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a. \(A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
\(A=\dfrac{\left(\sqrt{15}-\sqrt{12}\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(A=\dfrac{5\sqrt{3}+2\sqrt{15}-2\sqrt{15}-4\sqrt{3}}{5-4}-\dfrac{2+\sqrt{3}}{4-3}\)
\(A=\sqrt{3}-2-\sqrt{3}=-2\)
b.
\(B=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(B=\left[\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right].\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right).\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(B=\dfrac{-8\sqrt{a}}{a-4}.\dfrac{a-4}{\sqrt{a}}\)
\(B=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
\(a.A=\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}=\sqrt{3}-\dfrac{1}{2-\sqrt{3}}=\dfrac{2\sqrt{3}-4}{2-\sqrt{3}}\)
\(b.\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}.\dfrac{a-4}{\sqrt{a}}=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\left(a>0;a\ne4\right)\)
\(A=\dfrac{a-\sqrt{a}-6}{4-a}-\dfrac{1}{\sqrt{a}-2}=\dfrac{a+2\sqrt{a}-3\sqrt{a}-6}{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}-\dfrac{1}{\sqrt{a}-2}=\dfrac{\sqrt{a}-3}{2-\sqrt{a}}+\dfrac{1}{2-\sqrt{a}}=\dfrac{\sqrt{a}-2}{2-\sqrt{a}}=-1\) \(F=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1-\sqrt{a}}{1-a}.\dfrac{1-\sqrt{a}}{1-a}=\left(a+2\sqrt{a}+1\right).\dfrac{\left(1-\sqrt{a}\right)^2}{\left(\sqrt{a}+1\right)^2\left(1-\sqrt{a}\right)^2}=1\)
A=(\(\dfrac{\left(x+4\right)}{3\left(x+2\right)}-\dfrac{1}{\left(x+2\right)^2}\))(\(\dfrac{x+5+x-1}{x+5}\))
A=\(\dfrac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}\cdot\dfrac{2x+2}{x+5}\)
A=\(\dfrac{x^2+2x+4x+8-3}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+5}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+9-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{\left(x+3\right)^2-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{2\left(x+3-2\right)\left(x+3+2\right)}{3\left(x-2\right)\left(x+5\right)}\)
A=\(\dfrac{2\left(x+1\right)}{3\left(x-2\right)}\)
a: ĐKXĐ: a>=0; a<>1
b: \(A=\left(\dfrac{a+3\sqrt{a}+2}{3\sqrt{a}-2}-\dfrac{\sqrt{a}}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}-1+\sqrt{a}+1}{a-1}\)
\(=\left(\dfrac{\left(a-1\right)\left(\sqrt{a}+2\right)-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{a-1}{2\sqrt{a}}\)
\(=\dfrac{a\sqrt{a}+2a-\sqrt{a}-2-3a+2\sqrt{a}}{\left(3\sqrt{a}-2\right)}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(=\dfrac{\left(a\sqrt{a}-a+\sqrt{a}-2\right)}{3\sqrt{a}-2}\cdot\dfrac{\sqrt{a}+1}{2\sqrt{a}}\)
\(B=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{-8\sqrt{a}}{\sqrt{a}}=-8\)
a: \(=\dfrac{-a-2\sqrt{a}+a-2\sqrt{a}-4a-2\sqrt{a}+4}{a-4}:\dfrac{-2\sqrt{a}+2+\sqrt{a}}{\sqrt{a}\left(2-\sqrt{a}\right)}\)
\(=\dfrac{-4a-6\sqrt{a}+4}{a-4}\cdot\dfrac{-\sqrt{a}\left(\sqrt{a}-2\right)}{-\sqrt{a}+2}\)
\(=\dfrac{4a+6\sqrt{a}-4}{\sqrt{a}+2}\cdot\dfrac{\sqrt{a}}{2-\sqrt{a}}=\dfrac{\sqrt{a}\left(4a+6\sqrt{a}-4\right)}{4-a}\)
b: Để \(A=\sqrt{a}+2\) thì \(4a\sqrt{a}+6a-4\sqrt{a}=\left(\sqrt{a}+2\right)\left(4-a\right)=4\sqrt{a}-a\sqrt{a}+8-2a\)
=>\(5a\sqrt{a}+8a-8\sqrt{a}-8=0\)
=>\(5a\cdot\sqrt{a}+10a-2a-4\sqrt{a}-4\sqrt{a}-8=0\)
