1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)

a.

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

ĐKXĐ: \(x\ge1\)

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

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

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)

b.

\(\sqrt{4x^2-20x+25}=5-2x\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)

\(\Leftrightarrow\left|5-2x\right|=5-2x\)

\(\Leftrightarrow5-2x\ge0\)

\(\Leftrightarrow x\le\dfrac{5}{2}\)

c.

ĐKXĐ: \(x\ge3\)

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

\(\Rightarrow x^2-x-6=x-3\)

\(\Leftrightarrow x^2-2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=3\end{matrix}\right.\)

d.

ĐKXĐ: \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

\(\sqrt{x^2-x}=\sqrt{3-x}\)

\(\Rightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2=3\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (thỏa mãn)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

Sửa đề: căn x-5/căn x-3

a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

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

b: x-5căn x+6=0

=>căn x=2 hoặc căn x=3

=>x=9(loại) hoặc x=4(nhận)

Khi x=4 thì A=5/(2+3)=5/5=1

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

=>\(x^2+9-12\sqrt{x^2-25}=13x+5-12\sqrt{x^2-25}\)

<=> \(x^2-13x+4=0\)

........

\(=>x^2+11-12\sqrt{x^2-25}=13x+25-12\sqrt{x^2-25}\)

\(< =>x^2-13x-14=0\)

\(< =>\left(x+1\right)\left(x-14\right)=0\)

..............

`sqrt{x^2-25}-6=3sqrt{x+5}-2sqrt{x-5}(x>=5)`

`<=>sqrt{(x-5)(x+5)}+2sqrt{x-5}=3sqrt{x+5}+6`

`<=>sqrt{x-5}(sqrt{x+5}+2)=3(sqrt{x+5}+2)`

`<=>(sqrt{x+5}+2)(sqrt{x-5}-3)=0`

Vì `sqrt{x+5}+2>0`

`<=>sqrt{x-5}-3=0`

`<=>sqrt{x-5}=3`

`<=>x-5=9<=>x=14(tm)`

Vậy `x=14`

\(\sqrt{x^2-25}-6=3\sqrt{x+5}-2\sqrt{x-5}\\ \Leftrightarrow\sqrt{\left(x-5\right)\left(x+5\right)}-6-3\sqrt{x+5}+2\sqrt{x-5}=0\\ \Leftrightarrow\left(2\sqrt{x-5}+\sqrt{\left(x-5\right)\left(x+5\right)}\right)-\left(3\sqrt{x+5}+6\right)=0\Leftrightarrow\sqrt{x-5}\left(2+\sqrt{x+5}\right)-3\left(2+\sqrt{x+5}\right)=0\\ \Leftrightarrow\left(\sqrt{x-5}-3\right)\left(2+\sqrt{x-5}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x-5}=3\\\sqrt{x-5}=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=9\\x\in\varnothing\end{matrix}\right.\Leftrightarrow x=14\)

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)

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

\(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\sqrt{x}-5}-\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}\)

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

a: Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}+\sqrt{2}\right)\)

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

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

=3-1

=2

b: Ta có: \(\dfrac{x-25}{\sqrt{x}-5}-\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}\)

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

=3