`(x-1)^3-(x-1)(x^2+x+1)-3(x-1)x`

`=x^3-3x^2+3x-1-x^3+1-3x^2+x`

`=-6x^2+4x`

   `->` Bạn xem lại đề!

Answer 

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#hoctot

If this ans is correct pls tick!

\(\left(2a+3\right)^2=\left(2a\right)^2+2.2a.3+3^2=4a^2+12a+9\)

\(A=\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)+\left(x-4\right)^2\\ =x^2+4x+4-\left(x^2-4\right)+x^2-8x+16=x^2-4x+24\\ \cdot x=-2=>A=\left(-2\right)^2-4.\left(-2\right)+24=36\\ \cdot x=0=>A=0^2-4.0+24=24\\ \cdot x=2=>A=2^2-4.2+24=20\\ A=\left(x-2\right)^2+20>0\left(DPCM\right)\)

\(\left[{}\begin{matrix}x+3=0\\2x+4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=5\end{matrix}\right.\)

Ta có: \(\left(x+3\right)\left(2x+4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+4=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\\x=5\end{matrix}\right.\)

Vậy....

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

đặt \(\left(x^2+x+2\right)^2=a\left(a>0\right)\)và \(x^2=b\)

\(=a^2-3ab+2b^2\)

\(=\left(a-b\right).\left(a-2b\right)\)

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

bạn tự pt nốt nhé

Sửa đề: \(A=\left(x^2+x+2\right)^4-3x^2\left(x^2+x+2\right)^2+2x^4\)

- Đặt \(\left\{{}\begin{matrix}\left(x^2+x+2\right)^2=a\\x^2=b\end{matrix}\right.\)

\(\Rightarrow A=a^2-3ab+2b^2=a^2-ab-2ab+2b^2=a\left(a-b\right)-2b\left(a-b\right)=\left(a-b\right)\left(a-2b\right)\)

\(\Rightarrow A=\left[\left(x^2+x+2\right)^2-x^2\right]\left[\left(x^2+x+2\right)^2-2x^2\right]\)

\(=\left(x^2+2\right)\left(x^2+2x+2\right)\left(x^2+x-x\sqrt{2}+2\right)\left(x^2+x+x\sqrt{2}+2\right)\)