Question

Tìm x biết: (x^2-25)^2-(x+5)^2=0

Answer 1

\(\left(x^2-25\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left[x^2-5^2\right]^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left[\left(x+5\right)\left(x-5\right)\right]^2-\left(x+5\right)^2=0\)

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

\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left[\left(x-5\right)+1\right]\left[\left(x-5\right)-1\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\x-4=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x=4\\x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=4\\x=6\end{matrix}\right.\)

Vậy: \(S=\left\{-5;6;4\right\}\)

Answer 2

Ta có ( x² - 25 )² - ( x + 5 )² = 0

Vì ( x² - 25 )² ≥ 0 ; ( x + 5 )² ≥ 0 

⇒ ( x² - 25 )² - ( x + 5 )² ≥ 0

Dấu " = " xảy ra khi 

\(\left[{}\begin{matrix}\left(x^2-25\right)^2=0\\\left(x+5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm5\\x=-5\end{matrix}\right.\Rightarrow x=-5\)

Vậy x = 5 

Answer 3

\(\left(x^2-25\right)^2-\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x^2-25\right)^2=\left(x+5\right)^2\)

\(\Leftrightarrow x^2-25=x+5\) hay \(\Leftrightarrow x^2-25=-\left(x+5\right)\)

\(\Leftrightarrow x^2-x-30=0\) hay \(\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+5\right)=0\) hay \(\left(x-4\right)\left(x+5\right)=0\)

\(\Rightarrow x\in\left\{-5;4;6\right\}\)