Giải:
\(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left[3\left(x+1\right)\right]^2=0\)
\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\)
\(\Leftrightarrow\left[4x-\left(3x+3\right)\right]\left[4x+\left(3x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0+3\\7x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-\dfrac{3}{7}\).
Chúc bạn học tốt!
16x2-9(x+1)2=0
=>(4x)2-[3(x+1)]2=0
=>(4x)2-(3x+3)2=0
=>[4x-(3x+3)][4x+(3x+3)]=0
=>(4x-3x-3)(4x+3x+3)=0
=>(x-3)(7x+3)=0
=>x-3=0 hoặc 7x+3=0
=>x=3 hoặc 7x=-3=>x=-\(\dfrac{3}{7}\)
vậy x=3 hoặc x=-\(\dfrac{3}{7}\)
\(=>\left(4x\right)^2-3^2\left(x+1\right)^2=0\)
\(=>\left(4x\right)-\left(3x+3\right)^2=0\)
\(=>\left(4x-3x-3\right)\left(4x+3x+3\right)\)
\(=>\left(x-3\right)\left(7x+3\right)=0\)
\(< =>\left[{}\begin{matrix}x-3=0\\7x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
16x2-9(x+1)2=0
(4x)2-(3(x+1))2=0
(4x)2-(3x+3)2=0
(4x-3x-3)(4x+3x+3)=0
=>TH1: 4x-3x-3=0 TH2:4x+3x+3=0
x=3 x=-3/7
Vậy x=3 hoặc x=-3/7
