`#` `\text{dkhanhqlv}`
`(x+1)(x+4)=(2-x)(2+x)`
`<=>(x+1)(x+4)-(2-x)(2+x)=0`
`<=>x^2+x+4x+4-2^2+x^2=0`
`<=>x(2x+5)=0`
`<=>x=0` hoặc `2x+5=0`
`@TH1:x=0`
`@TH2:2x+5=0<=>x=-2,5`
Vậy ...
`(x+1)(x+4)=(2-x)(2+x)`
`<=> x^2 +4x+x+4=4-x^2`
`<=> x^2 +x^2 +4x+x=4-4`
`<=> 2x^2 +5x=0`
`<=> x(2x+5)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
x^2 + x + 4x + 4 = 4 - x^2
`<=> x^2 + 5x + 4 = 4 - x^2`
`<=> 2x^2 + 5x = 0 `
`<=> x(2x + 5) = 0`
`<=> x = 0` hoặc `2x + 5 = 0`
`<=> x = 0` hoặc `x = -5/2`
\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)
\(\Rightarrow x^2+4x+x+4=4-x^2\)
\(\Rightarrow2x^2+5x=0\)
\(\Rightarrow x\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)
\(\Leftrightarrow x^2+5\text{x}+4=4-x^2\)
\(\Leftrightarrow2\text{x}^2+5\text{x}=0\)
\(\Leftrightarrow x\left(2\text{x}+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)
\(\Leftrightarrow x^2+4x+x+4=4-x^2\)
\(\Leftrightarrow x^2+4x+x+4-4+x^2=0\)
\(\Leftrightarrow2x^2+5x=0\)
\(\Leftrightarrow x\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5}{2}\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0;\dfrac{-5}{2}\right\}\)
