Question

\(a.\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}-1\) tìm x nhé

b.\(\sqrt{x+8+2\sqrt{x+7}}=2\)

Answer 1

cả 2 câu đều tìm x

Answer 2

Lời giải:

a) ĐKXĐ: \(x\geq 1\)

Ta có : \(\sqrt{x+2\sqrt{x-1}}=2\sqrt{x-1}-1\)

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

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+1)^2}=2\sqrt{x-1}-1\)

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

\(\Leftrightarrow \sqrt{x-1}=2\Rightarrow x-1=4\Rightarrow x=5\) (thỏa mãn)

Vậy.........

b) ĐKXĐ: \(x\geq -7\)

\(\sqrt{x+8+2\sqrt{x+7}}=2\)

\(\Leftrightarrow \sqrt{x+7+1+2\sqrt{x+7}}=2\)

\(\Leftrightarrow \sqrt{(\sqrt{x+7}+1)^2}=2\)

\(\Leftrightarrow \sqrt{x+7}+1=2\)

\(\Leftrightarrow \sqrt{x+7}=1\Rightarrow x+7=1\Rightarrow x=-6\) (thỏa mãn)

Vậy..............