Không có dấu '' = '' để tìm x nhé.

\([4.\left(x-y\right)^5+2.\left(x-y\right)^3-3.\left(x-y\right)^2];\left(y-x\right)^2\)

\(=[4.\left(x-y\right)^5+2.\left(x-y\right)^3-3.\left(x-y\right)^2]:\left(x-y\right)^2\)

\(=4.\left(x-y\right)^3+2.\left(x-y\right)-3\)

\(=4.\left(x^3-3x^2y+3xy^2-y^3\right)+2x-2y-3\)

\(=4x^3-12x^2y+12xy^2-y^3+2x-2y-3\)

\(x.\left(x^2-4\right)-3x+6\)

\(=x.\left(x+2\right).\left(x-2\right)-3.\left(x-2\right)\)

\(=\left(x^2+2x\right).\left(x-2\right)-3.\left(x-2\right)\)

\(=\left(x-2\right).\left(x^2+2x-3\right)\)

\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)

\(=\left(x-2\right).[x.\left(x-1\right)+3.\left(x-1\right)]\)

\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)

còn 7 con đúng không

10 con

em cần câu C thôi ạ

\(\left(3x-5\right)^2-\left(x-1\right)^2=0\)

\(\Rightarrow\left(3x-5-x+1\right).\left(3x-5+x-1\right)=0\)

\(\Rightarrow\left(2x-4\right).\left(4x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-4=0\\4x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\4x=6\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

\(16.\left(2-3x\right)+x^2.\left(3x-2\right)=0\)

\(\Rightarrow16.\left(2-3x\right)-x^2.\left(2-3x\right)=0\)

\(\Rightarrow\left(16-x^2\right).\left(2-3x\right)=0\)

\(\Rightarrow\left(4-x\right).\left(4+x\right).\left(2-3x\right)=0\)

Trường hợp 1: \(4-x=0\Rightarrow x=4\)

Trường hợp 2: \(4+x=0\Rightarrow x=-4\)

Trường hợp 3: \(2-3x=0\Rightarrow3x=2\Rightarrow x=\frac{2}{3}\)