Question

lm câu 3 câu 4 thôi ạ

Answer 1

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

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

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

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

\(\Leftrightarrow\left(x+3\right)\left(x+1\right)\left(x+2+\sqrt{5}\right)\left(x+2-\sqrt{5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+1=0\\x+2+\sqrt{5}=0\\x+2-\sqrt{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\\x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)

Vậy: Phương trình có tập nghiệm \(S=\left\{-3;-1;-\sqrt{5}-2;\sqrt{5}-2\right\}\)

Answer 2

lm cho em câu 4 nữa đc ko

Answer 3

Answer 4

em cảm ơn

Answer 5

a) \(x^{4^{ }}-x^2-2=x^4-2x^2+x^2-2=0\)

⇔\(x^4-2x^2+x^2-2=0\)

⇔\(\left(x^2-2\right)\left(x^2+1\right)=0\) nhưng x² + 1 > 0 với mọi x vì x² ≥ 0

⇔\(x^2-2=0\) 

⇔\(x^2=2\)

⇔\(x=2^2=4\)Vậy x = 4