\(a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

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

    x\(^2\)- x -x\(^{^2}\)-2x +x+2=0

     -2x+2=0

      -2x=0+2

       -2x=2

         x=-1

Vậy x bằng -1

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

a) 4x(x+1)=8(x+1)

<=>4x(x+1)-8(x+1)=0

<=>(4x-8)(x+1)=0

<=>\(\left[\begin{array}{} 4x-8=0\\ x+1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=2\\ x=-1 \end{array} \right.\)

Vậy...

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

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

<=>\(\left[\begin{array}{} x+2=0\\ x-1=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-2\\ x=1 \end{array} \right.\)

Vậy...

c)5x(x-2)-(2-x)=0

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

<=>\(\left[\begin{array}{} 5x+1=0\\ x-2 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=-1/5\\ x=2 \end{array} \right.\)

d)5x(x-200)-x+200=0

<=>(5x-1)(x-200)=0

<=>\(\left[\begin{array}{} 5x-1=0\\ x-200=0 \end{array} \right.\)

<=>\(\left[\begin{array}{} x=1/5\\ x=200 \end{array} \right.\)

e)\(x^3+4x=0 \)

\(\Leftrightarrow x(x^2+4)=0 \)

\(\Leftrightarrow \left[\begin{array}{} x=0\\ x^2+4=0 (loại vì x^2+4>=0 với mọi x) \end{array} \right.\)

Vậy x=0

f)\((x+1)=(x+1)^2\)

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

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

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

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

Vậy....

b) \(\Leftrightarrow5x^2-6x+1=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow x=\frac{1}{5}\) hoặc x = 1

c) \(\Leftrightarrow x^2+4x-21-x^2-4x+5=0\Leftrightarrow-16=0\) (vô lí) => PT vô nghiệm

d) \(\Leftrightarrow x^2+3x-10=0\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\)x = 2 hoặc x = -5

e) \(\Leftrightarrow x\left(x-2\right)=0\)<=> x = 0 hoặc x = 2

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

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

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

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

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

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

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

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

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

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

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

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

vậy giỏi zữ vậy

\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)

\(\Rightarrow\left(x^2+5x\right)^2-2\left(x^2+5x\right).1+1-25=0\)

\(\Rightarrow\left(x^2-5x+1\right)^2-25=0\)

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

\(\Rightarrow\left(x^2-5x+6\right)\left(x^2-5x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-5x+6=0\\x^2-5x-4=0\end{cases}}\)

TH1 : \(x^2-5x+6=0\Rightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

Th2 : \(x^2-5x+4=0\Rightarrow\left(x-4\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}}\)

Cậu ghi gì thế ? @@ 

Bài làm:

a) \(x+5x^2=0\)

\(\Leftrightarrow x\left(1+5x\right)=0\)

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

b) \(x\left(x-1\right)=x-1\)

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

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

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

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

c) \(5x\left(x-1\right)=1-x\)

\(\Leftrightarrow5x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+1\right)=0\)

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

d) \(\left(3x-4\right)^2-\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(2x-5\right)\left(4x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{4}\end{cases}}\)

\(a,x+5x^2=0< =>x\left(5x+1\right)=0\)

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

\(b,x\left(x-1\right)=x-1< =>x^2-x=x-1\)

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

\(< =>\left(x-1\right)\left(x-1\right)=0< =>x=1\)

\(c,5x\left(x-1\right)=1-x< =>5x^2-5x=1-x\)

\(< =>5x^2-5x+x-1=0< =>5x^2-4x-1=0\)

\(< =>5x^2-5x+x-1=0< =>5x\left(x-1\right)+x-1=0\)

\(< =>\left(5x+1\right)\left(x-1\right)=0< =>\orbr{\begin{cases}5x+1=0\\x-1=0\end{cases}}\)

\(< =>\orbr{\begin{cases}5x=-1\\x=1\end{cases}< =>\orbr{\begin{cases}x=-\frac{1}{5}\\x=1\end{cases}}}\)

\(d,\left(3x-4\right)^2-\left(x+1\right)^2=0\)

\(< =>9x^2-24x+16-x^2-2x-1=0\)

\(< =>8x^2-26x+15=0< =>8\left(x^2-\frac{13}{4}x+\frac{169}{64}\right)-\frac{2082}{64}=0\)

\(< =>\left(x-\frac{13}{8}\right)^2=\frac{2082}{512}=\frac{2082}{16\sqrt{2}}\)

\(< =>\orbr{\begin{cases}x-\frac{13}{8}=\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x-\frac{13}{8}=-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{13}{8}+\frac{\sqrt{2082}}{4\sqrt[4]{2}}\\x=\frac{13}{8}-\frac{\sqrt{2082}}{4\sqrt[4]{2}}\end{cases}}\)(nghiệm vô tỉ)

a) x + 5x² = 0

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

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

b) x( x - 1 ) = x - 1

<=> x² - x - x + 1 = 0

<=> x² - 2x + 1 = 0

<=> ( x - 1 )² = 0

<=> x - 1 = 0 

<=> x = 1

c) 5x( x - 1 ) = 1 - x

<=> 5x² - 5x - 1 + x = 0

<=> 5x² - 4x - 1 = 0

<=> 5x² + x - 5x - 1 = 0

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

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

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

d) ( 3x - 4 )² - ( x + 1 )² = 0

<=> [ ( 3x - 4 ) - ( x + 1 ) ][ ( 3x - 4 ) + ( x + 1 ) ] = 0

<=> [ 3x - 4 - x - 1 ][ 3x - 4 + x + 1 ] = 0

<=> [ 2x - 5 ][ 4x - 3 ] = 0

<=> \(\orbr{\begin{cases}2x-5=0\\4x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{4}\end{cases}}\)