a) $\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{\frac{sinx-sinx.cosx}{cosx}}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx(\frac{1-cosx}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.(\frac{sin^2\frac{x}{2}}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.\frac{1}{cosx}(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{4cosx}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\frac{1}{4cosx}=\frac{1}{4}$

a) $\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{\frac{sinx-sinx.cosx}{cosx}}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx(\frac{1-cosx}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.(2\frac{sin^2\frac{x}{2}}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.\frac{2}{cosx}(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{2.cosx}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\frac{1}{2cosx}=\frac{1}{2}$

a) $\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx-sinx.cosx}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx(1-cosx)}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.sin^2(\frac{x}{2})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{4}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\frac{1}{4}$

a) $\mathop {\lim }\limits_{x \to 0}\frac{tanx-sinx}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{\frac{sinx-sinx.cosx}{cosx}}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx(\frac{1-cosx}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{sinx.(\frac{sin^2\frac{x}{2}}{cosx})}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x.\frac{sinx}{x}.\frac{1}{cosx}(\frac{x}{2})^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\mathop {\lim }\limits_{x \to 0}\frac{x^3\frac{sinx}{x}.\frac{1}{4cosx}.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2}{x^3}$=$\frac{1}{4cosx}=\frac{1}{4}$

a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$c. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$

a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$c. $y'=3sin^2(2x+1)[sin(2x+1)]'=3sin^2(2x+1)[cos(2x+1).(2x+1)']=6sin^2(2x+1).cos(2x+1)=3sin(4x+2).sin(2x+1)$

a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$c. $y'=2sin(2x+1)[sin(2x+1)]=2sin(2x+1)[cos(2x+1).(2x+1)']=4sin(2x+1).cos(2x+1)=2sin(4x+2)$

a. $y'=2(\frac{sinx}{1+cosx})(\frac{sinx}{1+cosx})'=2(\frac{sinx}{1+cosx})[\frac{(1+cosx)}{(1+cosx)^2}]=\frac{2sinx}{(1+cosx)^2}$b. $y'=x'.cosx+(cosx)'.x=cosx-x.sinx$c. $y'=3sin^2(2x+1)[sin(2x+1)]'=2sin^2(2x+1)[cos(2x+1).(2x+1)']=4sin^2(2x+1).cos(2x+1)=2sin(4x+2).sin(2x+1)$

Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$Ta đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$Ta có$\mathop {\lim }\limits_{x \to \frac{\pi}{6}}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{1}{2(\frac{1}{12}+\frac{x}{2})}=\frac{6}{1+\pi}$

Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$Ta đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$$u_{n} =\frac{sin2t}{cos\frac{\pi}{6}-cos(2t+\frac{\pi}{6})}=\frac{sin2t}{2sin(\frac{\pi}{6}+t)sin2t}=\frac{1}{2sin(\frac{\pi}{6}+t)}$Ta có$\mathop {\lim }\limits_{x \to \frac{\pi}{6}}u_{n}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{(\frac{\pi}{6}+t)}{2(\frac{\pi}{6}+t).sin(\frac{\pi}{6}+t)}=\mathop {\lim }\limits_{x \to \frac{\pi}{6}}\frac{1}{2(\frac{\pi}{12}+\frac{x}{2})}=\frac{3}{\pi}$

$4(sin^4x+cos^4x)+sin4x-2=0$<=>$4(sin^2x+cos^2x)^2-8sin^2x.cos^2x+sin4x-2=0$<=>$1-2sin^2x+sin4x+1=0$<=>$cos4x+sin4x=-1$<=>$cos(\frac{\pi}{4}-4x)=-\frac{\sqrt{2}}{2}$Dạng cơ bản rồi!

$4(sin^4x+cos^4x)+sin4x-2=0$<=>$4(sin^2x+cos^2x)^2-8sin^2x.cos^2x+sin4x-2=0$<=>$1-2sin^22x+sin4x+1=0$<=>$cos4x+sin4x=-1$<=>$cos(\frac{\pi}{4}-4x)=-\frac{\sqrt{2}}{2}$Dạng cơ bản rồi!

