Bài 1:

a: \(\sqrt{27}+\dfrac{1}{2}\sqrt{48}-\sqrt{108}\)

\(=3\sqrt{3}+\dfrac{1}{2}\cdot4\sqrt{3}-6\sqrt{3}\)

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

b: \(\left(\sqrt{14}-\sqrt{10}\right)\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{2}\cdot\sqrt{6+\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{12+2\sqrt{35}}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)=7-5=2\)

c: \(\dfrac{\sqrt{15}+\sqrt{3}}{1+\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)

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

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

Bài 2:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)

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

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

\(=\dfrac{x-5+\sqrt{x}-1+2\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

b: A=2

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

=>\(2\sqrt{x}-2=\sqrt{x}\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

c: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1+1⋮\sqrt{x}-1\)

=>\(\sqrt{x}-1\inƯ\left(1\right)\)

=>\(\sqrt{x}-1\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{2;0\right\}\)

=>\(x\in\left\{4;0\right\}\)

a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:

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

b: Ta có: P=A:B

\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)

\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

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

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

Bài 2:

Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)

\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)

\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)

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

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

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

=1

câu 1. đkxđ: \(x\ge\frac{1}{2}\)
\(A\sqrt{2}=\sqrt{2x+2\sqrt{2x-1}}-\sqrt{2x-2\sqrt{2x-1}}\)

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

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

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

nếu \(\left|\sqrt{2x-1}-1\right|=\sqrt{2x-1}-1\) với \(\sqrt{2x-1}\ge1\Leftrightarrow x\ge1\)

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

=> A=\(\sqrt{2}\)

nếu \(\left|\sqrt{2x-1}-1\right|=1-\sqrt{2x-1}\) với \(\frac{1}{2}\le x< 1\)

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

=> A= \(\sqrt{4x-2}\)

câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)

\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)

=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)

\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)

\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)

= 4

1/Em không chắc nha, nhất là câu c ý, nó sai sai hay là em làm sai nhỉ?

a) ĐK \(x\ge0\). Bình phương hai vế:

\(x+5=x+2\sqrt{x}+1\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4\) (TMĐK)

b)ĐK \(0\le x\le1\) . Bình phương hai vế:

\(2\sqrt{x\left(1-x\right)}=0\Leftrightarrow x\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\left(TMĐK\right)\)

c) ĐK: \(\left\{{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\Leftrightarrow5\le x\le3\) (vô lí))

Vậy không tồn tại x thỏa mãn đề bài.

a) \(\sqrt{x-1}+\sqrt{2x-1}=5\)

\(\Leftrightarrow3x-2+2\sqrt{\left(x-1\right)\left(2x-1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=25-3x+2\)

\(\Leftrightarrow2\sqrt{\left(x-1\right)\left(2x-1\right)}=-3x+27\)

Bình phương 2 vế, ta được:

\(\Leftrightarrow4\left(x-1\right)\left(2x-1\right)=9\left(x-9\right)^2\)

\(\Leftrightarrow8x^2-4x-8x+4=9x^2-162x+729\)

\(\Leftrightarrow8x^2-12x+4-9x^2+162x-729=0\)

\(\Leftrightarrow-x^2+150x-725=0\)

\(\Leftrightarrow x^2-150x+725=0\)

\(\Leftrightarrow\left(x-145\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-145=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=145\left(ktm\right)\\x=5\left(tm\right)\end{cases}}\)

\(\Rightarrow x=5\)

b) \(x+\sqrt{2x-1}-2=0\)

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

Bình phương 2 vế, ta được:

\(\Leftrightarrow2x-1=4-4x^2+x^2=0\)

\(\Leftrightarrow2x-1-4+4x-x^2=0\)

\(\Leftrightarrow6x-5-x^2=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(ktm\right)\\x=1\left(tm\right)\end{cases}}\)

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)

\(=\dfrac{20-6}{2}=7\)

1.\(\sqrt{1+2\sqrt{5}+5}=\sqrt{\left(1+\sqrt{5}\right)^2}=1+\sqrt{5}\)

2.\(\sqrt{10-4\sqrt{6}}=\sqrt{4-4\sqrt{6}+6}=\sqrt{\left(2-\sqrt{6}\right)^2}=\left|2-\sqrt{6}\right|=\sqrt{6}-2\)       \(\sqrt{15-6\sqrt{6}}=\sqrt{9+6\sqrt{6}+6}=\sqrt{\left(3+\sqrt{6}\right)^2}=3+\sqrt{6}\)

=>\(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)=\(3+\sqrt{6}-\sqrt{6}+2\)=5

3. Tương tự bằng :\(8-3\sqrt{6}\)

1) \(\sqrt{6+2\sqrt{5}}\) = \(\sqrt{1+2.1.\sqrt{5}+\sqrt{5}^2}\) =  \(\sqrt{\left(1+\sqrt{5}\right)^2}\)

2) \(\sqrt{15-6\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)

= \(\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{2^2.2.2.\sqrt{6}+\sqrt{6}^2}\)

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

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

= 3 + \(\sqrt{6}\) - 2 + \(\sqrt{6}\)

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

3) \(\sqrt{31-10\sqrt{6}}-\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\sqrt{5^2-2.5.\sqrt{6}+\sqrt{6}^2}\) - \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\sqrt{\left(5-\sqrt{6}\right)^2}\) - \(\sqrt{\left(3-2\sqrt{6}\right)^2}\)

= \(\left|5-\sqrt{6}\right|\) -  \(\left|3-2\sqrt{6}\right|\)

= 5 - \(\sqrt{6}-3-2\sqrt{6}\)

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

 Chúc bạn học tốt