Question

[Nga Văn]

câu1 ; gp trình

a, \(\sqrt{x-x}+x=3\)

b, \(\sqrt{x^2-4x+4}=x-2\)

câu 2 rút gọn

A = \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)

B= \(\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}\)

Answer 1

[Nhiên An Trần]

Bài 1:

a, Sai đề

b, \(\sqrt{x^2-4x+4}=x-2\)

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

\(\Leftrightarrow\left|x-2\right|=x-2\)(*)

TH1: \(x\ge2\Rightarrow\left|x-2\right|=x-2\)

(*)\(\Leftrightarrow x-2=x-2\)

\(\Leftrightarrow0x=0\)\(\Rightarrow\)PT có vô số nghiệm

TH2: \(x< 2\Rightarrow\left|x-2\right|=2-x\)

(*)\(\Leftrightarrow2-x=x-2\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\)

Bài 2:

a, \(A=\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)

\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}\)

\(=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}\)

b, \(B=\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}\)\(\left(x\ge\dfrac{5}{2}\right)\)

\(=\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}\)

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

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

\(=\sqrt{2x-5}+3+\sqrt{2x-5}-1\)

\(=2\sqrt{2x-5}+2\)

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

Sai thì nhớ báo nhé bạn.

Answer 2

[Nga Văn]

câu a là \(\sqrt{x-2}+x=3\)