$4(sin^4x+cos^4x)+sin4x-2=0$<=>$4(sin^2x+cos^2x)^2-8sin^2x.cos^2x+sin4x-2=0$<=>$2(1-sin^22x)+sin4x=0$<=>$2cos^24x+sin4x=0$<=>$2(cos2x-sin2x)^2+2sin2x.cos2x=0$Dạng cơ bản rồi! Đặt $t=cos2x-sin2x$

$4(sin^4x+cos^4x)+sin4x-2=0$<=>$4(sin^2x+cos^2x)^2-8sin^2x.cos^2x+sin4x-2=0$<=>$1-2sin^2x+sin4x+1=0$<=>$cos4x+sin4x=-1$<=>$cos(\frac{\pi}{4}-4x)=-\frac{\sqrt{2}}{2}$Dạng cơ bản rồi!

Câu d*$\mathop {\lim }\limits_{x \to +\infty }\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}}=\mathop {\lim }\limits_{x \to +\infty }\frac{\sqrt{x}(\sqrt{1+\sqrt{\frac{1}{x}}})}{\sqrt{x}(\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}})}=1$Còn $x\rightarrow -\infty $ tương tự câu trên lưu ý đặt dấu trừ trước căn nhé!

Câu d*$\mathop {\lim }\limits_{x \to +\infty }\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}+\sqrt{x}}}=\mathop {\lim }\limits_{x \to +\infty }\frac{\sqrt{x}(\sqrt{1+\sqrt{\frac{1}{x}}})}{\sqrt{x}(\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}+1})}=\frac{1}{2}$Còn $x\rightarrow -\infty $ tương tự câu trên lưu ý đặt dấu trừ trước căn nhé!

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(sin\frac{x}{2}-cos\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

Câu a.$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{1-sinx}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{(cos\frac{x}{2}-sin\frac{x}{2})^2}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{2sin^2[\frac{1}{2}(\frac{\pi}{2}-x)]}{(\frac{\pi}{2}-x)^2}$=$\mathop {\lim }\limits_{x \to \frac{\pi}{2}}\frac{\frac{1}{2}(\frac{\pi}{2}-x)^2.(\frac{sin[\frac{1}{2}(\frac{\pi}{2}-x]}{\frac{1}{2}(\frac{\pi}{2}-x)})^2}{(\frac{\pi}{2}-x)^2}$=$\frac{1}{2}$

$4(sin^2\frac{x}{2}+cos^2\frac{x}{2})^2-2sin^2\frac{x}{2}.cos^2\frac{x}{2}-\sqrt{3}sin2x=4$$<=>4-\frac{1}{2}.sin^2x-2\sqrt{3}sinx.cosx=4$$<=>sinx(\frac{1}{2}sinx+2\sqrt{3}cosx)=0$Đơn giản rồi, bạn tự giải nhé

$4(sin^2\frac{x}{2}+cos^2\frac{x}{2})^2-8sin^2\frac{x}{2}.cos^2\frac{x}{2}-\sqrt{3}sin2x=4$$<=>4-2.sin^2x-2\sqrt{3}sinx.cosx=4$$<=>2sinx(sinx+\sqrt{3}cosx)=0$Đơn giản rồi, bạn tự giải nhé

Câu 2$cos^2x-cos^24x-sinx.cos4x=\frac{1}{4}$$<=>(cosx-cos4x)(cosx+cos4x)-sinx.cos4x=\frac{1}{4}$$<=>-2sin\frac{5x}{2}.cos\frac{5x}{2}.2sin\frac{3x}{2}.cos\frac{3x}{2}-\frac{1}{2}sin5x-\frac{1}{2}sin3x-\frac{1}{4}=0$$<=>(sin5x.sin3x+\frac{1}{2}sin5x)+(\frac{1}{2}sin3x+\frac{1}{4})=0 $$<=>sin5x(sin3x+\frac{1}{2})+\frac{1}{2}(sin3x+\frac{1}{2})=0$$<=>(sin3x+\frac{1}{2})(sin5x+\frac{1}{2})=0$Đơn giản rồi bạn tự giải nhé

