Question

Rút gọn các biểu thức chứa căn

A= \(x-4-\sqrt{16-8x^2+x^4}\left(x< 4\right)\)

Answer 1

Lời giải:

\(A=x-4-\sqrt{16-8x^2+x^4}=x-4-\sqrt{4^2-2.4.x^2+(x^2)^2}\)

\(=x-4-\sqrt{(x^2-4)^2}=x-4-|x^2-4|\)

Nếu \(2\leq x< 4\):

\(|x^2-4|=|x-2||x+2|=(x-2)(x+2)=x^2-4\)

\(\Rightarrow A=x-4-(x^2-4)=x-x^2\)

Nếu \(-2\leq x< 2\):

\(|x^2-4|=|x-2||x+2|=(2-x)(x+2)=4-x^2\)

\(\Rightarrow A=x-4-(4-x^2)=x^2+x-8\)

Nếu $x< -2$:

\(|x^2-4|=|x+2||x-2|=(-x-2)(2-x)=x^2-4\)

\(\Rightarrow A=x-4-(x^2-4)=x-x^2\)

Answer 2

Lời giải:

\(A=x-4-\sqrt{16-8x^2+x^4}=x-4-\sqrt{4^2-2.4.x^2+(x^2)^2}\)

\(=x-4-\sqrt{(x^2-4)^2}=x-4-|x^2-4|\)

Nếu \(2\leq x< 4\):

\(|x^2-4|=|x-2||x+2|=(x-2)(x+2)=x^2-4\)

\(\Rightarrow A=x-4-(x^2-4)=x-x^2\)

Nếu \(-2\leq x< 2\):

\(|x^2-4|=|x-2||x+2|=(2-x)(x+2)=4-x^2\)

\(\Rightarrow A=x-4-(4-x^2)=x^2+x-8\)

Nếu $x< -2$:

\(|x^2-4|=|x+2||x-2|=(-x-2)(2-x)=x^2-4\)

\(\Rightarrow A=x-4-(x^2-4)=x-x^2\)