Bài 3:

b. $B=(x+y)(2x-y)+(xy^4-x^2y^2):(xy^2)$

$=(2x^2-xy+2xy-y^2)+(y^2-x)$

$=2x^2+xy-y^2+y^2-x=2x^2+xy-x$

Bài 4:
a. $25x^3-10x^2+x=x(25x^2-10x+1)=x(5x-1)^2$
b. $x^2-9x+9y-y^2=(x^2-y^2)-(9x-9y)=(x-y)(x+y)-9(x-y)=(x-y)(x+y-9)$

c. $16-x^2-4y^2-4xy=16-(x^2+4y^2+4xy)$

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

Bài 5:

a. $36-x^3=100$

$x^3=36-100=-64=(-4)^3$

$\Rightarrow x=-4$

b.

$(x+2)(x-2)-(x+1)^2=7$

$\Leftrightarrow (x^2-4)-(x^2+2x+1)=7$

$\Leftrightarrow -2x-5=7$

$\Leftrightarrow -2x=12$

$\Leftrightarrow x=-6$

nhìu giữ cha !!!!

a.

$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.

$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.

$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$

d.

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

e.

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

$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$

f.

$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$

g.

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

h. Không phân tích được thành nhân tử

i.

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

j.

$x^2+4x-12=(x^2-2x)+(6x-12)=x(x-2)+6(x-2)=(x-2)(x+6)$

k.

$x^2-8x-9=(x^2+x)-(9x+9)=x(x+1)-9(x+1)=(x+1)(x-9)$

l.

$x^2+x-6=(x^2+3x)-(2x+6)=x(x+3)-2(x+3)=(x-2)(x+3)$

Phân tích đa thức thành nhân tử

1: \(x^2-x-y^2-y\)

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

\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

2: \(x^2-y^2+x-y\)

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

\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+1\right)\)

3: \(3x-3y+x^2-y^2\)

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

\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y+3\right)\)

4: \(5x-5y+x^2-y^2\)

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

\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(5+x+y\right)\)

5: \(x^2-5x-y^2-5y\)

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

\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-5\right)\)

6: \(x^2-y^2+2x-2y\)

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

\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y+2\right)\)

7: \(x^2-4y^2+x+2y\)

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

\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y+1\right)\)

8: \(x^2-y^2-2x-2y\)

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

\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-2\right)\)

9: \(x^2-4y^2+2x+4y\)

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

\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y+2\right)\)

Lời giải:

a.

$x^2-x=y^2-1$
$\Leftrightarrow x^2-x+1=y^2$

$\Leftrightarrow 4x^2-4x+4=4y^2$

$\Leftrightarrow (2x-1)^2+3=(2y)^2$

$\Leftrightarrow 3=(2y)^2-(2x-1)^2=(2y-2x+1)(2y+2x-1)$

Đến đây xét các TH:

TH1: $2y-2x+1=1; 2y+2x-1=3$

TH2: $2y-2x+1=-1; 2y+2x-1=-3$

TH3: $2y-2x+1=3; 2y+2x-1=1$

TH4: $2y-2x+1=-3; 2y+2x-1=-1$

b.

$x^2+12x=y^2$

$\Leftrightarrow (x+6)^2=y^2+36$

$\Leftrightarrow 36=(x+6)^2-y^2=(x+6-y)(x+6+y)$

Đến đây xét trường hợp tương tự phần a.

c.

$x^2+xy-2y-x-5=0$

$\Leftrightarrow x^2+xy=x+2y+5$
$\Leftrightarrow 4x^2+4xy=4x+8y+20$

$\Leftrightarrow (2x+y)^2=4x+8y+20+y^2$

$\Leftrightarrow (2x+y)^2-2(2x+y)+1=y^2+6y+21$

$\Leftrightarrow (2x+y-1)^2=(y+3)^2+12$
$\Leftrightarrow (2x+y-1)^2-(y+3)^2=12$

$\Leftrightarrow (2x+y-1-y-3)(2x+y-1+y+3)=12$

$\Leftrightarrow (2x-4)(2x+2y+2)=12$

$\Leftrightarrow (x-2)(x+y+1)=3$

Đến đây đơn giản rồi.

a) \(x^2-x=y^2-1\)

\(\Rightarrow x^2-x+1=y^2\)

\(\Rightarrow4x^2-4x+4=4y^2\)

\(\Rightarrow4x^2-4x+1+3=\left(2y\right)^2\)

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

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

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}\left(2x-2y+1\right)\left(2x+2y+1\right)\in Z\\\left(2x-2y+1\right)\left(2x+2y+1\right)\inƯ\left(7\right)\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(0;1\right);\left(-1;-1\right);\left(-1;-1\right);\left(0;-1\right)\right\}\)

a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=x^3+8y^3-x^3+y^3\)

\(=9y^3\)

b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)

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

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

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

\(=-x^2-x-7\)

11: \(2x^2-12xy+18y^2\)

\(=2\left(x^2-6xy+9y^2\right)\)

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

12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)

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

Bài 1: 

a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

a: \(5^{\left(x-2\right)\left(x+3\right)}=1\)

=>\(\left(x-2\right)\left(x+3\right)=0\)

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

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)