của biến trở khi đó ? / b,phải chỉnh biến trở có điện trở bao nhiêu để có A chỉ 3A? 11/1/2022 9:41:58 PM
Ngothikhuyen886: đây ạ 11/1/2022 9:42:03 PM
Ngothikhuyen886: giúp mik với 11/1/2022 9:42:09 PM
Ngothikhuyen886: lớp 9 11/1/2022 9:42:11 PM
Ngothikhuyen886: 11/1/2022 9:44:19 PM
truongthithanhnhan99: hí ae 11/10/2022 7:32:16 AM
vanhieu21061979: hello 11/14/2022 7:58:01 PM
vanhieu21061979: anh em ơi 11/14/2022 7:58:18 PM
loll: giúp em sẽ gầy vs 11/23/2022 2:58:58 PM
loll: onichan 11/23/2022 3:00:55 PM
loll: yamate 11/23/2022 3:01:26 PM
loll: =00 11/23/2022 3:01:32 PM
loll: 11/23/2022 3:01:35 PM
Hiusegay: Hê lô kitty 11/23/2022 8:46:07 PM
kimyoungran227: 1/25/2023 8:14:22 PM

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