b) \(x\left(2x+5\right)=0\)

TH1: \(x=0\)

TH2: \(2x+5=0\Leftrightarrow2x=-5\Leftrightarrow x=\dfrac{-5}{2}\)

\(a,\Rightarrow5x\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x-3\right)\left(5x-x-3\right)=0\\ \Rightarrow\left(x-3\right)\left(4x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{4}\end{matrix}\right.\\ b,\Rightarrow x\left(2x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow-3x+5=0\)

hay \(x=\dfrac{5}{3}\)

b: Ta có: \(x^2-6x=0\)

\(\Leftrightarrow x\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

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

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

=> x-2=0           hoặc x+1=0

=> x=2              hoặc x=-1

b. 5x(x-3)-x+3=0

=> 5x(x-3)-(x-3)=0

=> (x-3).(5x-1)=0

=> x-3=0         hoặc 5x-1=0

=> x=3            hoặc x=1/5

a) x = 2

b) x = 3

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

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

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

<=>x-2=0 hoặc x+1=0

<=>x=2 hoặc x=-1

b) 5x(x - 3) - x + 3 = 0

<=>5x.(x-3)-(x-3)=0

<=>(x-3)(5x-1)=0

<=>x-3=0 hoặc 5x-1=0

<=>x=3 hoặc x=1/5

x(x-2) + x-2= x(x-2) + (x-2).1=(x-2)(x+1)

à như thế này

 x (x-2) + 1( x - 2) = 0

 (x - 2)( x + 1) = 0 

a)  \(x\left(x-2\right)+x-2=0\)

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

<=>  \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy...

b)  \(5x\left(x-3\right)-x+3=0\)

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

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

<=>  \(\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

Vậy...

a, x.( x - 2 ) + 2x - 4 = 0

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

<=> x=2 V x=-2

b, 5x.(x - 3 ) - x + 3 = 0

<=> (x-3)(5x-1)=0

<=> x=3 V x=1/5

a ) \(x.\left(x-2\right)+2x-4=0\)

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

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

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)

b ) \(5x.\left(x-3\right)-x+3=0\)

\(\Leftrightarrow5x.\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-\frac{1}{5}\end{array}\right.\)

Vậy ............

a, x.( x - 2 ) + 2x - 4 = 0

=>x²+2x+2x-4=0

=>x²+4x-4=0

=>(x-2)²=0

=>x=2

b, 5x.(x - 3 ) - x + 3 = 0

=>5x²-15x-x+3=0

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

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

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=3\end{array}\right.\)

- Help Me :((

\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)

\(25\left(x^2+6x+9\right)+1-25x^2=0\)

\(25x^2+150x+225+1-25x^2=0\)

\(150x=-226\)

\(x=-\frac{113}{75}\)

25 ( x + 3 )² + ( 1 - 5x )( 1 + 5x ) = 0

25 ( x² + 6x + 9 ) + 1 + 5x - 5x - 25x² = 0

25x² + 150x + 225 + 1 + 5x - 5x - 25x² = 0

150x + 226 = 0

150x = -226

x = -226/150