b) \(\left(x+3\right)^2-5x-15=0\\ \Leftrightarrow\left(x+3\right)^2-5\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+3-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-3;2\right\}\)

c) \(2x^5-4x^3+2x=0\\ \Leftrightarrow2x\left(x^4-2x^2+1\right)=0\\ \Leftrightarrow2x\left(x^2-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\\left(x^2-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

Vậy tập nghiệm của pt là : \(S=\left\{0;1;-1\right\}\)

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

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

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

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

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

\(\Leftrightarrow\)\(\orbr{\begin{cases}\left(x-1\right)^2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}}\)

Vậy \(x=1\) hoặc \(x=-1\)

Chúc bạn học tốt ~ 

a) x³-x²-x+1 = 0 \(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Leftrightarrow x=1\)

\(c)\)\(x^4+2x^3-6x-9=0\)

\(\Leftrightarrow\)\(\left(x^4-9\right)+\left(2x^3-6x\right)=0\)

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

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

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

Hoặc \(x^2+3+2x=0\)

\(\Leftrightarrow\)\(x^2=3\)

Hoặc \(x\left(x+2\right)=-3\)

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

Hoặc \(x;\left(x-2\right)\inƯ\left(-3\right)\)

Ta có bảng : 

Vậy \(x\in\left\{1;-1;3;-3;5\right\}\)

Chúc bạn học tốt ~ 

a) \(\Rightarrow x\left(x+3\right)=0\)

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

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

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

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

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

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

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

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

\(d,x\left(x-3\right)-7x+21=0\)

\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)

\(a,2x\left(x-7\right)+5x-35=0\)

 \(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)

\(c,4x^2+12x+9=0\)

\(\Leftrightarrow4x^2+6x+6x+9=0\)

\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)

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

\(\Leftrightarrow2x+3=0\)

\(\Leftrightarrow x=-\frac{3}{2}\)

a) 2x(x-7)+5x-35=0
<=> 2x(x-7)+5(x-7)=0
<=>(2x+5)(x-7)=0
<=> (2x+5)=0 <=> x=-5/2
hoặc <=> x-7=0 <=> x=7

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

<=> ( 2x - 3 ) + ( 2x - 3 )( x - 1 ) = 0

<=> ( 2x - 3 )( 1 + x - 1 ) = 0

<=> x( 2x - 3 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy .....

a, 25x^2 - 1 - (5x -1)(x+2)=0

=> (5x)^2 - 1 + (5x-1)(x+2) = 0

=> (5x-1)(5x+1) + (5x-1)(x+2) = 0

=> (5x-1)(5x+1+x+2) = 0

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

a)     25x² - 1 - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 ) - ( 5x - 1 )( x + 2 ) = 0

<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0

<=> ( 5x - 1 )( 4x - 1 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)

Vậy .......

a)(2x-3)(x+5)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy x=3/2 hoặc x=-5

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};-5\right\}\)

b) \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{2;\dfrac{7}{2}\right\}\)

c) \(5x\left(2x-3\right)-6x+9=0\)

\(\Leftrightarrow5x\left(2x-3\right)-3\left(2x-3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\5x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\dfrac{3}{2};\dfrac{3}{5}\right\}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-5\end{matrix}\right.\)

b: Ta có: \(3x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(3x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)

c: Ta có: \(5x\left(2x-3\right)-6x+9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

a)

(2x-1)²-(5x-5)²=0

<=>(2x-1-5x+5)(2x-1+5x-5)=0

<=>(-3x+4)(7x-6)=0

<=>\(\orbr{\begin{cases}-3x+4=0\\7x-6=0\end{cases}}\)

<=>\(\orbr{\begin{cases}-3x=-4\\7x=6\end{cases}}\)

<=>\(\orbr{\begin{cases}x=\frac{-4}{-3}=\frac{4}{3}\\x=\frac{6}{7}\end{cases}}\)

b)

(2x+1)²-4(x+3)²=0

<=>(2x+1)²-[2(x+3)]²=0

<=>(2x+1)²-(2x+6)²=0

<=>(2x+1-2x-6)(2x+1+2x+6)=0

<=>-5(4x+7)=0

<=>4x+7=0

<=>4x=-7

<=>\(x=-\frac{7}{4}\)