Đề 1:Bài 1: Giải và biện luận các phương trình sau( gồm cả phương trình lượng giác):1) A32x−1+A42x−2=3x2+P22) 2sin(6x+π3)cos(2x)=tan(3x−π4)tan(3x−π6)3)...
Trả lời 06-05-20 11:59 AM
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lim
Trả lời 15-03-17 03:27 AM
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Tính : \mathop {\lim }\limits_{x \to 1}\frac{x^{2016}-2016x+2015}{(x-1)^2}
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\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt {1 + x} .\sqrt[3]{{1 + \frac{x}{2}}}.\sqrt[4]{{1 + \frac{x}{3}}} - \sqrt[4]{{1 - x}}}}{{\frac{3}{2}\sqrt {4 + x} - \sqrt[3]{{3 - x}} - \sqrt[4]{{1 + x}}}}
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g(x)=\left\{ \begin{array}{l} \frac{x^3-8}{\sqrt{x+2}-2} ,khi x > 2\\ 20x+8 ,khi x <2\\ m^2-5m+52, khi x = 2 \end{array} \right. Tại x_0 = 2
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L=\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{cosx}-\sqrt[3]{cosx}}{sin^2x}
Trả lời 21-05-14 12:15 PM
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\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x^2-2x+1}
Trả lời 11-05-14 02:07 AM
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\mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x^2-2x+1}
Trả lời 10-05-14 01:00 PM
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1) Tìm giới hạn của hàm số sau:\mathop {\lim }\limits_{x \to 0}\frac{\sqrt[3]{6x+1}-\sqrt{4x+1}}{1-cos2x}.
Trả lời 07-05-14 10:17 PM
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a)\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1} b)\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}c)\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}
Trả lời 28-04-14 09:18 PM
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a)\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1} b)\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}c)\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}
Trả lời 28-04-14 09:08 PM
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a)\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[4]{2x-1}+\sqrt[5]{x-2}}{x-1} b)\mathop {\lim }\limits_{x \to 1}\frac{x^3-\sqrt{3x-2}}{x-1}c)\mathop {\lim }\limits_{x \to 0}\frac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}
Trả lời 28-04-14 09:00 PM
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tìm các giới hạn sau.\mathop {\lim \frac{1-cos6x}{x^{2}cosx} }\limits_{x \to 0}\mathop {\lim \frac{1+sinx-cosx}{1-sinx-cosx}}\limits_{x \to 0}
Trả lời 18-04-14 09:36 PM
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1) Tính các giới hạn sau:a/\mathop {\lim }\limits_{x \to 2}\frac{(2x-1)^5-243}{x-2} b/\mathop {\lim }\limits_{x \to 4}\frac{\sqrt{(x-3)^3}-1}{16(2x-7)^5-x^2}*giải kĩ hộ em với, có giải nhưng khó hiêu quá!
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1) Tính các giới hạn sau:a/\mathop {\lim }\limits_{x \to 2}\frac{(2x-1)^5-243}{x-2} b/\mathop {\lim }\limits_{x \to 4}\frac{\sqrt{(x-3)^3}-1}{16(2x-7)^5-x^2}*giải kĩ hộ em với, có giải nhưng khó hiêu quá!
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1) \mathop {\lim }\limits_{x \to 1}\frac{x^m-1}{x^n-1}2) \mathop {\lim }\limits_{x \to 1}\frac{x^n-nx+n-1}{x-1}.
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Tìm \mathop {\lim }\limits_{x \to 1} \frac{x^{100}-1}{x^{99}-1}
Trả lời 20-03-14 08:34 PM
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4) \mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+x}\sqrt[3]{1+\frac{x}{2}}\sqrt[4]{1+\frac{x}{3}}-\sqrt[4]{1-x}}{\frac{3}{2}\sqrt{4+x}-\sqrt[3]{8-x}-\sqrt[4]{1-x}}5)\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[3]{6x^2+2}-2\sqrt{x}}{x^3-x^2-x+1}
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4) \mathop {\lim }\limits_{x \to 0}\frac{\sqrt{1+x}\sqrt[3]{1+\frac{x}{2}}\sqrt[4]{1+\frac{x}{3}}-\sqrt[4]{1-x}}{\frac{3}{2}\sqrt{4+x}-\sqrt[3]{8-x}-\sqrt[4]{1-x}}5)\mathop {\lim }\limits_{x \to 1}\frac{\sqrt[3]{6x^2+2}-2\sqrt{x}}{x^3-x^2-x+1}
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1) \mathop {\lim }\limits_{x \to 0} \frac{(1+x)(1+2x)(1+3x)(1+4x)-1}{x}2)\mathop {\lim }\limits_{x \to 0} \frac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[4]{1+4x}-1}{x}3)\mathop {\lim }\limits_{x \to 0}\frac{(x^2+2014)\sqrt[7]{1-2x}-2014}{x}cách làm bài...
Trả lời 17-03-14 07:07 AM
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