Câu 2:
a: \(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+1\right)^2=0\)
hay \(x\in\left\{-5;-1;3\right\}\)
b: \(\Leftrightarrow\left(x^2-3x\right)^2-4x\left(x-3\right)+2x\left(x-3\right)-8=0\)
\(\Leftrightarrow x^2\left(x-3\right)^2-4x\left(x-3\right)+2x\left(x-3\right)-8=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x^2-3x-4\right)+2\left(x^2-3x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)=0\)
hay \(x\in\left\{4;-1;1;2\right\}\)
c: \(\Leftrightarrow\left(x^2+x\right)^2+3\left(x^2+x\right)=0\)
=>x(x+1)=0
=>x=0 hoặc x=-1
\(a,\left(x^2+2x\right)^2-14\left(x^2+2x\right)-15=0\\ \Rightarrow\left[\left(x^2+2x\right)^2-15\left(x^2+2x\right)\right]+\left[\left(x^2+2x\right)-15\right]=0\\ \Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)+\left(x^2+2x-15\right)=0\\ \Rightarrow\left(x^2+2x-15\right)\left(x^2+2x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2+2x-15=0\\x^2+2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-1\end{matrix}\right.\)
giúp mình câu 4 với câu 2 phần 2 với
giải giúp mình câu 2 với
