x(x+1)-(x-2)(x+1)=0

\(\left(x+1\right)\left(x-x+2\right)=0\\ \left(x+1\right)\cdot2=0\\ =>x+1=0\\ x=0-1\\ x=-1\)

=>(x+1)(x-x+2)=0

=>x+1=0

=>x=-1

\(\Leftrightarrow\left(x-2021\right)\left(x-5\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\)

a)  \(x\left(x+4\right)\left(x-4\right)-\left(x^2-1\right)\left(x^2+1\right)\)

\(=x\left(x^2-16\right)-\left(x^4-1\right)\)

\(=x^3-16x-x^4+1\)

bạn ktra lại đề

b)  \(x^4+2x^3+5x^2+4x-12\)

\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)

\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

Ủa pạn có thể giải ại cái bước thứ 2 đc ko ạk

\(a.2x+7x+x=-270\)

\(10x=-270\)

\(x=-27\)

\(b,\left(x-1\right)\left(x-9\right)=0\)

\(=>x-1=0\) \(=>x=1\)

       \(x-9=0=>x=9\)

Vậy \(x\in\left\{1;9\right\}\)

\(\left(x+3\right)^2-\left(4-x\right)\left(4+x\right)=10\)

<=>   \(x^2+6x+9-\left(16-x^2\right)=10\)

<=>   \(2x^2+6x-17=0\)

<=>  \(x^2+3x-\frac{17}{2}=0\)

<=>  \(\left(x+\frac{3}{2}\right)^2-\frac{43}{4}=0\)

<=>  \(\left(x+\frac{3}{2}+\frac{\sqrt{43}}{2}\right)\left(x+\frac{3}{2}-\frac{\sqrt{43}}{2}\right)=0\)

<=>  \(\orbr{\begin{cases}x+\frac{3}{2}+\frac{\sqrt{43}}{2}=0\\x+\frac{3}{2}-\frac{\sqrt{43}}{2}=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=\frac{-3-\sqrt{43}}{2}\\x=\frac{\sqrt{43}-3}{2}\end{cases}}\)

Vậy...

\((x+3)^2-(4-x)(4+x)=10\)

\(\Rightarrow x^2+6x+9-(16+4x-4x+x^2)=10\)

\(\Rightarrow x^2+6x+9-16-x^2=10\)

\(\Rightarrow6x+9=26\)

\(\Rightarrow6x=17\)

\(\Rightarrow x\in\varnothing\)

DỂ QUÁ!!!!!!!!!!!!!!!!

TUI HK BIẾT LÀM

\(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\)   \(3\)

<=> \(-4\left(x^2-2x+1\right)+4x^2-1=-3\)

<=> \(-4x^2+8x-4+4x^2-1=-3\)

<=> \(8x-5=-3\)

<=>  \(8x=2\)

<=> \(x=\frac{1}{4}\)

        \(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\)

\(\Rightarrow-4\left(x^2-2x+1\right)+4x^2-1=-3\)

\(\Rightarrow-4x^2+8x-4+4x^2-1=-3\)

\(\Rightarrow8x-5=-3\)

\(\Rightarrow8x=2\Rightarrow x=0,25\)

Vậy \(x=0,25\)

Chúc bạn học tốt.

\(2x^2-16=0\)

\(\Leftrightarrow2x^2=16\)

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow x=\pm\sqrt{8}\)

`2x^2-16=0`

`<=>2x^2=0+16`

`<=>2x^2=16`

`<=>x^2=16:2`

`<=>x^2=8`

`<=>x=+-`\(\sqrt{\text{8}}\)