Bài 1: Căn bậc hai

\(\left(\dfrac{x}{x-4}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

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

\(=\left(\dfrac{x-x-2\sqrt{x}}{x-4}\right):\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{-2\sqrt{x}}{x-4}:\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(x-4\right)}\)

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

\(=\dfrac{x-\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

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

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

3:

a: \(=\left(\sqrt{6a}+\dfrac{\sqrt{6a}}{3}+\sqrt{6a}\right):\sqrt{6a}\)

=1+1/3+1

=7/3

b: \(=\dfrac{2}{3a-1}\cdot\sqrt{3}\cdot a\cdot\left|3a-1\right|\)

\(=\dfrac{2\sqrt{3}\cdot a\left(1-3a\right)}{3a-1}=-2a\sqrt{3}\)

8:

a: ĐKXĐ: \(x\in R\)

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

=>|x-3|=3-x

=>x-3<=0

=>x<=3

b: ĐKXĐ: \(x\in R\)

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

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

=>2x-5<=0

=>x<=5/2

c: ĐKXĐ: \(x\in R\)

\(\sqrt{36x^2-12x+1}=5\)

=>|6x-1|=5

=>6x-1=5 hoặc 6x-1=-5

=>6x=6 hoặc 6x=-4

=>x=-2/3 hoặc x=1

9:

a: ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\1-x>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{5}{2}\\x< =1\end{matrix}\right.\)

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

=>2x+5=1-x

=>3x=-4

=>x=-4/3(nhận)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x^2-x>=0\\3-x>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< =3\\\left[{}\begin{matrix}x>=1\\x< =0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}1< =x< =3\\x< =0\end{matrix}\right.\)

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

=>x²-x=3-x

=>x²=3

=>\(x=\sqrt{3}\left(nhận\right);x=-\sqrt{3}\left(nhận\right)\)

c: ĐKXĐ: \(\left\{{}\begin{matrix}2x^2-3>=0\\4x-3>=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2>=\dfrac{3}{2}\\x>=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=\sqrt{\dfrac{3}{2}}\\x< =-\dfrac{\sqrt{3}}{2}\end{matrix}\right.\\x>=\dfrac{3}{4}\end{matrix}\right.\Leftrightarrow x>=\dfrac{3}{4}\)

\(\sqrt{2x^2-3}=\sqrt{4x-3}\)

=>2x^2-3=4x-3

=>2x^2-4x=0

=>2x(x-2)=0

=>x=2(nhận) hoặc x=0(loại)

a: \(A=\left(\dfrac{x-1}{2\sqrt{x}}\right)\cdot\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

b: A>-6

=>2căn x<6

=> căn x<3

=>0<x<9 và x<>1

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

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

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

\(=3\)

a) \(x-2\sqrt{x}\)

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

b) \(3x-2\sqrt{x}\)

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

c) \(4x+2\sqrt{x}\)

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

d) \(x-2\sqrt{x}+1\)

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

e) \(x\sqrt{x}-1\)

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

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

g) \(x\sqrt{x}+27\)

\(=\left(\sqrt{x}\right)^3+3^3\)

\(=\left(\sqrt{x}+3\right)\left(x+3\sqrt{x}+9\right)\)

h) \(x-y\)

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

i) \(x\sqrt{x}-y\sqrt{y}\)

\(=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3\)

\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)

k) \(4x-4\sqrt{x}+1\)

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

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

\(\sqrt{ab}-\sqrt{a}-\sqrt{b}+1\)

\(=\left(\sqrt{ab}-\sqrt{a}\right)-\left(\sqrt{b}-1\right)\)

\(=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)\)

\(=\left(\sqrt{b}-1\right)\left(\sqrt{a}-1\right)\)

1) \(5\sqrt{20}-3\sqrt{12}+5\sqrt{\dfrac{1}{5}}-2\sqrt{27}\)

\(=5\cdot2\sqrt{5}-3\cdot2\sqrt{3}+\sqrt{25\cdot\dfrac{1}{5}}-2\cdot3\sqrt{3}\)

\(=10\sqrt{5}-6\sqrt{3}+\sqrt{5}-6\sqrt{3}\)

\(=11\sqrt{5}\)

2) \(\dfrac{9}{\sqrt{10}-1}+\dfrac{5\sqrt{2}-\sqrt{5}}{\sqrt{5}}\)

\(=\dfrac{9\left(\sqrt{10}+1\right)}{\left(\sqrt{10}-1\right)\left(\sqrt{10}+1\right)}+\dfrac{\sqrt{5}\left(\sqrt{10}-1\right)}{\sqrt{5}}\)

\(=\dfrac{9\left(\sqrt{10}+1\right)}{10-1}+\sqrt{10}-1\)

\(=\sqrt{10}+1+\sqrt{10}-1\)

\(=2\sqrt{10}\)

3) \(\left(\dfrac{\sqrt{7}-\sqrt{21}}{\sqrt{3}+1}-\dfrac{\sqrt{5}-\sqrt{15}}{\sqrt{3}-1}\right)\cdot\sqrt{12-2\sqrt{35}}\)

\(=\left(\dfrac{\sqrt{7}\left(1+\sqrt{3}\right)}{\sqrt{3}+1}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{\sqrt{3}-1}\right)\sqrt{12-2\sqrt{35}}\)

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

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

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

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

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

\(=7-5\)

\(=2\)

1: \(=5\cdot2\sqrt{5}-3\cdot2\sqrt{3}-2\cdot3\sqrt{3}+\sqrt{5}\)

=11căn 5-12căn 3

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

3: \(=\left(\dfrac{\sqrt{7}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)

=(căn 7+căn 5)(căn 7-căn 5)

=7-5

=2