bài tiếp nha : 1 )[TEX] sin^23x+sin^24x= sin^25x+sin^26x[/TEX]
[TEX]\Leftrightarrow \frac{1-cos6x}{2}+\frac{1-cos8x}{2}=\frac{1-cos10x}{2}+\frac{1-cos12x}{2} \Leftrightarrow cos6x+cos8x=cos10x+cos12x \Leftrightarrow \left[cosx=0 \\ cos7x=cos11x[/TEX]
2)[TEX] sin^3x(1+cotgx)+cos^3x(1+tgx)=2\sqrt[]{sinxcosx}[/TEX]
[TEX]\Leftrightarrow \frac{sin^3x(sinx+cosx)}{sinx}+\frac{cos^3x(sinx+cosx)}{cosx}=2\sqrt{sinxcosx} \Leftrightarrow sinx+cosx=2\sqrt{sinxcosx} -->ok[/TEX]
3)[TEX]\frac{sin3x}{cos2x}+\frac{cos3x}{sin2x}=\frac{2}{sin3x}[/TEX]
[TEX]\Leftrightarrow \frac{cosx}{cos2x.sin2x}=\frac{2}{sin3x} \Leftrightarrow 4cos2xsinx=sin3x \Leftrightarrow \left[sinx=0 \ (Loai) \\ 4(1-2sin^2x)=3-4sin^2x -->ok[/TEX]
4 )[TEX] sin^22x-tg^2x=\frac{9}{2}cos2x[/TEX]
[TEX]\Leftrightarrow tan^2x(4cos^4x-1)=\frac{9}{2}(2cos^2x-1) \Leftrightarrow \left[ 2cos^2x-1=0 \\ tan^2x(2cos^2x+1)=\frac{9}{2} \Leftrightarrow 2sin^2x+\frac{sin^2x}{cos^2x}=\frac{9}{2} \Leftrightarrow sin2x=10cos^2x-1=5cos2x+4-->ok[/TEX]
5)[TEX] (cos^2x+\frac{1}{cos^2x})^2+(sin^2x+\frac{1}{sin^2x})^2=12+\frac{1}{2}siny[/TEX]
[TEX]VP \leq \frac{25}{2}[/TEX]
[TEX]dat \ sin^2x=x, \ cos^2x=y \Rightarrow x+y=1[/TEX]
[TEX]VT=(x+\frac{1}{x})^2+(y+\frac{1}{y})^2 =(x^2+y^2)+(\frac{1}{x^2}+\frac{1}{y^2})+2 \geq \frac{1}{2}(x+y)^2+\frac{1}{2}(\frac{1}{x}+\frac{1}{y})^2+2 \geq \frac{1}{2}(x+y)^2+\frac{1}{2}.\frac{4^2}{(x+y)^2}+2=\frac{1}{2}+8+2=\frac{25}{2} \geq VT \Rightarrow "=" \Leftrightarrow ...[/TEX]
hoặc chuyển về tanx, cotx, rồi đánh giá