dễ mak

Bài 1 :

1) a² - 4 + y ( a - 2 )

= ( a + 2 ) ( a - 2 ) + y ( a - 2 )

= ( a - 2 ) ( a + 2 + y )

2) ( x - 2 )² - 9y²

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

Bài 2 :

1) 3 ( x + 4 ) - 2x = 5

=> 3x + 12 - 2x = 5

=> x + 12 = 5

=> x = 5 - 12 = - 7

Vậy x = - 7

2) x ( x - 2 ) - x² - 6 = 0

=> x² - 2x - x² - 6 = 0

=> - 2x - 6 = 0

=> 2x = - 6

=> x = \(-\frac{6}{2}=3\)

Vậy x = 3

3 ) x² - 3x = 0

=> x ( x - 3 ) = 0

=> \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy \(x\in\left\{0;3\right\}\)

4) 5 - 3 ( x - 6 ) = 4

=> 5 - 3x + 18 = 4

=> 3x = 5 + 18 - 4

=> 3x = 19

=> x = \(\frac{19}{3}\)

Vậy \(x=\frac{19}{3}\)

b: =>4x^2+8x-8x^2+5x-10=0

=>-4x^2+13x-10=0

=>x=2 hoặc x=5/4

c: =>2x^2-5x+6x-15=2x^2+8x

=>x-15=8x

=>-7x=15

=>x=-15/7

d: =>3x^2+15x-2x-10-3x^2-12x=5

=>x-10=5

=>x=15

e: =>x^2-3x+2x^2+2x=3x^2-12

=>-x=-12

=>x=12

a) \(\dfrac{2}{x+3}+\dfrac{1}{x}\) MTC: \(x\left(x+3\right)\)

\(=\dfrac{2x}{x\left(x+3\right)}+\dfrac{x+3}{x\left(x+3\right)}\)

\(=\dfrac{2x+x+3}{x\left(x+3\right)}\)

\(=\dfrac{3x+3}{x\left(x+3\right)}\)

b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}\)

\(=\dfrac{x+1}{2\left(x-1\right)}+\dfrac{-2x}{\left(x-1\right)\left(x+1\right)}\) MTC: \(2\left(x-1\right)\left(x+1\right)\)

\(=\dfrac{\left(x+1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}+\dfrac{-2x.2}{2\left(x-1\right)\left(x+1\right)}\)

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

\(=\dfrac{\left(x+1\right)-4x}{2\left(x-1\right)}\)

\(=\dfrac{x+1-4x}{2\left(x-1\right)}\)

\(=\dfrac{1-3x}{2\left(x-1\right)}\)

c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}\)

\(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\) MTC: \(6y\left(y-6\right)\)

\(=\dfrac{y\left(y-12\right)}{6y\left(y-6\right)}+\dfrac{6.6}{6y\left(y-6\right)}\)

\(=\dfrac{y\left(y-12\right)+6^2}{6y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+6^2}{6y\left(y-6\right)}\)

\(=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}\)

\(=\dfrac{y-6}{6y}\)

Bạn Nguyễn Nam làm sai câu b rồi , làm lại cho tất nè

a) \(\dfrac{2}{x+3}+\dfrac{1}{x}=\dfrac{2x+x+3}{x\left(x+3\right)}=\dfrac{3x+3}{x\left(x+3\right)}\)

b) \(\dfrac{x+1}{2x-2}+\dfrac{-2x}{x^2-1}=\dfrac{x+1}{2\left(x-1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)

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

\(=\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\)

c) \(\dfrac{y-12}{6y-36}+\dfrac{6}{y^2-6y}=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{\left(y-6\right)^2}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

d) \(\dfrac{6x}{x+3}+\dfrac{3}{2x+6}=\dfrac{6x}{x+3}+\dfrac{3}{2\left(x+3\right)}=\dfrac{12x}{2\left(x+3\right)}\)( sửa đề )

a: -2x(x+3)+x(2x-1)=10

=>-2x^2-6x+2x^2-x=10

=>-7x=10

=>x=-10/7

b: Sửa đề: 2/3x(9/2x+1/4)-(3x^2+2)=3

=>3x^2+1/6x-3x^2-2=3

=>1/6x-2=3

=>x=30

sao sửa, đề nó vậy á

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2+5x+6-x^2-3x+10=6\)

\(\Leftrightarrow2x=-10\)

hay x=-5

b: Ta có: \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\)

hay \(x=-\dfrac{1}{2}\)

c: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-2x-3x+1\right)-\left(18x^2-2x-27x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+27x-3=0\)

hay x=0

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: \(2x^2+4x+2-2y^2\)

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

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

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

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

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)