\(x^2-2x-4y^2-4y\)

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

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

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

Trả lời:

a) \(\frac{1}{4}x^2y+5x^3-x^2y^2=x^2\left(\frac{1}{4}y+5x-y^2\right)\)

 b) 5x ( x - 1 ) - 3y ( 1 - x ) = 5x ( x - 1 ) + 3y ( x - 1 ) = ( x - 1 )( 5x + 3y )

 c) 4x² - 25 = ( 2x )² - 5² = ( 2x - 5 )( 2x + 5 )

 d) 6x- 9x² = 3x ( 2 - 3x )

giúp mk vs

bài 1: ... phá hết ra

bài 2

câu a, tách -2x^2 thành -x^2-x^2 rồi tự giải quyết

câu b, thêm bớt 1 để tạo hằng đẳng thức

câu c, đổi z-x thành -x-z

câu d là hằng đẳng thức đó má nội

mình rất muốn làm hết nhưng cái tật lười nó ko cho mình làm, mong bạn thông cảm

7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)

8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)

9, ĐK x >= 0 

\(x-2\sqrt{x}-3=x-3\sqrt{x}+\sqrt{x}-3\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)

10, \(-4x^2-4x+10=-\left(4x^2+4x+1\right)+11\)

\(=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)

11;12 xem lại đề

13, \(-x^3+6xy^2-12xy^2+8y^3=-\left(x^3-6xy^2+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)

Trả lời:

7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)

8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)

9, \(x-2\sqrt{x}-3\left(ĐK:x\ge0\right)\)

\(=x-3\sqrt{x}+\sqrt{x}-3=\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)

10, \(10-4x-4x^2=-\left(4x^2+4x-10\right)=-\left(4x^2+4x+1-11\right)=-\left[\left(2x+1\right)^2-11\right]\)

\(=-\left(2x+1\right)^2+11=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)

11,sửa đề:  \(15x\left(x-3y\right)+20y\left(3y-x\right)=15x\left(x-3y\right)-20y\left(x-3y\right)=5\left(x-3y\right)\left(3x-4y\right)\)

12, \(25x^2-2=\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)\)

13, sửa đề: \(-x^3+6x^2y-12xy^2+8y^3=-\left(x^3-6x^2y+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)

Bài 1 : 

a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)

b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)

Bài 2 : tự kết luận nhé, ngại mà lười :( 

a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)

\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)

\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)

\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)

b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)

\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)

\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)

Vậy phương trình có vô số nghiệm 

c, \(\left|2x-3\right|=4\)

Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)

Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)

d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)

Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)

\(\left(x+5\right)\left(x^2-5x+25\right)\)

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

\(=x^3+5^3\)

\(=x^3+125\)

3) \(27-y^3\)

\(=3^3-y^3\)

\(=\left(3-y\right)\left(9-3y+y^2\right)\)

1. 4-32x³

= 4.(1-8x³)

= 4.[1³-(2x)³ ]

= 4.(1-2x).(1+2x+4x²)

2. b. \(\left(\frac{x}{xy-y^2}-\frac{2x-y}{xy-x^2}\right):\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\left[\frac{x}{y\left(x-y\right)}+\frac{2x-y}{x\left(x-y\right)}\right]:\left(\frac{y}{xy}+\frac{x}{xy}\right)\)

\(=\left[\frac{x.x}{y\left(x-y\right).x}+\frac{\left(2x-y\right).y}{x\left(x-y\right).y}\right]:\left(\frac{x+y}{xy}\right)\)

\(=\left[\frac{x^2+2xy-y^2}{xy\left(x-y\right)}\right]:\left(\frac{x+y}{xy}\right)\)

\(=\left[\frac{-\left(x-y\right)^2}{xy\left(x-y\right)}\right].\frac{xy}{x+y}\)

\(=\frac{-\left(x-y\right)}{xy}.\frac{xy}{x+y}\)

\(=\frac{y-x}{x+y}\)