g) \(x^2-4x+4=16\)
\(\Leftrightarrow x^2-2.2x+2^2=16\)
\(\Leftrightarrow\left(x-2\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
KL:....
h) \(\left(x+1\right)^2=x+1\)
Xét 2 TH:
TH1: \(x+1=0\Leftrightarrow x=-1\)
Thử lại ta thấy x=-1 t/m
TH2: \(x+1\ne0\Leftrightarrow x\ne-1\)
Khi đó: \(\left(x+1\right)^2=x+1\)
\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)
KL:....
i) \(4x\left(x-3\right)-x\left(3-x\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)+x\left(x-3\right)=0\)
\(\Leftrightarrow5x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
KL:....
j) \(25x^2+10x+1=16\)
\(\Leftrightarrow\left(5x\right)^2+2.5x.1+1^2=16\)
\(\Leftrightarrow\left(5x+1\right)^2=16\Leftrightarrow\left[{}\begin{matrix}5x+1=16\\5x+1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=15\\5x=-17\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=3\\x=-\dfrac{17}{5}\end{matrix}\right.\)
g: =>(x-2)^2-16=0
=>(x-2-4)(x-2+4)=0
=>(x-6)(x+2)=0
=>x=6 hoặc x=-2
h: =>(x+1)^2-(x+1)=0
=>(x+1)(x+1-1)=0
=>x(x+1)=0
=>x=0 hoặc x=-1
i: =>4x(x-3)+x(x-3)=0
=>5x(x-3)=0
=>x=0 hoặc x=3
j: =>(5x+1)^2=16
=>5x+1=4 hoặc 5x+1=-4
=>x=-1 hoặc x=3/5