=>\(\left(\sqrt{a}+2\right)\left(5a-2\sqrt{a}-4\right)=0\)
=>\(5a-2\sqrt{a}-4=0\)
=>\(a=\dfrac{22+2\sqrt{21}}{25}\)
\(D=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(=\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\cdot\left(\dfrac{a-4}{\sqrt{a}}\right)\)
\(=\dfrac{a-2\sqrt{a}+4-a-2\sqrt{a}-4}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}\)
\(=\dfrac{-4\sqrt{a}\cdot\left(a-4\right)}{\sqrt{a}\cdot\left(a-4\right)}=-4\)
\(D=\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
\(D=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ D=\dfrac{-8\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ D=-\dfrac{8\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\dfrac{a-4}{\sqrt{a}}\\ D=-\dfrac{8}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)=-8\)
Vậy $D=-8$
Câu 2:
a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)
b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)
\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)
1/ đkxđ: a > 0; a khác 1
a/ A= (\(\dfrac{\sqrt{a}}{2\sqrt{a}}-\dfrac{1}{2\sqrt{a}}\))\(\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{\sqrt{a}-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{a-1}\)
\(=\dfrac{1}{2\sqrt{a}}\cdot\dfrac{-4a}{a-1}=-\dfrac{2\sqrt{a}}{a-1}=\dfrac{2\sqrt{a}}{a+1}\)
b/+) A = 4
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}=4\)\(\Leftrightarrow2\sqrt{a}=4a+4\)
=> Không có gt a nào t/m
+) \(A>-6\)
\(\Leftrightarrow\dfrac{2\sqrt{a}}{a+1}>-6\)
\(\Leftrightarrow2\sqrt{a}>-6a-6\)
\(\Leftrightarrow6a+2\sqrt{a}+6>0\) (luôn đúng vì a > 0)
=> bpt có nghiệm với mọi a > 0
vậy........
c/ \(a^2-3=0\Leftrightarrow\left[{}\begin{matrix}a=\sqrt{3}\left(tm\right)\\a=-\sqrt{3}\left(ktmđkxđ\right)\end{matrix}\right.\)
Với a = \(\sqrt{3}\) ta có:
\(A=\dfrac{2\sqrt{3}}{\sqrt{3}+1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3-1}=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{2}=\sqrt{3}\left(\sqrt{3}-1\right)=3-\sqrt{3}\)
Ta có:\(A=\left(\dfrac{a+4\sqrt{a}+4}{a+2\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{a-2\sqrt{a}}-\dfrac{3\sqrt{a}+6}{4-a}\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}\left(\sqrt{a}+2\right)}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right]:\left[\dfrac{\sqrt{a}-4}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{3\left(\sqrt{a}+2\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right]\)
\(=\dfrac{a-4-a-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}:\dfrac{\sqrt{a}-4+3\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\dfrac{-4-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}.\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{4\sqrt{a}-4}=\dfrac{-2-\sqrt{a}}{2\sqrt{a}-2}\)
Ta có: \(A=\left(\dfrac{a+4\sqrt{a}+4}{a+2\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{a-2\sqrt{a}}-\dfrac{3\sqrt{a}+6}{4-a}\right)\)
\(=\left(\dfrac{\sqrt{a}+2}{\sqrt{a}}-\dfrac{\sqrt{a}}{\sqrt{a}-2}\right):\left(\dfrac{\sqrt{a}-4}{\sqrt{a}\left(\sqrt{a}-2\right)}+\dfrac{3}{\sqrt{a}-2}\right)\)
\(=\dfrac{a-4-a}{\sqrt{a}\left(\sqrt{a}-2\right)}:\dfrac{\sqrt{a}-4+3\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-2\right)}\)
\(=\dfrac{-4}{4\left(\sqrt{a}+1\right)}=\dfrac{-1}{\sqrt{a}+1}\)