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a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
a) 7x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=07x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=0
⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3
c) x2−10x=−25⇒x2−10x+

\(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\)
=> x -5 =0
=> x =5

\(x^2-10x=-25\)
\(x^2-10x+25=0\)
\(\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(x=5\)

x2 - 10x = -25
x2 - 10x + 25 = 0
x2 - 2 * 5x + 52 = 0
( x - 5 )2 = 0
=> x - 5 = 0
x = 5

\(\left(2x-3\right)^2-x^2+10x-25=0\)
\(\left(2x-3\right)^2-\left(x-5\right)^2=0\)
\(\left(2x-3+x-5\right)\left(2x-3-x+5\right)=0\)
\(\left(3x-8\right)\left(x+2\right)=0\)
\(\Rightarrow3x-8=0\)hoặc \(x+2=0\)
=> \(x=\frac{8}{3}\) hoặc \(x=-2\)

x2+25=10x
25+x2=10x
25+(-10x)+x2=10x+(-10x)
25+(-10x)+x2=0
[5+(-1x)][5+(-1x)]=0
=>5+(-1x)=0
\(\Rightarrow x\in\left\{-5;5\right\}\)

Ta có : x2 - 10x = -25
=> x2 - 2.x.5 + 52 = 0
<=> (x - 5)2 = 0
=> x - 5 = 0
=> x = 5

x2 - 10x = - 25
x2 - 10x + 25 = 0
(x - 5)2 = 0
x - 5 = 0
x = 5