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: \(2x^2+7x+3=0\)

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

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

a)(3x-1)(4x-8)=0

⇔3x-1=0 hoặc 4x-8=0

1.3x-1=0⇔3x=1⇔x=1/3

2.4x-8=0⇔4x=8⇔x=2

phương trình có 2 nghiệm:x=1/3 và x=2

b)(x-2)(1-3x)=0

⇔x-2=0 hoặc 1-3x=0

1.x-2=0⇔x=2

2.1-3x=0⇔-3x=1⇔x=-1/3

phương trình có 2 nghiệm:x=2 và x=-1/3

c)(x-3)(x+4)-(x-3)(2x-1)=0

⇔(x+4)(2x-1)=0

⇔x+4=0 hoặc 2x-1=0

1.x+4=0⇔x=-4

2.2x-1=0⇔2x=1⇔x=1/2

phương trình có hai nghiệm:x=-4 và x=1/2

d)(x+1)(x+2)=2x(x+2)

⇔(x+1)(x+2)-2x(x+2)=0

⇔2x(x+1)=0

⇔2x=0 hoặc x+1=0

1.2x=0⇔x=0

2.x+1=0⇔x=-1

phương trình có 2 nghiệm:x=0 và x=-1

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

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

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

c: Ta có: \(4\left(x-1\right)\left(x+5\right)-\left(x+5\right)\left(x+2\right)-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4\left(x^2+4x-5\right)-\left(x^2+7x+10\right)-3\left(x^2+x-2\right)=0\)

\(\Leftrightarrow4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0\)

\(\Leftrightarrow6x=24\)

hay x=4

d: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-5+1=-4\)

hay \(x=\dfrac{2}{7}\)

\(a)\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1.\left(x\ne2;4\right).\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1.\\ \Rightarrow x^2-4x-3x+12+x^2-4x+4+x^2-4x-2x+8=0.\\ \Leftrightarrow3x^2-17x+24=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}.\\x=3.\end{matrix}\right.\) (TM).

\(b)3x+12=0.\\ \Leftrightarrow3x=-12.\\ \Leftrightarrow x=-4.\)

\(c)5+2x=x-5.\\ \Leftrightarrow2x-x=-5-5.\\ \Leftrightarrow x=-10.\)

\(d)2x\left(x-2\right)+5\left(x-2\right)=0.\\ \Leftrightarrow\left(2x+5\right)\left(x-2\right)=0.\)

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

\(e)\dfrac{3x-4}{2}=\dfrac{4x+1}{3}.\\ \Rightarrow3\left(3x-4\right)-2\left(4x+1\right)=0.\\ \Leftrightarrow9x-12-8x-2=0.\\ \Leftrightarrow x=14.\)

\(f)\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1.\left(x\ne\pm1\right).\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x}{x^2-1}=1.\\ \Leftrightarrow x^2+3x-x^2+1=0.\\ \Leftrightarrow3x+1=0.\\ \Leftrightarrow x=\dfrac{-1}{3}.\)

\(g)\dfrac{2x}{x-1}+\dfrac{3-2x}{x+2}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\left(x\ne1;-2\right).\\ \Leftrightarrow\dfrac{2x^2+4x+\left(3-2x\right)\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{6}{\left(x-1\right)\left(x+2\right)}.\\ \Rightarrow2x^2+4x+3x-3-2x^2+2x-6=0.\\ \Leftrightarrow9x=9.\)

\(\Leftrightarrow x=1\left(koTM\right).\)

\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

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

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

2: 

a: =>-2x=10

=>x=-5

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

=>x=3 hoặc x=-5/2