Toán CMR: `3(sin^4x+cos^4x)-2(sin^6x+cos^6x)=1` 31/08/2021 By Quinn CMR: `3(sin^4x+cos^4x)-2(sin^6x+cos^6x)=1`
$VT= 3(\sin^4x + \cos^4x)-2(\sin^2x+\cos^2x)(\sin^4x-\sin^2x.\cos^2x+ \cos^4x)$ $= 3\sin^4x+3\cos^4x-2\sin^4x+2sin^2x.\cos^4x-2\cos^4x$ $= \sin^4x+2\sin^2x.\cos^2x+\cos^4x$ $=(\sin^2x+\cos^2x)^2$ $= 1^2=1=VP$ Trả lời
Đáp án: Tham khảo Giải thích các bước giải: $ 3(sin^4x+cos^4x)-2(sin^6+cos^6x)=1$ $sin^4x+cos^4x=(sin²x+cos²x)²-2sin²x.cos²x=1-2sin²x.cos²x$ $sin^6x+cos^6=(sin²+cos²x)³-3sin².cos²x(sin²x+cos²x)=1-3sin²x.cosx$ $⇒3(sin^4x+cos^4)-2(sin^6+cos^6+cos^6)$ $=3(1-2sin²x.cos²x)-2(1-3sin²x.cos²x)$ $=3-2=1=VP (đpcm)$ Trả lời
$VT= 3(\sin^4x + \cos^4x)-2(\sin^2x+\cos^2x)(\sin^4x-\sin^2x.\cos^2x+ \cos^4x)$
$= 3\sin^4x+3\cos^4x-2\sin^4x+2sin^2x.\cos^4x-2\cos^4x$
$= \sin^4x+2\sin^2x.\cos^2x+\cos^4x$
$=(\sin^2x+\cos^2x)^2$
$= 1^2=1=VP$
Đáp án:
Tham khảo
Giải thích các bước giải:
$ 3(sin^4x+cos^4x)-2(sin^6+cos^6x)=1$
$sin^4x+cos^4x=(sin²x+cos²x)²-2sin²x.cos²x=1-2sin²x.cos²x$
$sin^6x+cos^6=(sin²+cos²x)³-3sin².cos²x(sin²x+cos²x)=1-3sin²x.cosx$
$⇒3(sin^4x+cos^4)-2(sin^6+cos^6+cos^6)$
$=3(1-2sin²x.cos²x)-2(1-3sin²x.cos²x)$
$=3-2=1=VP (đpcm)$