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\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2}{a-b}\)
=\(\frac{a+b+2\sqrt{ab}+a+b-2\sqrt{ab}}{a-b}=\frac{2\left(a+b\right)}{a-b}\)
b/\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
=\(\sqrt{a}+\sqrt{b}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)=\(\frac{a+b+2\sqrt{ab}+a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}=\frac{2a+2b+3\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
a) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)
Để A nguyên thì 4 ⋮ √x - 2
\(\Rightarrow\sqrt{x}-2\inƯ\left(4\right)\)
\(\Rightarrow\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)
Mà x \(\sqrt{x}\ge0\)
=> x thuộc {9; 1; 16; 0; 36}
b)
a: \(P=-5\sqrt{\dfrac{160}{90}}=-5\cdot\dfrac{4}{3}=-\dfrac{20}{3}\)
b: \(Q=\sqrt{a}-\sqrt{b}+2\sqrt{b}=\sqrt{a}+\sqrt{b}\)
\(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab-b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
\(=\left(\frac{\sqrt{b}}{\sqrt{a}\sqrt{a}-\sqrt{a}\sqrt{b}}-\frac{\sqrt{a}}{\sqrt{a}\sqrt{b}-\sqrt{b}\sqrt{b}}\right).\left(\sqrt{a}\sqrt{a}\sqrt{b}-\sqrt{b}\sqrt{b}\sqrt{a}\right)\)
\(=\left(\frac{\sqrt{b}}{\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\frac{\left(\sqrt{b}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\left(\sqrt{a}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}\sqrt{b}\right)}\right).\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{b}\right)^2-\left(\sqrt{a}\right)^2}{\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)}.\sqrt{a}\sqrt{b}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{b}\right)^2-\left(\sqrt{b}\right)^2\)
\(=b-a\)
\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}\)
\(=2\sqrt{b}\)
\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\)
\(D=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{-b+\sqrt{a}.\sqrt{b}}{\sqrt{b}}\)
\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right].\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right).\sqrt{b}}-\frac{\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)
\(D=\frac{\left[\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}\right]-\left(\sqrt{a}.\sqrt{b}-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)
\(D=\frac{2b.\sqrt{a}+2b.\sqrt{b}}{\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)}\)
\(D=\frac{2b.\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}\)
\(D=2\sqrt{b}\)
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2\) \(ĐKXĐ:\hept{\begin{cases}a\ge0\\b\ge0\\a\ne b\end{cases}}\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\left(\left(a+\sqrt{ab}+b\right)+\sqrt{ab}\right)\left(\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)
\(=1\)
\(\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\sqrt{ab}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}+\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\sqrt{a}-b\sqrt{b}+a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{a\left(\sqrt{a}+\sqrt{b}\right)-b\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(a-b\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}\right)\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\left(\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2.\)
\(=\left(\sqrt{a}+\sqrt{b}\right)^2\cdot\frac{1}{\left(\sqrt{a}+\sqrt{b}\right)^2}.\)\(=1\)
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
a: ta có: \(M=\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{a\left(\sqrt{ab}-a\right)+b\left(\sqrt{ab}+b\right)}{\left(\sqrt{ab}+b\right)\left(\sqrt{ab}-a\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)\cdot\sqrt{a}\cdot\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\)
\(=\dfrac{-\sqrt{ab}\left(a+b\right)+\left(a-b\right)\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}-\dfrac{a^2-b^2}{\sqrt{ab}\left(a-b\right)}\)
\(=\dfrac{-\sqrt{ab}}{\sqrt{ab}\left(a-b\right)}\)
\(=-\dfrac{1}{a-b}\)
b: Thay \(a=\sqrt{5}+1\) và \(b=\sqrt{5}-1\) vào M, ta được:
\(M=\dfrac{-1}{\sqrt{5}+1-\sqrt{5}+1}=\dfrac{-1}{2}\)
a)
\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)
\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)
\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)
b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)
Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)
\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)
\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)
Vậy...