\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(b,\Rightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Rightarrow5\left(x+2\right)=0\\ \Rightarrow x=-2\\ c,\Rightarrow2x\left(x^2-2x+1\right)=0\\ \Rightarrow2x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ d,\Rightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Rightarrow3x\left(-x-2\right)=0\\ \Rightarrow-3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}-3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

a)thiếu dấu

b)(x+2)² -(x+2)(x-3)=0

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

(x+2)5=0

x+2=0

x=-2

c)2x³-4x²+2x=0

2x(x²-2x+1)=0

2x(x-1)²

suy ra 2 trường hợp

x=0

x-1=0=>x=1

d)(x-1)²-(2x+1)²=0

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

(x-2)3x=0

x=0

x=2

e: ta có: \(4x^2+4x-6=2\)

\(\Leftrightarrow4x^2+4x-8=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

f: Ta có: \(2x^2+7x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

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

\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

hay x=-2

a) \(\text{5x(x-2)+(2-x)=0}\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\text{x(2x-5)-10x+25=0}\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)

\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)

\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)

a) \(5x\left(x-2\right)+\left(2-x\right)=0\)

\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(x\left(2x-5\right)-10x+25=0\)

\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)

\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)

c) \(\dfrac{25}{16}-4x^2+4x-1=0\)

\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)

\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)

d) \(x^4+2x^2-8=0\)

\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)

`#3107.101107`

`1.`

`a,`

`(2x - 3)^2 = |3 - 2x|`

`=> (2x - 3)^2 = |2x - 3|`

`=>`\(\left[{}\begin{matrix}2x-3=\left(2x-3\right)^2\\2x-3=-\left(2x-3\right)^2\end{matrix}\right.\)

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

`=>`\(\left[{}\begin{matrix}\left(2x-3\right)\left(1-2x+3\right)=0\\\left(2x-3\right)\left(1+2x-3\right)=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}2x-3=0\\4-2x=0\\2x-2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\\x=1\end{matrix}\right.\)

Vậy, `x \in {3/2; 2; 1}`

`b,`

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

`=>`\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(2x-1\right)^2=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy, `x \in {1; 1/2}`

`c,`

`5 - x^2 = 1`

`=> x^2 = 4`

`=> x^2 = (+-2)^2`

`=> x = +-2`

Vậy, `x \in {-2; 2}`

`d,`

`x - 2\sqrt{x} = 0`

`=> x^2 - (2\sqrt{x})^2 = 0`

`=> x^2 - 4x = 0`

`=> x(x - 4) = 0`

`=>`\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

Vậy, `x \in {0; 4}`

`g,`

`(x - 1) + 1/7 = 0`

`=> x - 1 + 1/7 = 0`

`=> x - 6/7 = 0`

`=> x = 6/7`

Vậy, `x = 6/7.`