cho x = a+1 - \(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}\) (a>0)
P=\(\frac{\sqrt{x}+\sqrt{x^{ }-2\sqrt{x}+1}+1}{\sqrt{x^{2^{ }}-2x+1}}\)
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24 tháng 7 2019
ĐKXĐ: Bạn tự làm nha
\(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-2\sqrt{x}-1+2\sqrt{x}+2\)
\(=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}+1\)
\(=\frac{x^2-\sqrt{x}+x+\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\frac{x^2+x+1}{x+\sqrt{x}+1}\)
\(B=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1\left(\sqrt{a}-1\right)-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)}{\sqrt{a}}.\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-1-2}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(a-1\right)}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
NV
13 tháng 7 2016
ĐKXĐ: \(x\ge4\)
a/ \(A=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left[\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\left(\frac{x-4-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(-3\right)}\)
\(=\frac{\sqrt{x}-2}{-3\sqrt{x}}\)
b/ A = 0 \(\Rightarrow\frac{\sqrt{x}-2}{-3\sqrt{x}}=0\Rightarrow\sqrt{x}-2=0\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
+) \(P=\frac{\sqrt{x}+\sqrt{x^2-2x+1}+1}{\sqrt{x^2-2x+1}}=\frac{\sqrt{x}+\left|x-1\right|+1}{\left|x-1\right|}\)
+) \(x=a+1-\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}\)
\(=a+1-\sqrt{\left(a+1\right)^2-2a+\frac{a^2}{\left(a+1\right)^2}}\)
\(=a+1-\sqrt{\left(a+1-\frac{a}{a+1}\right)^2}\) vì a > 0 => \(a+1-\frac{a}{a+1}=\frac{a^2+a+1}{a+1}>0\)
\(=a+1-\left(a+1-\frac{a}{a+1}\right)=\frac{a}{a+1}\)
=> \(\left|x-1\right|=\left|\frac{a}{a+1}-1\right|=\left|-\frac{1}{a+1}\right|=\frac{1}{a+1}\)
=> \(P=\frac{\sqrt{\frac{a}{a+1}}+\frac{1}{a+1}+1}{\frac{1}{a+1}}=\sqrt{a\left(a+1\right)}+a+2\)