4. Gợi ý : Đặt $t=\frac{2x}{5}$Pt $\Leftrightarrow 2cos^23t-1-cos2t+2cos4t-2=0$$\Leftrightarrow cos6t-cos2x-4sin^22t=0$$\Leftrightarrow 4sin^22t.cos2t+4sin^22t=0$$\Leftrightarrow sin^22tcos^2t=0$

4. Gợi ý : Đặt $t=\frac{2x}{5}$Pt $\Leftrightarrow 2cos^23t-1-cos2t+2cos4t-2=0$$\Leftrightarrow cos6t-cos2x-4sin^22t=0$$\Leftrightarrow 4sin^22t.cos2t+4sin^22t=0$$\Leftrightarrow sin^22tcos^2t=0$$\Leftrightarrow t=0=> x= ...$

Ta có :$f'(x)=2.2cos(4x-1).[cos(4x-1)]'=-8.2cos(4x-1)sinx(4x-1)=-8sin[2(4x-1)]$Suy ra $\left| {f'(x)} \right|=8|sin[2(4x-1)]|\leq 8$ (Do $0\leq \left| {sinx} \right|\leq 1$)Dấu đẳng thức xảy ra khi và chỉ khi$sin[2(4x-1)]=\pm 1$ $\Leftrightarrow 2(4x-1)=\frac{\pi}{2}+k\pi$$\Leftrightarrow x=\frac{1}{16}(\pi+4+k2\pi)$ $(k\in Z)$

Ta có :$f'(x)=2.2cos(4x-1).[cos(4x-1)]'=-8.2cos(4x-1)sin(4x-1)=-8sin[2(4x-1)]$Suy ra $\left| {f'(x)} \right|=8|sin[2(4x-1)]|\leq 8$ (Do $0\leq \left| {sinx} \right|\leq 1$)Dấu đẳng thức xảy ra khi và chỉ khi$sin[2(4x-1)]=\pm 1$ $\Leftrightarrow 2(4x-1)=\frac{\pi}{2}+k\pi$$\Leftrightarrow x=\frac{1}{16}(\pi+4+k2\pi)$ $(k\in Z)$

Ta có :$f'(x)=2.2cos(4x-1).[cos(4x-1)]'=-8.2cos(4x-1)sinx(4x-1)=-8sin[2(4x-1)]$Suy ra $\left| {f'(x)} \right|=8sin[2(4x-1)]\leq 8$ (Do $0\leq \left| {sinx} \right|\leq 1$)Dấu đẳng thức xảy ra khi và chỉ khi$sin[2(4x-1)]=\pm 1$ $\Leftrightarrow 2(4x-1)=\frac{\pi}{2}+k\pi$$\Leftrightarrow x=\frac{1}{16}(\pi+4+k2\pi)$ $(k\in Z)$

Ta có :$f'(x)=2.2cos(4x-1).[cos(4x-1)]'=-8.2cos(4x-1)sinx(4x-1)=-8sin[2(4x-1)]$Suy ra $\left| {f'(x)} \right|=8|sin[2(4x-1)]|\leq 8$ (Do $0\leq \left| {sinx} \right|\leq 1$)Dấu đẳng thức xảy ra khi và chỉ khi$sin[2(4x-1)]=\pm 1$ $\Leftrightarrow 2(4x-1)=\frac{\pi}{2}+k\pi$$\Leftrightarrow x=\frac{1}{16}(\pi+4+k2\pi)$ $(k\in Z)$

c. $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{tan^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2\frac{x}{2}.cos^2x}{sin^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{(\frac{1}{4})x^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{sin^2x(1+\sqrt{cosx})}=\frac{1}{8}$

c. $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{tan^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2\frac{x}{2}.cos^2x}{sin^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}(\frac{x}{sinx})^2.\frac{(\frac{1}{4}).(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{1+\sqrt{cosx}}=\frac{1}{8}$

c. $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1-cos^2x}{tan^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2x.cos^2x}{sin^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{cos^2x}{1+\sqrt{cosx}}=\frac{1}{2}$

c. $\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{tan^2x}$=$\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{tan^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{sin^2\frac{x}{2}.cos^2x}{sin^2x(1+\sqrt{cosx})}$=$\mathop {\lim }\limits_{x \to 0}\frac{(\frac{1}{4})x^2.(\frac{sin\frac{x}{2}}{\frac{x}{2}})^2cos^2x}{sin^2x(1+\sqrt{cosx})}=\frac{1}{8}$

