b) Ta có: \(x^2y+xy+x+1\)

\(=xy\left(x+1\right)+\left(x+1\right)\)

\(=\left(x+1\right)\left(xy+1\right)\)

c) Ta có:  \(ax+by+ay+bx\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

d) Ta có: \(x^2-\left(a+b\right)x+ab\)

\(=x^2-ax-bx+ab\)

\(=x\left(x-a\right)-b\left(x-a\right)\)

\(=\left(x-a\right)\left(x-b\right)\)

A(x)+B(x)-C(x)

=x^3+2x^2+3x+1-x^3+x+1-2x^2+1=0

=>4x+3=0

=>x=-3/4

\(\left(2x+x^2\right)\left(x^2-3x+2\right)=0\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=2\end{matrix}\right.\\ A=\left\{-2;0;1;2\right\}\)

\(3\le x^3\le27\Leftrightarrow x\in\left\{2;3\right\}\\ B=\left\{2;3\right\}\)

\(\Leftrightarrow A\cup B=\left\{-2;0;1;2;3\right\}\)

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

câu b thiếu bn ơi

ok

a) ${}^{}-25=0$

${}^{}=0+25=25$

${}^{}={}^{}={}^{}$

$\Rightarrow [\begin{array}{c}2x-1=5\\ 2x-1=-5\end{array}\Rightarrow [\begin{array}{c}2x=6\\ 2x=-4\end{array}\Rightarrow [\begin{array}{c}x=3\\ x=-2\end{array}$

b) $8{}^{}-50x=0$

$2x(4{}^{}$

a: Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

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

\(\left(x+3\right)^{2022}+\left(\sqrt{y-2}-1\right)^{2023}=0\)    \(\left(ĐKXĐ: y\ge2\right)\)

Xét \(\left(x+3\right)^{2022}\ge0\forall x\)

\(\Rightarrow\left(\sqrt{y-2}-1\right)^{2023}\le0\)

\(\Leftrightarrow\sqrt{y-2}-1\le0\)

\(\Leftrightarrow\sqrt{y-2}\le1\) 

\(\Leftrightarrow y-2\le1\)

\(\Rightarrow y\le3\)

\(\Rightarrow2\le y\le3\) mà \(y\in Z\)

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

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

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

3.

$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$

$=(x-1)(x^2+3x+1)$

a) \(4x^2-6x=2x\left(2x-3\right)\)

b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(x^2+y\right)\)

c) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)

\(2x-3\text{​​}\text{ ⋮ }x-1\)

\(\Rightarrow2\left(x-1\right)-1\text{ ⋮ }x-1\)

\(\text{mà 2(x-1) ⋮ x-1 }\)\(\Rightarrow1\text{ ⋮ }x-1\)

\(\Rightarrow\)\(x-1\inƯ\left(1\right)=\left\{\pm1\right\}\)

Vậy \(x\in\left\{0;2\right\}\)