Câu 2$cos^2x-cos^24x-sinx.cos4x=\frac{1}{4}$$<=>(cosx-cos4x)(cosx+cos4x)-sinx.cos4x=\frac{1}{4}$$<=>2sin\frac{5x}{2}.cos\frac{5x}{2}.2sin\frac{3x}{2}.cos\frac{3x}{2}-\frac{1}{2}sin5x+\frac{1}{2}sin3x-\frac{1}{4}=0$$<=>(sin5x.sin3x-\frac{1}{2}sin5x)+(\frac{1}{2}sin3x-\frac{1}{4})=0 $$<=>sin5x(sin3x-\frac{1}{2})+\frac{1}{2}(sin3x-\frac{1}{2})=0$$<=>(sin3x-\frac{1}{2})(sin5x-\frac{1}{2})=0$Đơn giản rồi bạn tự giải nhé

3.Đặt $t=\left| {sinx+cosx} \right|=> sin2x=1-t^2$ ĐK:$0\leq t\leq \sqrt{2}$(3)=>$4t^2-t-3=0$<=> $t=1\vee t=-\frac{3}{4}(loại)$<=> $sin(x-\frac{\pi}{4})=\pm 2$Giải nghiệm và hợp nghiệm => $x=\frac{k\pi}{2}(k\in Z)$

3.Đặt $t=\left| {sinx+cosx} \right|=> sin2x=1-t^2$ ĐK:$0\leq t\leq \sqrt{2}$(3)=>$4t^2-t-3=0$<=> $t=1\vee t=-\frac{3}{4}(loại)$<=> $sin(x-\frac{\pi}{4})=\pm \frac{\sqrt{2}}{2}$Giải nghiệm và hợp nghiệm => $x=\frac{k\pi}{2}(k\in Z)$

e) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos5x.cos7x}{sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1- cos12x + 1 - cos 2x}{2sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{2cos^26x+2cos^2x}{2sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{(6x)^2.(\frac{cos6x}{6x})^2+x^2.(\frac{cosx}{x})^2}{(11x)^2.(\frac{sin11x}{11x})^2}$=$\mathop {\lim }\limits_{x \to 0}\frac{6.(\frac{cos6x}{6x})^2+(\frac{cosx}{x})^2}{11.(\frac{sin11x}{11x})^2}$= $\frac{6+1}{11}$=$\frac{7}{11}$ ( vì ta có $\mathop {\lim }\limits_{x \to 0}\frac{sinx}{x}=1$)

e) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos5x.cos7x}{sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1- cos12x + 1 - cos 2x}{2sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{2cos^26x+2cos^2x}{2sin^211x}$=$\mathop {\lim }\limits_{x \to 0}\frac{(6x)^2.(\frac{cos6x}{6x})^2+x^2.(\frac{cosx}{x})^2}{(11x)^2.(\frac{sin11x}{11x})^2}$=$\mathop {\lim }\limits_{x \to 0}\frac{36.(\frac{cos6x}{6x})^2+(\frac{cosx}{x})^2}{121.(\frac{sin11x}{11x})^2}$= $\frac{36+1}{121}$=$\frac{37}{121}$ ( vì ta có $\mathop {\lim }\limits_{x \to 0}\frac{sinx}{x}=1$)

c) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos^22x}{x.sinx}$= $\mathop {\lim }\limits_{x \to 0}\frac{sin^22x}{x.sinx}$=$\mathop {\lim }\limits_{x \to 0}\frac{(2x)^2.(\frac{sin2x}{2x})^2}{x^2.\frac{sinx}{x}}$= 2

c) $\mathop {\lim }\limits_{x \to 0}\frac{1-cos^22x}{x.sinx}$= $\mathop {\lim }\limits_{x \to 0}\frac{sin^22x}{x.sinx}$=$\mathop {\lim }\limits_{x \to 0}\frac{(2x)^2.(\frac{sin2x}{2x})^2}{x^2.\frac{sinx}{x}}$= 4

Số cách chọn 3 nam trong 10 nam là C^{3}_{10} Số cách chọn 3 nữ trong 6 nữ là C^{3}_{6} Từ 3 nam và 3 nữ được chọn ta chia thành 3 cặp nên sẽ có 3! cáchVậy có 3!.C^{3}_{10}.C^{3}_{6}=14400 cách

Số cách chọn 3 nam trong 10 nam là $C^{3}_{10}$Số cách chọn 3 nữ trong 6 nữ là $C^{3}_{6}$ Từ 3 nam và 3 nữ được chọn ta chia thành 3 cặp nên sẽ có 3! cáchVậy có $3!.C^{3}_{10}.C^{3}_{6}$=14400 cách