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a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{x-4}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)
b) Với \(x\ge0;x\ne1;x\ne4\):
Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:
\(P=\dfrac{\sqrt{3+2\sqrt{2}}+2}{\sqrt{3+2\sqrt{2}}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}+2}{\sqrt{\left(\sqrt{2}+1\right)^2}-1}\)
\(=\dfrac{\sqrt{2}+1+2}{\sqrt{2}+1-1}\)
\(=\dfrac{\sqrt{2}+3}{\sqrt{2}}\)
\(=\dfrac{2+3\sqrt{2}}{2}\)
c) Với \(x\ge0;x\ne1;x\ne4\),
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+3}{\sqrt{x}-1}=1+\dfrac{3}{\sqrt{x}-1}\)
Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên
\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)
\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)
\(\Rightarrow x\in\left\{4;16;0\right\}\)
Kết hợp với ĐKXĐ của \(x\), ta được:
\(x\in\left\{0;16\right\}\)
Vậy: ...
\(\text{#}Toru\)
P = (\(\dfrac{1}{\sqrt{x}-1}\) - \(\dfrac{1}{\sqrt{x}}\)) : (\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) - \(\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)) với 0 < \(x\) ≠ 1; 4
P = \(\dfrac{\sqrt{x}-\left(\sqrt{x}-1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): (\(\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right).\left(\sqrt{x-2}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\))
P = \(\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\): \(\dfrac{x-1-\left(x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) : \(\dfrac{3}{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}\)
P = \(\dfrac{1}{\sqrt{x}.\left(\sqrt{x}-1\right)}\) \(\times\) \(\dfrac{\left(\sqrt{x}-2\right).\left(\sqrt{x}-1\right)}{3}\)
P = \(\dfrac{\sqrt{x}-2}{3.\sqrt{x}}\)
P = \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\)
b, P = \(\dfrac{1}{4}\)
⇒ \(\dfrac{\sqrt{x}.\left(\sqrt{x}-2\right)}{3x}\) = \(\dfrac{1}{4}\)
⇒4\(x\) - 8\(\sqrt{x}\) = 3\(x\)
⇒ 4\(x\) - 8\(\sqrt{x}\) - 3\(x\) = 0
\(x\) - 8\(\sqrt{x}\) = 0
\(\sqrt{x}\).(\(\sqrt{x}\) - 8) = 0
\(\left[{}\begin{matrix}x=0\\\sqrt{x}=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=64\end{matrix}\right.\)
\(x=0\) (loại)
\(x\) = 64
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\a< >1\end{matrix}\right.\)
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: 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) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
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: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
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