Chọn D

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 ???

\(\sqrt{43-14\sqrt{3}}\) nha bạn. Mình nhầm =))))

\(a.2\sqrt{x-2}=16\left(ĐK:x\ge2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)

\(b.\sqrt{x-1}>3\left(ĐK:x\ge1\right)\Leftrightarrow x-1>9\Leftrightarrow x>10\)

\(c.-5\sqrt{2x+4}\le-10\left(ĐK:x\ge2\right)\\ \Leftrightarrow\sqrt{2x+4}\ge2\\ \Leftrightarrow2x+4\ge4\\ \Leftrightarrow2x\ge0\Leftrightarrow x\ge0\)

\(a.2\sqrt{x-2}=16\left(ĐK:x>2\right)\Leftrightarrow\sqrt{x-2}=8\Leftrightarrow x-2=64\Leftrightarrow x=66\)

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 \(x\ge2\)

mn ơi giải giúp mik bài não cũng đc a

mình cảm ơn mn nhiều ạ =))

tớ nghĩ tớ giải đc 1-2 bài gì đó nhưng tớ ko bít bấm can lm sao giải cho cậu đc

Bài 2 :

a) Sửa đề :

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

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

\(A=-1\)

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

\(B=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(B=\sqrt{2}+1-\sqrt{2}+1\)

\(B=2\)

c) \(C=\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

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

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

\(C=4\)

d) \(D=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)

\(D=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)

\(D=4+\sqrt{7}-\sqrt{7}\)

\(D=4\)

Bài 1 :

a) Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) có nghĩa

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)

TH1 :\(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow x\ge3}\)

TH2 : \(\hept{\begin{cases}x-1\le0\\x-3\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\le3\end{cases}\Leftrightarrow}x\le1}\)

Vậy để biểu thức có nghĩa thì \(\orbr{\begin{cases}x\ge3\\x\le1\end{cases}}\)

b) Để \(\sqrt{\frac{1-x}{x+2}}\)có nghĩa

\(\Leftrightarrow\frac{1-x}{x+2}\ge0\)

TH1 : \(\hept{\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-2\end{cases}\Leftrightarrow}-2\le x\le1}\)

TH2 : \(\hept{\begin{cases}1-x\le0\\x+2\le0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\le-2\end{cases}\Leftrightarrow x\in\varnothing}\)

Vậy để biểu thức có nghĩa thì \(-2\le x\le1\)