Question

[Phương Nguyễn]

( x + 2 )( x + 4 )²( x + 8 ) = 63x²

Answer 1

[TNA Atula]

[(x+2)(x+8)].(x²+8x+16)=63x²

=> (x²+10x+16).(x²+8x+16)=63x²

Dat x²+9x+16=a nen : (a+x)(a-x)=63x²

=> a²-x²-63x²=0

=> a²-64x²=0

=> (a-8x).(a+8x)=0

=> \(\left\{{}\begin{matrix}a-8x=0\\a+8x=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a=8x\\a=-8x\end{matrix}\right.\)

Voi a=8x thi : x²+9x+16=8x

=> x²+x+16=0

Vi x²+x+16>0 => pt vo nghiem

Voi a=-8x thi : x²+9x+16=-8x

=> x²+17x+16=0

=> x²+x+16x+16=0

=> (x+1)(x+16)=0

=> \(\left\{{}\begin{matrix}x+1=0\\x+16=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=-1\\x=-16\end{matrix}\right.\)

Answer 2

[Trần Quốc Lộc]

\(\left(x+2\right)\left(x+4\right)^2\left(x+8\right)=63x^2\\ \Leftrightarrow\left(x^2+2x+8x+16\right)\left(x^2+8x+16\right)=63x^2\\ \Leftrightarrow\left(x^2+8x+16\right)^2+2x\left(x^2+8x+16\right)=63x^2\\ \Leftrightarrow\left(x^2+8x+16\right)^2+2x\left(x^2+8x+16\right)-63x^2=0\\ \Leftrightarrow\left[\left(x^2+8x+16\right)^2+2x\left(x^2+8x+16\right)+x^2\right]-64x^2=0\\ \Leftrightarrow\left(x^2+8x+16+x\right)^2-64x^2=0\\ \Leftrightarrow\left(x^2+9x+16+8x\right)\left(x^2+9x+16-8x\right)=0\\ \Leftrightarrow\left(x^2+17x+16\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left(x^2+16x+x+16\right)\left(x^2+x+\dfrac{1}{4}+\dfrac{63}{4}\right)=0\\ \Leftrightarrow\left[\left(x^2+16x\right)+\left(x+16\right)\right]\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{63}{4}\right]=0\\ \Leftrightarrow\left[x\left(x+16\right)+\left(x+16\right)\right]\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+16\right)=0\left(\text{Vì }\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-16\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{-1;-16\right\}\)