a) x²-xz-9y²+3yz

=(x²-9y²)-(xz-3yz)

=(x-3y)(x+3y)-z(x-3y)

=(x-3y)(x+3y-z)

b)x³-x²-5x+125

=x³-6x²+25x+5x²-30x+125

=x(x²-6x+25)+5(x²-6x+25)

=(x+5)(x²-6x+25)

c.x³+2x²-6x-27

=x³+5x²+9x-3x²-15x-27

=x(x²+5x+9)-3(x²+5x+9)

=(x-3)(x²+5x+9)

d. 12x³+4x²-27x-9

=12x³+4x²-27x-9

=4x²(3x+1)-9(3x+1)

=(4x²-9)(3x+1)

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

e.x⁴-25x²+20x-4

=x⁴+5x³-2x²-5x²-25x+10+2x²+10x-4

=x²(x²+5x-2)-5(x²+5x-2)+2(x²+5x-2)

=(x²-5x+2)(x²+5x-2)

f.x²(x²-6)-x²+9

=x⁴+x³-3x²-x³-x²+3x-3x²-3x+9

=x²(x²+x-3)-x(x²+x-3)-3(x²+x-3)

=(x²-x-3)(x²+x-3)

mà bạn ơi sao câu b bạn tách ra nv ?

a) x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

b) 10x – 25 – x² = -(-10x + 25 +x²) = -(25 – 10x + x²)

                         = -(5² – 2 . 5 . x – x²) = -(5 – x)²

c) 8x³ - 1/8 = (2x)³ – (1/2)³ = (2x - 1/2)[(2x)² + 2x . 12 + (1/2)²]

                    ^{= (2x - 1/2)(4x2 + x + 1/4)} 

d)1/25x² – 64y² = (1/5x)2(1/5x)2- (8y)² = (1/5x + 8y)(1/5x - 8y)

\(a,=2x^3y+2x^2y^2-6xy^3\\ b,=3x^3+6x^2-4x-8\\ c,=\left(4x^2+16x-20x-80+76\right):\left(x+4\right)\\ =\left[\left(x+4\right)\left(4x-20\right)+76\right]:\left(x+4\right)\\ =4x-20\left(dư.76\right)\\ d,=\left(x^4-x^2-x^3+x-2x^2+2\right):\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2-x-2\right):\left(x^2-1\right)\\ =x^2-x-2\)

em mới lp 7 =)))

b: 4x^2-20x+25=(x-3)^2

=>(2x-5)^2=(x-3)^2

=>(2x-5)^2-(x-3)^2=0

=>(2x-5-x+3)(2x-5+x-3)=0

=>(3x-8)(x-2)=0

=>x=8/3 hoặc x=2

c: x+x^2-x^3-x^4=0

=>x(x+1)-x^3(x+1)=0

=>(x+1)(x-x^3)=0

=>(x^3-x)(x+1)=0

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

=>\(x\in\left\{0;1;-1\right\}\)

d: 2x^3+3x^2+2x+3=0

=>x^2(2x+3)+(2x+3)=0

=>(2x+3)(x^2+1)=0

=>2x+3=0

=>x=-3/2

a: =>x^2(5x-7)-3(5x-7)=0

=>(5x-7)(x^2-3)=0

=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)

a) x² - 9

= x² - 3²

= (x - 3)(x + 3)

b) 4x² - 1

= (2x)² - 1²

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

c) x⁴ - 16

= (x²)² - 4²

= (x² - 4)(x² + 4)

= (x² - 2²)(x² + 4)

= (x - 2)(x + 2)(x + 4)

d) x² - 4x + 4

= x² - 2.x.2 + 2²

= (x - 2)²

e) x³ - 8

= x³ - 2³

= (x - 2)(x² + 2x + 4)

f) x³ + 3x² + 3x + 1

= x³ + 3.x².1 + 3.x.1² + 1³

= (x + 1)³

a) \(4x^2-16+\left(3x+12\right)\left(4-2x\right)\)

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

\(=\left(2x-4\right)\left(2x+4-3x-12\right)\)

\(=-\left(2x-4\right)\left(x+8\right)\)

b) \(x^3+x^2y-15x-15y\)

\(=x^2\left(x+y\right)-15\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-15\right)\)

c) \(3\left(x+8\right)-x^2-8x\)

\(=3\left(x+8\right)-x\left(x+8\right)\)

\(=\left(x+8\right)\left(3-x\right)\)

d) \(x^3-3x^2+1-3x\)

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

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

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

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

d) \(5x^2-5y^2-20x+20y\)

\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y-4\right)\)

1.

a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)

\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)

\(\Leftrightarrow x^3+3x^2+2x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)

b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)

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

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

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

1c/

\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

Vậy pt có nghiệm duy nhất \(x=-1\)

2.

a. \(x^4-x^3+x^2+x^2-x+1=0\)

\(\Leftrightarrow x^2\left(x^2-x+1\right)+x^2-x+1=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

b.

\(x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x^2-x+1\right)+x^2=0\)

Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn

a. 

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

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

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

d.

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

$=(x+1)(x^2-4x+1)$

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

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

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

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

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

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

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

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