a) \(\left(2x-5\right)^2-x^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{3}\end{matrix}\right.\)

b) \(\left(x^2-2x+1\right)-9x^2=0\)

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

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

\(\left[{}\begin{matrix}-2x=1\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{4}\end{matrix}\right.\)

a) Ta có: \(x^2-2x-3=0\)

\(\Leftrightarrow x^2-3x+x-3=0\)

\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)

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

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

Vậy: \(S_1=\left\{3;-1\right\}\)(1)

Ta có: \(\left(x+1\right)\left(x+3\right)=0\)

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

Vậy: \(S_2=\left\{-3;-1\right\}\)(2)

Từ (1) và (2) suy ra \(S_1\ne S_2\)

hay Hai phương trình \(x^2-2x-3=0\) và \(\left(x+1\right)\left(x+3\right)=0\) không tương đương với nhau

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

\(a,x\left(x+9\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Rightarrow x\left(x^2+4x+4\right)=0\\ \Rightarrow x\left(x+2\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(x-5-4\right)\left(x-5+4\right)=0\\ \Rightarrow\left(x-9\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\\ d,\Rightarrow3\left(x+2\right)-x\left(x+2\right)=0\\ \Rightarrow\left(x+2\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\\ e,\Rightarrow x^3+6x^2+12x+8-x^3-6x^2=4\\ \Rightarrow12x=-4\Rightarrow x=-\dfrac{1}{3}\\ g,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

b: =>1/4x+4/5-x-5=1/3x+1-1/2x+1

=>-3/4x+1/6x=2+5-4/5=24/5

=>x=-288/35

c: =>6x^2+3x-30x-15=6x^2+10x-21x-35

=>-27x-15=-11x-35

=>-16x=-20

=>x=5/4

a)\(9x^2-6x-3=0\)

\(\Leftrightarrow\)\(3x^2-2x-1=0\)

\(\Leftrightarrow\)\(3x^2-3x+x-1=0\)

\(\Leftrightarrow\)\((3x-1)(x-1)=0\)

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

a) \(9x^2-6x-3=0\)

\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)

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

b) \(x^3+9x^2+27x+19=0\)

\(\Leftrightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\)

\(\Leftrightarrow x=-1\)( do \(x^2+8x+19=\left(x+4\right)^2+3>0\))

c) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)

\(\Leftrightarrow x\left(x^2-25\right)-x^3-8=3\)

\(\Leftrightarrow x^3-25x-x^3=8\Leftrightarrow-25x=11\Leftrightarrow x=-\dfrac{11}{25}\)

a) \(9x^2-6x-3=0\\ \Rightarrow\left(9x^2-9x\right)+\left(3x-3\right)=0\\ \Rightarrow9x\left(x-1\right)+3\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(9x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\9x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) \(x^3+9x^2+27x+19=0\\ \Rightarrow\left(x^3+x^2\right)+\left(8x^2+8x\right)+\left(19x+19\right)=0\\ \Rightarrow x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2+8x+19\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\x^2+8x+19=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x^2+8x+16\right)+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\\left(x+4\right)^2+3=0\left(vôlí\right)\end{matrix}\right.\)

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

Lời giải:

a. $9x^2-16-(3x-4)(2x+5)=0$

$\Leftrightarrow [(3x)^2-4^2]-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4)-(3x-4)(2x+5)=0$

$\Leftrightarrow (3x-4)(3x+4-2x-5)=0$

$\Leftrightarrow (3x-4)(x-1)=0$

$\Leftrightarrow 3x-4=0$ hoặc $x-1=0$

$\Leftrightarrow x=\frac{4}{3}$ hoặc $x=1$.

b.

$x^2+4x=12$

$\Leftrightarrow x^2+4x-12=0$

$\Leftrightarrow (x^2-2x)+(6x-12)=0$

$\Leftrightarrow x(x-2)+6(x-2)=0$

$\Leftrightarrow (x-2)(x+6)=0$

$\Leftrightarrow x-2=0$ hoặc $x+6=0$

$\Leftrightarrow x=2$ hoặc $x=-6$

c.

$x^2-2x=35$

$\Leftrightarrow x^2-2x-35=0$

$\Leftrightarrow (x^2+5x)-(7x+35)=0$

$\Leftrightarrow x(x+5)-7(x+5)=0$

$\Leftrightarrow (x+5)(x-7)=0$

$\Leftrightarrow x+5=0$ hoặc $x-7=0$

$\Leftrightarrow x=-5$ hoặc $x=7$

\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)

\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)