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a, ĐK: \(a\ge0;\) a khác 4
b,\(A=\frac{a+3\sqrt{a}+2+2a-4\sqrt{a}-5\sqrt{a}-2}{\left(\sqrt{a}-2\right)\cdot\left(\sqrt{a}+2\right)}=\frac{3a-6\sqrt{a}}{\left(\sqrt{a}+2\right)\cdot\left(\sqrt{a}-2\right)}=\frac{3\sqrt{a}}{\sqrt{a}+2}\)
c, để A= 2 KHI \(\frac{3\sqrt{a}}{\sqrt{a}+2}=2\)
<=>\(3\sqrt{a}=2\sqrt{a}+4\)
<=>\(\sqrt{a}=4\)
<=>a=16
tick nha
1,
\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)
\(=\frac{\left(a\sqrt{a}-1\right)\left(a+\sqrt{a}\right)-\left(a\sqrt{a}+1\right)\left(a-\sqrt{a}\right)}{a^2-a}.\frac{a-2}{a+2}\)
\(=\frac{a^2\sqrt{a}+a^2-a-\sqrt{a}-\left(a^2\sqrt{a}-a^2+a-\sqrt{a}\right)}{a\left(a-1\right)}.\frac{a-2}{a+2}\)
\(=\frac{2a\left(a-1\right)\left(a-2\right)}{a\left(a-1\right)\left(a+2\right)}=\frac{2\left(a-2\right)}{a+2}\)
Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)
2,
a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)
b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)
\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)
\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)
c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)
Bài 2:
\(\Leftrightarrow3\sqrt{x+5}-2\sqrt{x+5}=7\)
\(\Leftrightarrow\sqrt{x+5}=7\)
=>x+5=25
hay x=18
a: ĐKXĐ: \(x>0\)
b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}\)
\(=2\sqrt{x}-1\)
a)\(\hept{\begin{cases}a\ge0\\\sqrt{a}-2>0\Leftrightarrow\\\sqrt{a}+2>0\end{cases}a>4}\)
b)\(\frac{\sqrt{a}\left(\sqrt{a}+2\right)+\sqrt{a}\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}.\frac{a-4}{2\sqrt{a}}\) \(=\frac{2a}{a-4}.\frac{a-4}{2\sqrt{a}}=\sqrt{a}\)
c)\(\sqrt{a}>3\Leftrightarrow a>9\)
a: ĐKXĐ: x>0; x<>1
b: \(A=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2}{x-1}\)
c: A nguyên
=>x-1 thuộc {1;-1;2;-2}
=>x thuộc {2;3}
a) đk: \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)
Ta có:
\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)
\(C=\frac{a-2\cdot\left(\sqrt{a}+4\right)-2\cdot\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(C=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(C=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)
b) Ta có: \(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow\sqrt{a}=\sqrt{5}-2\)
Khi đó: \(C=\frac{\sqrt{5}-2}{\sqrt{5}-2+4}=\frac{\sqrt{5}-2}{\sqrt{5}+2}=\frac{\left(\sqrt{5}-2\right)^2}{1}=9-4\sqrt{5}\)
\(C=\frac{a}{a-16}-\frac{2}{\sqrt{a}-4}-\frac{2}{\sqrt{a}+4}\)
a) ĐKXĐ : \(\hept{\begin{cases}a\ge0\\a\ne16\end{cases}}\)
\(=\frac{a}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}+4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}-\frac{2\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{a-2\sqrt{a}-8-2\sqrt{a}+8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{a-4\sqrt{a}}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-4\right)}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}=\frac{\sqrt{a}}{\sqrt{a}+4}\)
b) Với \(a=9-4\sqrt{5}\)( tmđk )
\(C=\frac{\sqrt{a}}{\sqrt{a}+4}=1-\frac{4}{\sqrt{a}+4}\)
\(C=1-\frac{4}{\sqrt{9-4\sqrt{5}}+4}\)
\(=1-\frac{4}{\sqrt{5-4\sqrt{5}+4}+4}\)
\(=1-\frac{4}{\sqrt{\left(\sqrt{5}-2\right)^2}+4}\)
\(=1-\frac{4}{\left|\sqrt{5}-2\right|+4}\)
\(=1-\frac{4}{\sqrt{5}-2+4}\)
\(=1-\frac{4}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}+2-4}{\sqrt{5}+2}\)
\(=\frac{\sqrt{5}-2}{\sqrt{5}+2}\)
\(=\frac{\left(\sqrt{5}-2\right)\left(\sqrt{5}-2\right)}{1}=9-4\sqrt{5}\)
