Question

Answer 1

Ta có : \(PT\Leftrightarrow\sin^4x+2\sin^2x\cos^2x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2\)

\(\Leftrightarrow\left(\sin^2x+\cos^2x\right)^2=\left(\sin^2x+\cos^2x\right)^2\)

\(\Leftrightarrow0=0\) ( Lđ )

Vậy ...

Answer 2

Ta có:`sin^4x+cos^4x=(sin^2x)^2+2sin^{2}x.cos^{2}x+(cos^2x)^2-2sin^{2}x.cos^{2}x=(sin^2x+cos^2x)^2-2sin^{2}x.cos^{2}x`