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a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)
b) Ta có: \(M=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)^2-\sqrt{a}\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}\left[\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2\right]}{2\sqrt{a}}\)
\(=\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{2}\)
\(=\dfrac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
c) Để M=-4 thì \(-2\sqrt{a}=-4\)
\(\Leftrightarrow\sqrt{a}=2\)
hay a=4(thỏa ĐK)
Sửa đề: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{1;4\right\}\end{matrix}\right.\)
Ta có: \(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}-1}{\sqrt{a}+2}\right)\)
\(=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a+3\sqrt{a}+2-a+3\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{6\sqrt{a}}\)
\(=\dfrac{a-4}{6a\left(\sqrt{a}-1\right)}\)
c) Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{5-4\sqrt{5}}{6\left(9-4\sqrt{5}\right)\left(\sqrt{5}-3\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(9\sqrt{5}-27-20+12\sqrt{5}\right)}\)
\(=\dfrac{5-4\sqrt{5}}{6\left(21\sqrt{5}-47\right)}\)
\(=\dfrac{\left(5-4\sqrt{5}\right)\left(21\sqrt{5}+47\right)}{-24}\)
\(=\dfrac{105\sqrt{5}+235-420-188\sqrt{5}}{-24}\)
\(=\dfrac{-83\sqrt{5}-185}{-24}=\dfrac{83\sqrt{5}+185}{24}\)
Lời giải:
ĐK: $x>0; a\neq 1; a\neq 4$
a)
$M=\frac{\sqrt{a}-(\sqrt{a}-1)}{\sqrt{a}(\sqrt{a}-1)}:\frac{(\sqrt{a}+1)(\sqrt{a}-1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-2)(\sqrt{a}-1)}$
$=\frac{1}{\sqrt{a}(\sqrt{a}-1)}:\frac{3}{(\sqrt{a}-2)(\sqrt{a}-1)}=\frac{1}{\sqrt{a}(\sqrt{a}-1)}.\frac{(\sqrt{a}-2)(\sqrt{a}-1)}{3}=\frac{\sqrt{a}-2}{3\sqrt{a}}$
b)
$M>\frac{-1}{2}\Leftrightarrow \frac{\sqrt{a}-2}{3\sqrt{a}}+\frac{1}{2}>0$
$\Leftrightarrow \frac{5\sqrt{a}-4}{6\sqrt{a}}>0$
$\Leftrightarrow 5\sqrt{a}-4>0$
$\Leftrightarrow a>\frac{16}{25}$
Kết hợp với ĐKXĐ thì $a>\frac{16}{25}; a\neq 1; a\neq 4$
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
3:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)
\(M=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{6}{3\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\)
b: M>1/3
=>M-1/3>0
=>\(\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
=>\(\dfrac{6-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}>0\)
=>\(3-\sqrt{x}>0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
c: \(\sqrt{x}+3>=3\) với mọi x thỏa mãn ĐKXĐ
=>\(M=\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\) với mọi x thỏa mãn ĐKXĐ
Dấu = xảy ra khi x=0
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
\(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{a-1}\)
\(=\dfrac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-a-\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b: \(P=\sqrt{a}+7\)
=>\(2\left(a+2\sqrt{a}+1\right)=a+7\sqrt{a}\)
=>\(2a+4\sqrt{a}+2-a-7\sqrt{a}=0\)
=>\(a-3\sqrt{a}+2=0\)
=>\(\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[{}\begin{matrix}a=1\left(loại\right)\\a=4\left(nhận\right)\end{matrix}\right.\)
c: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}=\dfrac{2a-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+\dfrac{1}{4}+\dfrac{3}{4}\right)}{\sqrt{a}}=\dfrac{2\left[\left(\sqrt{a}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]}{\sqrt{a}}>0\)
=>P>6
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a\ne1\end{matrix}\right.\)
b: Sửa đề: \(C=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left[1:\dfrac{a+\sqrt{1}-\sqrt{a}}{\sqrt{a}+1}\right]\cdot\left[\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right]\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\dfrac{a-1}{a+1}\)
c: Để C là số nguyên thì \(a-1⋮a+1\)
=>\(a+1-2⋮a+1\)
=>\(-2⋮a+1\)
=>\(a+1\in\left\{1;-1;2;-2\right\}\)
=>\(a\in\left\{0;-2;1;-3\right\}\)
Kết hợp ĐKXĐ, ta được: a=0
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\notin\left\{1;4\right\}\end{matrix}\right.\)
\(A=\left(\dfrac{1}{x-4}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{\sqrt{x}-1}{x+2\sqrt{x}}\)
\(=\left(\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{1+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}}{\sqrt{x}-1}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Để A là số nguyên thì \(\sqrt{x}⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\inƯ\left(2\right)\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2\right\}\)
=>\(\sqrt{x}\in\left\{3;1;4;0\right\}\)
=>\(x\in\left\{9;1;16;0\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{9;16\right\}\)
c: A<0
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4 và x<>1
a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)
b) Với \(a>0;a\ne1;a\ne4\), ta có:
\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).
Lời giải:
a) ĐKXĐ: \(a>0; a\neq 1\)
\(M=\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)\left(\frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\frac{a-1}{2\sqrt{a}}.\frac{(a-\sqrt{a})(\sqrt{a}-1)-(a+\sqrt{a})(\sqrt{a}+1)}{(\sqrt{a}+1)(\sqrt{a}-1)}\)
\(=\frac{a-1}{2\sqrt{a}}.\sqrt{a}.\frac{(\sqrt{a}-1)^2-(\sqrt{a}+1)^2}{a-1}\)
\(=\frac{(\sqrt{a}-1)^2-(\sqrt{a}+1)^2}{2}=\frac{a+1-2\sqrt{a}-(a+1+2\sqrt{a})}{2}=\frac{-4\sqrt{a}}{2}=-2\sqrt{a}\)
b)
Để \(M=-4\Leftrightarrow -2\sqrt{a}=-4\Leftrightarrow \sqrt{a}=2\Rightarrow a=4\)
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