Question

(x+1)(x+4)=(2-x)(2+x)

Answer 1

(x+1)(x+4)=(2-x)(2+x)

x²+5x+4=4-x²

2x²+5x=0

x(2x+5)=0

x=0   ;  2x+5=0=>x=-5/2

Answer 2

x²+ 5x + 4 = 4 - x² 

<=> 2x² + 5x = 0 

<=> x(2x+5) = 0 

=> x = 0 hoặc x  -5/2

Answer 3

(x+1)(x+4)=(2-x)(2+x)

x²+5x+4=4-x²

2x²+5x=0

x(2x+5)=0

x=0   ;  2x+5=0=>x=

Answer 4

x² + 5x + 4 = 4 - x² 

<=> 2x² + 5x = 0 

<=> x. (2x+5) = 0 

=> x = 0 hoặc \(x=-\dfrac{5}{2}\)

Answer 5

\(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)

\(\Leftrightarrow x^2+5x+4=2^2-x^2\)

\(\Leftrightarrow2x^2+5x=0\)

\(\Leftrightarrow x\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Answer 6

(x+1)(x+4)=(2-x)(2+x)

<=> \(x^2+x+4x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+5x=0\)

\(\Leftrightarrow x\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;-\dfrac{5}{2}\right\}\)

Answer 7

a) (x + 1)(x + 4) = (2 - x)(2 + x)

\(\Leftrightarrow\) (x + 1)x + 4(x + 1) = 4 - x²

\(\Leftrightarrow\) x² + x + 4x + 4 = 4 - x²

\(\Leftrightarrow\) 2x² + 5x=0

\(\Leftrightarrow\) x(2x + 5) = 0

Ta được :

+) TH1 : x = 0

+) TH2 : 2x + 5 = 0

\(\Leftrightarrow\) 2x = - 5

\(\Leftrightarrow\) x= \(\dfrac{-5}{2}\) = - 2,5

Answer 8

Ta có: \(\left(x+1\right)\left(x+4\right)=\left(2-x\right)\left(2+x\right)\)

\(\Leftrightarrow x^2+5x+4=4-x^2\)

\(\Leftrightarrow x^2+5x+4-4+x^2=0\)

\(\Leftrightarrow2x^2+5x=0\)

\(\Leftrightarrow x\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;\dfrac{-5}{2}\right\}\)