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a: ĐKXĐ: a>=0; b>=0; ab<>0; a<>1\(M=\dfrac{3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-1\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{1}{a-1}=\dfrac{1}{a-1}\)
b: M nguyên khi a-1 thuộc {1;-1}
=>a thuộc {2;0}
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)
= \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)
= \(\frac{2}{a-1}\)
b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1
=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 }
a: ĐKXĐ: x>1; x<>2
b: \(P=\left(\dfrac{\sqrt{x}+\sqrt{x-1}}{x-x+1}-\sqrt{x-1}-\sqrt{2}\right)\cdot\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
\(=\left(\sqrt{x}-\sqrt{2}\right)\cdot\dfrac{\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+\sqrt{2}}{\sqrt{x}}\)
c: Khi x=3+2căn 2 thì
P=(-căn 2-1+căn 2)/(căn 2+1)=căn 2-1
\(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+\sqrt{b}}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\left(đk:a\ne b,a\ge0,b\ge0\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+\sqrt{b}\right)}.\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{a-2\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\dfrac{2}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2.2}{\left(\sqrt{a}-\sqrt{b}\right)^2\left(a-1\right)}=\dfrac{2}{a-1}\in Z\)
\(\Rightarrow a-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Do \(a\ge0\)
\(\Rightarrow a\in\left\{0;2;3\right\}\)
Ta có: \(P=\left(\dfrac{3\sqrt{a}}{a+\sqrt{ab}+b}-\dfrac{3a}{a\sqrt{a}-b\sqrt{b}}+\dfrac{1}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2a+2\sqrt{ab}+2b}\right)\)
\(=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}\cdot\dfrac{2}{a-1}\)
\(=\dfrac{2}{a-1}\)
Để P là số nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
hay \(a\in\left\{2;0;3\right\}\)
Lời giải:
ĐKXĐ: $x>0; x\neq 4$
Sửa lại đề 1 chút.
\(A=\left(\frac{1}{\sqrt{x}+2}+\frac{1}{\sqrt{x}-2}\right).\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{\sqrt{x}-2+\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\frac{\sqrt{x}-2}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\frac{2}{\sqrt{x}+2}\)
\(B=\frac{7}{3}A=\frac{14}{3(\sqrt{x}+2)}\)
Với mọi $x>0$ thì hiển nhiên $B>0$. Mặt khác, $\sqrt{x}+2\geq 2$ nên $B=\frac{14}{3(\sqrt{x}+2)}\leq \frac{14}{6}=\frac{7}{3}$
Vậy $0< B\leq \frac{7}{3}$. $B$ đạt giá trị nguyên thì $B=1;2$
$B=1\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=1$
$\Leftrightarrow x=\frac{64}{9}$ (thỏa mãn)
$B=2\Leftrightarrow \frac{14}{3(\sqrt{x}+2)}=2$
$\Leftrightarrow x=\frac{1}{9}$ (thỏa mãn)
a: \(P=\dfrac{3a-3\sqrt{ab}-3a+a+\sqrt{ab}+b}{a\sqrt{a}-b\sqrt{b}}:\dfrac{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}{2\left(a+\sqrt{ab}+b\right)}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{2\left(a+\sqrt{ab}+b\right)}{\left(a-1\right)\left(\sqrt{a}-\sqrt{b}\right)}\)
=2/a-1
b: Để P nguyên thì \(a-1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{2;0;3\right\}\)