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T TÍNH MỖI CÁI LÀ RA, HAHA (>_<)$y'=x'.\tan x+.\tan'x=\tan x+\frac{x}{\cos^2x}$$y''=\frac{1}{\cos^2x}+x'.\frac{1}{\cos^2x}+x.(\frac{1}{\cos^2x})'=\frac{2}{\cos^2x}+\frac{2x.\sin x}{\cos^3x}$$VT=x^2.y''-2(x^2+y^2)(1+y)$$=x^2(\frac{2}{\cos^2x}+\frac{2x.\sin x}{\cos^3x})-2(x^2+x^2.\tan^2x)(1+x.\tan x)$$=\frac{2x^2}{\cos^2x}+\frac{2x^3.\sin x}{\cos^3x}-2(x^2+x^3.\tan x + x^2.\tan^2x+x^3.\tan^3x)$$=\frac{2x^2}{\cos^2x}+\frac{2x^3.\sin x}{\cos^3x}-2x^2-\frac{2x^3.\sin x}{\cos x}-\frac{2x^2 \sin^2x}{\cos^2x}-\frac{2x^3.\sin^3x}{\cos^3x}$$=\frac{2x^2}{\cos^2 x}(1-\sin ^2x)+\frac{2x^3.\sin x}{\cos^3 x}(1-\sin^2x)-2x^2-\frac{2x^3.\sin x}{\cos x}$$=2x^2+\frac{2x^3.\sin x}{\cos x}-2x^2-\frac{2x^3.\sin x}{\cos x}=0=VP$ $(đpcm)$@ CLICK V CHO T @

T TÍNH MỖI CÁI LÀ RA, HAHA (>_<)$y'=x'.\tan x+.\tan'x=\tan x+\frac{x}{\cos^2x}$$y''=\frac{1}{\cos^2x}+x'.\frac{1}{\cos^2x}+x.(\frac{1}{\cos^2x})'=\frac{2}{\cos^2x}+\frac{2x.\sin x}{\cos^3x}$$VT=x^2.y''-2(x^2+y^2)(1+y)$$=x^2(\frac{2}{\cos^2x}+\frac{2x.\sin x}{\cos^3x})-2(x^2+x^2.\tan^2x)(1+x.\tan x)$$=\frac{2x^2}{\cos^2x}+\frac{2x^3.\sin x}{\cos^3x}-2(x^2+x^3.\tan x + x^2.\tan^2x+x^3.\tan^3x)$$=\frac{2x^2}{\cos^2x}+\frac{2x^3.\sin x}{\cos^3x}-2x^2-\frac{2x^3.\sin x}{\cos x}-\frac{2x^2 \sin^2x}{\cos^2x}-\frac{2x^3.\sin^3x}{\cos^3x}$$=\frac{2x^2}{\cos^2 x}(1-\sin ^2x)+\frac{2x^3.\sin x}{\cos^3 x}(1-\sin^2x)-2x^2-\frac{2x^3.\sin x}{\cos x}$$=2x^2+\frac{2x^3.\sin x}{\cos x}-2x^2-\frac{2x^3.\sin x}{\cos x}=0=VP$ $(đpcm)$ @ CLICK " V " CHO T @