\(a,=\sqrt{5}-4\sqrt{5}-12\sqrt{5}=-15\sqrt{5}\\ b,=2\sqrt{3}-\dfrac{2+\sqrt{3}}{1}=2\sqrt{3}-2-\sqrt{3}=\sqrt{3}-2\\ c,=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}}\\ =\dfrac{2\left(x+1\right)}{2\sqrt{x}}=\dfrac{x+1}{\sqrt{x}}\)

a: \(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)-\sqrt{x^3}\)

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

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

b: \(\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\cdot\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)

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

\(=\left(\dfrac{1}{\sqrt{a}+1}\right)^2\cdot\left(a+\sqrt{a}+1+\sqrt{a}\right)\)

\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+3\sqrt{x}+9}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\dfrac{3x+9}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3x+9}{x+4\sqrt{x}+3}\)

b: Để A<-1 thì A+1<0

\(\Leftrightarrow\dfrac{3x+9+x+4\sqrt{x}+3}{x+4\sqrt{x}+3}< 0\)

\(\Leftrightarrow\dfrac{4x+4\sqrt{x}+12}{x+4\sqrt{x}+3}< 0\)

hay \(x\in\varnothing\)

help

  => đề sai rồi bạn

\(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) (ĐK: \(x>0;x\ne4\))

\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

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

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

1. ĐKXĐ: $x>0; x\neq 9$

\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)

2. ĐKXĐ: $x\geq 0; x\neq 4$

\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)

\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)

3. ĐKXĐ: $a\geq 0; a\neq 1$

\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)

\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)

a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)

\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)

\(=\dfrac{1}{2\sqrt{2}a}\)

\(=\dfrac{\sqrt{2}}{4a}\)

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

chịu đấy :v

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

\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)

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

\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)

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

\(=\dfrac{-x+1+x^2}{x-3}\)

d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)

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

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

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

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

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

e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

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

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

\(=4x-2\sqrt{x}+x\)

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

a) Ta có: \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left[2-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}\right]\left[2+\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right]\)\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^2-\left(\sqrt{3}\right)^2=4-3=1\) (đpcm)

b) Ta có \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)\(=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right].\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)\(=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Ta có đẳng thức : (2−3+√3√3+1).(2+3−√3√3−1)=1

xét vế trái ta có :(2−3+√3√3+1).(2+3−√3√3−1)  = 

a) ta co \(\left(2-\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right).\left(2+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)=\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)=1\)

b) ta co \(A=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)

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

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

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

Vay \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)