Bài 2 :

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

ĐKXĐ : \(x\ge1\)

Pt(1) tương đương :

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

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

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (*)

Xét \(x\ge2\Rightarrow\sqrt{x-1}-1\ge0\)

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

Khi đó pt (*) trở thành :

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

\(\Leftrightarrow2\sqrt{x-1}=2\)

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\) ( Thỏa mãn )

Xét \(1\le x< 2\) thì \(x\ge2\Rightarrow\sqrt{x-1}-1< 0\)

Nên : \(\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\). Khi đó pt (*) trở thành :

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

\(\Leftrightarrow2=2\) ( Luôn đúng )

Vậy tập nghiệm của phương trình đã cho là \(S=\left\{x|1\le x\le2\right\}\)

Bài 1 : 

a) ĐKXĐ : \(-1\le a\le1\)

Ta có : \(Q=\left(\frac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\frac{3}{\sqrt{1-a^2}}\right)\)

\(=\left(\frac{3+\sqrt{1-a}.\sqrt{1+a}}{\sqrt{1+a}}\right)\cdot\frac{\sqrt{1-a^2}}{3}\)

\(=\frac{3+\sqrt{\left(1-a\right)\left(1+a\right)}}{\sqrt{1+a}}\cdot\frac{\sqrt{\left(1-a\right)\left(1+a\right)}}{3}\)

\(=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\)

Vậy \(Q=\frac{\left(3+\sqrt{1-a^2}\right).\sqrt{1-a}}{3}\) với \(-1\le a\le1\)

b) Với \(a=\frac{\sqrt{3}}{2}\) thỏa mãn ĐKXĐ \(-1\le a\le1\)nên ta có :

\(\hept{\begin{cases}1-a=1-\frac{\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{\left(\sqrt{3}-1\right)^2}{2^2}\\1-a^2=1-\frac{3}{4}=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\sqrt{1-a}=\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2^2}}=\left|\frac{\sqrt{3}-1}{2}\right|=\frac{\sqrt{3}-1}{2}\\\sqrt{1-a^2}=\frac{1}{2}\end{cases}}\)

Do đó : \(Q=\frac{\left(3+\frac{1}{2}\right)\cdot\frac{\sqrt{3}-1}{2}}{3}=\frac{5\sqrt{3}-5}{12}\)

Tiếc quá 

mình chưa học đến

bik thì giúp cho

Bài 1:

a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\frac{5-\sqrt{5}}{\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)

\(=3\sqrt{5}-1\)

b) \(\sqrt{48}-6\sqrt{\frac{1}{3}}+\frac{\sqrt{3}-3}{\sqrt{3}}\)

\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}\)

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

c) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right)\div\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)\)

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

\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)

\(=-3\)

Bài 2:

đk: \(x\ge1\)

Ta có: \(\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\frac{x+1}{16}}=5\)

\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x-1}-2\sqrt{x+1}=5\)

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

\(\Rightarrow\sqrt{x-1}=-\frac{5}{3}\) (vô lý)

=> PT vô nghiệm

\(B=\frac{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}{3+\sqrt{5}}=3-\sqrt{5}\)

\(C=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)

\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\frac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\)

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

\(D=\frac{2}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}+3}\)

\(=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)

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

\(=\sqrt{3}-1-\sqrt{3}-2-3+\sqrt{3}=\sqrt{3}-6\)

Cảm ơn @Đường Quỳnh Gianh nhiều nhé <3 

a,=0

b,\(5\sqrt{5}\)

c=\(8\sqrt{7a}\)

d,=\(11\sqrt{3}\)

bạn lm ra luôn đc ko