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a)\(\dfrac{2-\sqrt{2}}{\sqrt{2}}\)
\(\Leftrightarrow\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
\(\Leftrightarrow\sqrt{2}-1\)
b)\(\sqrt{\dfrac{x-49}{\sqrt{x}-7}}\)
\(\Leftrightarrow\sqrt{\dfrac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}}\)
\(\Leftrightarrow\sqrt{\sqrt{x}+7}\)
c)\(\sqrt{7-2\sqrt{6}}\)
\(\Leftrightarrow\sqrt{6-2\sqrt{6}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{6}-1\right)^2}\)
\(\Leftrightarrow\sqrt{6}-1\)
d)\(\sqrt{4+2\sqrt{3}}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow\sqrt{3}+1\)
e)\(\sqrt{13-14\sqrt{3}}\)
Câu này có dư số 1 ở chỗ 14 phải k bn ???
\(a.2\sqrt{x-2}=16\left(ĐK:x\ge2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
\(b.\sqrt{x-1}>3\left(ĐK:x\ge1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}\le-10\left(ĐK:x\ge2\right)\\ \Leftrightarrow\sqrt{2x+4}\ge2\\ \Leftrightarrow2x+4\ge4\\ \Leftrightarrow2x\ge0\Leftrightarrow x\ge0\)
\(a.2\sqrt{x-2}=16\left(ĐK:x>2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)
b.\(\sqrt{x-1}>3\left(ĐK:x>1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)
\(c.-5\sqrt{2x+4}< -10\left(ĐK:x>-2\right)\\ \Leftrightarrow\sqrt{2x+4}>2\\ \Leftrightarrow2x+4>4\\ \Leftrightarrow2x>0\Leftrightarrow x>0\)
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)
1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha 
@Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé 
Hướng dẫn trả lời:
Ta có: √2+√x=32+x=3 . Vì hai vế đều dương, ta bình phương hai vế
(√2+√x)2=32⇔2+√x=9⇔√x=7⇔(√x)2=72⇔x=49(2+x)2=32⇔2+x=9⇔x=7⇔(x)2=72⇔x=49
Chọn đáp án D
(A) 1
(B) \(\sqrt{7}\)
(C) 7
(D) 49