a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

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

a) \(\left(2x-3\right)\left(2x+3\right)=0\)

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

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

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

c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)

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

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

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

2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)

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

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

a

\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a,x(x-2)+x-2=0

⇔ (x-2)(x+1)=0

⇔ x=2;x=-1

b,x³+x²+x+1=0

⇔ x²(x+1)+x+1=0

⇔ (x+1)(x²+1)=0

⇔ x=-1

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

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

a: 49x^2-25=0

=>(7x-5)(7x+5)=0

=>7x-5=0 hoặc 7x+5=0

=>x=5/7 hoặc x=-5/7

b: Đề thiếu vế phải rồi bạn

c: (3x-2)^2-9(x+4)(x-4)=2

=>9x^2-12x+4-9(x^2-16)=2

=>9x^2-12x+4-9x^2+144=2

=>-12x+148=2

=>-12x=-146

=>x=146/12=73/6

d: x^3-6x^2+12x-8=0

=>(x-2)^3=0

=>x-2=0

=>x=2

e: x^3-9x^2+27x-27=0

=>(x-3)^3=0

=>x-3=0

=>x=3

a) \(-25+49x^2=0\)

\(\Leftrightarrow49x^2-25=0\)

\(\Leftrightarrow\left(7x\right)^2-5^2=0\)

\(\Leftrightarrow\left(7x-5\right)\left(7x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7x-5=0\\7x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}7x=5\\7x=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-\dfrac{5}{7}\end{matrix}\right.\)

b) \(16x^2-25\left(x-2\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left[5\left(x-2\right)\right]^2=0\)

\(\Leftrightarrow\left(4x-5x+10\right)\left(4x+5x-10\right)=0\)

\(\Leftrightarrow\left(10-x\right)\left(9x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}10-x=0\\9x=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{10}{9}\end{matrix}\right.\)

c) \(\left(3x-2\right)^2-9\left(x+4\right)\left(x+4\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9\left(x^2+8x+16\right)=2\)

\(\Leftrightarrow9x^2-12x+4-9x^2-72x-144=2\)

\(\Leftrightarrow-84x-140=2\)

\(\Leftrightarrow-84x=142\)

\(\Leftrightarrow x=-\dfrac{142}{84}\)

\(\Leftrightarrow x=-\dfrac{71}{42}\)

d) \(x^3-6x^2+12x-8=0\)

\(\Leftrightarrow x^3-3\cdot2\cdot x^2+3\cdot2^2\cdot x-2^3=0\)

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

e) \(-27+27x-9x^2+x^3=0\)

\(\Leftrightarrow x^3-9x^2+27x-27=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Mình sửa lại câu c một chút nha:

c: (3x-2)^2-9(x+4)(x+4)=2

=>(3x-2)^2-9(x+4)^2=2

=>(3x-2)^2-(3x+12)^2=2

=>(3x-2-3x-12)(3x-2+3x+12)=2

=>-14*(6x+10)=2

=>6x+10=-1/7

=>6x=-71/7

=>x=-71/42

a) x= + - 5
b) x\(\in\)\(\left\{-1;-7\right\}\)

a/ \(x^2-25=0\)

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

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

b/ \(x\left(x+7\right)+x+7=0\)

\(x\left(x+7\right)+\left(x+7\right)=0\)

\(\left(x+7\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)

a) x^2 - 25 = 0 <=> (x + 5)(x - 5) = 0

<=> x = -5 hoặc x = 5

b) x(x + 7) + x + 7 = 0 <=> (x + 7)(x + 1) = 0

<=> x = -7 hoặc x = -1

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1