a: \(=4a^3b^3c\left(10xc+3bc-5ab^2x\right)\)

b: \(=\left(b-2c\right)\left(a-b\right)+\left(a+b\right)\left(b-2c\right)\)

=(b-2c)(2a)

. Ai đó giúp tôi đi mà ._.

bài khó quá bạn ạ

Bài 1

a) 5x²y - 20xy²

= 5xy(x - 4y)

b) 1 - 8x + 16x² - y²

= (1 - 8x + 16x²) - y²

= (1 - 4x)² - y²

= (1 - 4x - y)(1 - 4x + y)

c) 4x - 4 - x²

= -(x² - 4x + 4)

= -(x - 2)²

d) x³ - 2x² + x - xy²

= x(x² - 2x + 1 - y²)

= x[(x² - 2x+ 1) - y²]

= x[(x - 1)² - y²]

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

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

e) 27 - 3x²

= 3(9 - x²)

= 3(3 - x)(3 + x)

f) 2x² + 4x + 2 - 2y²

= 2(x² + 2x + 1 - y²)

= 2[(x² + 2x + 1) - y²]

= 2[(x + 1)² - y²]

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

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

Bài 2:

a: \(x^2\left(x-2023\right)+x-2023=0\)

=>\(\left(x-2023\right)\left(x^2+1\right)=0\)

mà \(x^2+1>=1>0\forall x\)

nên x-2023=0

=>x=2023

b: 

ĐKXĐ: x<>0

\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)

=>\(-x\left(x-4\right)+2x^2-4x-9=0\)

=>\(-x^2+4x+2x^2-4x-9=0\)

=>\(x^2-9=0\)

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

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

c: \(x^2+2x-3x-6=0\)

=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)

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

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

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

d: 3x(x-10)-2x+20=0

=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)

=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)

=>\(\left(x-10\right)\left(3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)

Câu 1:

a: \(5x^2y-20xy^2\)

\(=5xy\cdot x-5xy\cdot4y\)

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

b: \(1-8x+16x^2-y^2\)

\(=\left(16x^2-8x+1\right)-y^2\)

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

\(=\left(4x-1-y\right)\left(4x-1+y\right)\)

c: \(4x-4-x^2\)

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

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

d: \(x^3-2x^2+x-xy^2\)

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

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

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

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

e: \(27-3x^2\)

\(=3\left(9-x^2\right)\)

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

f: \(2x^2+4x+2-2y^2\)

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

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

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

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

Bài 2

a) x²(x - 2023) - 2023 + x = 0

x²(x - 2023) - (x - 2023) = 0

(x - 2023)(x² - 1) = 0

x - 2023 = 0 hoặc x² - 1 = 0

*) x - 2023 = 0

x = 2023

*) x² - 1 = 0

x² = 1

x = 1 hoặc x = -1

Vậy x = -1; x = 1; x = 2023

b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0

-x² + 4x + 2x² - 4x - 9 = 0

x² - 9 = 0

x² = 9

x = 3 hoặc x = -3

Vậy x = 3; x = -3

c) x² + 2x - 3x - 6 = 0

(x² + 2x) - (3x + 6) = 0

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

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

x + 2 = 0 hoặc x - 3 = 0

*) x + 2 = 0

x = -2

*) x - 3 = 0

x = 3

Vậy x = -2; x = 3

d) 3x(x - 10) - 2x + 20 = 0

3x(x - 10) - (2x - 20) = 0

3x(x - 10) - 2(x - 10) = 0

(x - 10)(3x - 2) = 0

x - 10 = 0 hoặc 3x - 2 = 0

*) x - 10 = 0

x = 10

*) 3x - 2 = 0

3x = 2

x = 2/3

Vậy x = 2/3; x = 10

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

Mấy câu dễ mình làm trước nhé. Mấy câu khó hơn mình trình bày sau :)

1) 2x² - 5xy - 3y² = 2x² + xy - 6xy - 3y² = x( 2x + y ) - 3y( 2x + y ) = ( 2x + y )( x - 3y )

2) 7x² + 3xy - 10y² = 7x² - 7xy + 10xy - 10y² = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )

3) x² + 5x - 2 = ( x² + 5x + 25/4 ) - 33/4 = ( x + 5/2 )² - \(\left(\frac{\sqrt{33}}{2}\right)^2\)= \(\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)

6) x⁴ + 324 = ( x⁴ + 36x² + 324 ) - 36x² = ( x² + 18 )² - ( 6x )² = ( x² - 6x + 18 )( x² + 6x + 18 )

4) x⁸ + x⁷ + 1

= x⁸ + x⁷ + x⁶ - x⁶ + 1

= x⁶( x² + x + 1 ) - ( x⁶ - 1 )

= x⁶( x² + x + 1 ) - ( x³ - 1 )( x³ + 1 )

= x⁶( x² + x + 1 ) - ( x - 1 )( x² + x + 1 )( x³ + 1 )

= ( x² + x + 1 )( x⁶ - ( x - 1 )( x³ + 1 ) ]

= ( x² + x + 1 )( x⁶ - x⁴ + x³ - x + 1 )

5) x⁷ + x⁵ + 1

= x⁷ + x⁶ - x⁶ + x⁵ + 1

= x⁵( x² + x + 1 ) - ( x⁶ - 1 )

= x⁵( x² + x + 1 ) - ( x³ - 1 )( x³ + 1 )

= x⁵( x² + x + 1 ) - ( x - 1 )( x² + x + 1 )( x³ + 1 )

= ( x² + x + 1 )[ x⁵ - ( x - 1 )( x³ + 1 ) ]

= ( x² + x + 1 )( x⁵ - x⁴ + x³ - x + 1 )

7) x⁵ - 5x³ + 4x

= x⁵ - x³ - 4x³ + 4x

= x³( x² - 1 ) - 4x( x² - 1 )

= ( x² - 1 )( x³ - 4x )

= ( x - 1 )( x + 1 )x( x² - 4 )

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

8) Xin hàng :)

sai đề thì sửa dùm mik nhé

giúp mik bài này với

CẦN GẤP

a: \(=6x^3-12x^2+x^2-2x+x-2\)

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

b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)

\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)

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

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

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

c: \(=4x^3+x^2+4x^2+x+4x+1\)

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