Question

Rút gọn các biểu thức :

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)

c) \(\sqrt{9x^2-2x}\) với \(x< 0\)

d) \(x-4+\sqrt{16-8x+x^2}\) với \(x>4\)

Answer 1

a, Ta có : \(4-2\sqrt{3}=3-2\sqrt{3}+1=\left(\sqrt{3}\right)^2-2\sqrt{3}\times1+1^2=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}\)

Ta có : \(\sqrt{3}>\sqrt{1}\)(vì 3>1)

\(\Leftrightarrow\sqrt{3}>1\Leftrightarrow\sqrt{3}-1>0\Rightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)

Ta có: \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)

Answer 2

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)=\(\sqrt{3}-1-\sqrt{3}=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) = \(\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)

= \(3+\sqrt{2}-3+\sqrt{2}\) = \(2\sqrt{2}\)

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

= \(x-4+\left|4-x\right|\) = \(x-4-4+x\) (vì \(x>4\))

= \(2x-8\)

Answer 3

a) \(\sqrt{4-2\sqrt{3}}-\sqrt{3}\) \(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)

\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)

\(=\sqrt{3}-1-\sqrt{3}\)

\(=-1\)

b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\) \(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\)

\(=\left|3+\sqrt{2}\right|-3+\sqrt{2}\)

\(=3+\sqrt{2}-3+\sqrt{2}\)

\(=2\sqrt{2}\)

c) \(\sqrt{9x^2}-2x\) với \(x<0\)

\(=\left|3x\right|-2x\)

\(= -3x - 2x\)

\(= -5x\) (vì \(x<0\) )

d) \(x-4+\sqrt{16-8x+x^2}\) với \(x>4\)

\(=x-4+\sqrt{\left(4-x\right)^2}\)

\(=x-4+\left|4-x\right|\)

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

\(=2x-8\) (vì \(x>4\) )