b. $\mathop {\lim }\limits_{x \to 0}\frac{1-cos3x}{1-cos5x}$= $\mathop {\lim }\limits_{x \to 0}\frac{2sin^2\frac{3x}{2}}{2sin^2\frac{5x}{2}}$= $\mathop {\lim }\limits_{x \to 0}\frac{(\frac{3x}{2})^2.(\frac{sin\frac{3x}{2}}{\frac{3x}{2}})^2}{(\frac{5x}{2})^2.(\frac{sin\frac{5x}{2}}{\frac{5x}{2}})^2}=\frac{3}{5}$

b. $\mathop {\lim }\limits_{x \to 0}\frac{1-cos3x}{1-cos5x}$= $\mathop {\lim }\limits_{x \to 0}\frac{2sin^2\frac{3x}{2}}{2sin^2\frac{5x}{2}}$= $\mathop {\lim }\limits_{x \to 0}\frac{(\frac{3x}{2})^2.(\frac{sin\frac{3x}{2}}{\frac{3x}{2}})^2}{(\frac{5x}{2})^2.(\frac{sin\frac{5x}{2}}{\frac{5x}{2}})^2}=\frac{9}{25}$

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

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

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}$

Câu b.Đặt $u_{n}=\frac{sin(x-\frac{\pi}{6})}{\frac{\sqrt{3}}{2}-cosx}$n \to \inftyTa đặt $2t=x-\frac{\pi}{6}=>x=2t+\frac{\pi}{6}$ (Khi $x \to \frac{\pi}{6} \Leftrightarrow t \to 0)$$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_{t \to 0}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)}=\frac{3}{\pi}$

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

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

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq -\frac{\pi}{6}+k\frac{2\pi}{3} (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!

Đk: $x\neq \frac{\pi}{3}+k\frac{2\pi}{3}(k\in Z)$Pt<=> $\frac{2cos^2x+2cos2x.cosx}{cos2x+cosx}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $\frac{2cosx(cosx+cos2x)}{(cos2x+cosx)}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $cosx+\frac{\sqrt{3}}{3}sinx=1$<=> $cos\frac{\pi}{6}.cosx+sin\frac{\pi}{6}.sinx=cos\frac{\pi}{6}$<=> $cos(x-\frac{\pi}{6}=cos\frac{\pi}{6}$Bạn tự giải tìm nghiệm giúp mình nhé!

Đk: $x\neq \frac{\pi}{3}+k\frac{2\pi}{3}(k\in Z)$Pt<=> $\frac{2cos^2x+2cos2x.cosx}{cos2x+cosx}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $\frac{2cosx(cosx+cos2x)}{(cos2x+cosx)}=\frac{2}{3}(3-\sqrt{3}sinx)$<=> $cosx+\frac{\sqrt{3}}{3}sinx=1$<=> $cos\frac{\pi}{6}.cosx+sin\frac{\pi}{6}.sinx=cos\frac{\pi}{6}$<=> $cos(x-\frac{\pi}{6})=cos\frac{\pi}{6}$Bạn tự giải tìm nghiệm giúp mình nhé!

Mình làm cách khác được ko bạn? Ta có: $tanA+tanB+tanC=tanA.tanB.tanC (*)$ - Bạn xem chứng minh ở đây http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/124018/giup-em-voi-sap-thi-roiÁp dụng BĐT Côsi:$tanA.tanB.tanC\leq (\frac{tanA+tanB+tanC}{3})^3$Từ (*) => $tanA.tanB.tanC\leq (\frac{tanA+tanB+tanC}{3})^3$<=> $(tanA+tanB+tanC)^2\geq 3\sqrt{3}$=> MinP=$3\sqrt{3}$

Mình làm cách khác được ko bạn? Ta có: $tanA+tanB+tanC=tanA.tanB.tanC (*)$ - Bạn xem chứng minh ở đây http://toan.hoctainha.vn/Hoi-Dap/Cau-Hoi/124018/giup-em-voi-sap-thi-roiÁp dụng BĐT Côsi:$tanA.tanB.tanC\leq (\frac{tanA+tanB+tanC}{3})^3$Từ (*) => $tanA.tanB.tanC\leq (\frac{tanA+tanB+tanC}{3})^3$<=> $(tanA+tanB+tanC)^2\geq 27$<=> $tanA+tanB + tan C\geq 3\sqrt{3}$=> MinP=$3\sqrt{3}$