a) \(5x^3-5x^2y-10x^2+10xy\)

\(=\left(5x^3-10x^2\right)-\left(5x^2y-10xy\right)\)

\(=5x^2\left(x-2\right)-5xy\left(x-2\right)\)

\(=\left(x-2\right)\left(5x^2-5xy\right)\)

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

b) \(2x^2-7xy+5y^2\)

\(=2x^2-2xy-5xy+5y^2\)

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

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

\(M+N=3x^2-5y^3+2x^2+y^3-1\)

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

\(=5x^3-4y^3-1\)

\(M-N=3x^2-5y^3-2x^2-y^3+1\)

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

\(=x^2-6y^3+1\)

`a)7x^3y^2+14x^2y^3+7xy^4`

`=7xy^2(x^2+2xy+y^2)`

`=7xy^2(x+y)^2`

______________________________________________

`b)x^2-xy+5x-5y`

`=x(x-y)+5(x-y)`

`=(x-y)(x+5)`

______________________________________________

`c)3x^2-6xy-12+3y^2`

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

`=3[(x-y)^2-4]`

`=3(x-y-2)(x-y+2)`

a)7x³y²+14x²y³+7xy⁴

=7xy²(x²+2xy+y²)

=7xy²(x+y)²

b)x²-xy + 5x - 5y

=x(x-y) + 5(x-y)

=(x-y) (x+5)

p

a) \(39x-39y=39\left(x-y\right)\)

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

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

c) \(16x^2+24xy+9y^2=\left(4x\right)^2+4x.3y.2+\left(3y\right)^2=\left(4x+3y\right)^2\)

d) \(25x^2-\frac{1}{25y^2}=\left(5x\right)^2-\left(\frac{1}{5y}\right)^2=\left(5x-\frac{1}{5y}\right)\left(5x+\frac{1}{5y}\right)\)

e) \(7x^2-7xy+5x-5y=7x\left(x-y\right)+5\left(x-y\right)=\left(x-y\right)\left(7x+5\right)\)

f) \(5x^2-45y^2-30y-5=5\left(x^2-9y^2-6y-1\right)=5\left[x^2-\left(9y^2+6y+1\right)\right]\)

\(=5\left[x^2-\left(3y+1\right)^2\right]=5\left(x-3y-1\right)\left(x+3y+1\right)\)

g) \(x^2+2x+1-y^2-4y-1=\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\) ( Chắc đề vậy :v ) 

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

h) \(4x^2+8x-5=4x^2-2x+10x-5=2x\left(2x-1\right)+5\left(2x-1\right)\)

\(=\left(2x-1\right)\left(2x+5\right)\)

a/ ĐKXĐ: ...

\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)

TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)

TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)

\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)

\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)

Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)

\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)

\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)

Vế trái luôn dương nên pt vô nghiệm

b/ ĐKXĐ: ...

\(2x^3-2y^3+5x-5y=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)

\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)

Thế vào pt dưới:

\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)

Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:

\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)

Đặt \(x+\frac{1}{x}+1=t\)

\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)

\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)

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

Đề có vẻ sai sai. Bạn xem lại đề xem có đúng không?

a) 6x² - 12x

= 6x(x - 2)

b) x² + 2x + 1 - y²

= (x² + 2x + 1) - y²

= (x + 1)² - y²

= (x + 1 - y)(x + 1 + y)

c) x + y + z + x² + xy + xz

= (x + x²) + (y + xy) + (z + xz)

= x(1 + x) + y(1 + x) + z(1 + x)

= (x + y + z)(x + 1)

d) xy + xz + y² + yz

= (xy + xz) + (y² + yz)

= x(y + z) + y(y + z)

= (x + y)(x + z)

e) x³ + x² + x + 1

= (x³ + x²) + (x + 1)

= x²(x + 1) + (x + 1)

= (x² + 1)(x + 1)

f) xy + y - 2x - 2

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

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

= (y - 2)(x + 1)

g) x³ + 3x - 3x² - 9

= (x³ - 3x²) + (3x - 9)

= x²(x - 3) + 3(x - 3)

= (x² + 3)(x - 3)

h) x² - y² - 2x - 2y

= (x² - y²) - (2x + 2y)

= (x + y)(x - y) - 2(x + y)

= (x + y)(x - y - 2)

i) 7x² - 7xy - 5x = 5y

mk thấy con này sai sai ý

i) 7x² - 7xy - 5x + 5y

= (7x² - 7xy) - (5x - 5y)

= 7x(x - y) - 5(x - y)

= (7x - 5)(x - y)

2: x^2+4x+3

=x^2+3x+x+3

=(x+3)(x+1)

3: 2x^2-7x+5

=2x^2-2x-5x+5

=(x-1)(2x-5)