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a: Sửa đề: \(E=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{x-1}+4\sqrt{x}\right):\dfrac{x-1}{\sqrt{x}}\)
\(=\left(\dfrac{4\sqrt{x}+4\sqrt{x}\left(x-1\right)}{x-1}\right)\cdot\dfrac{\sqrt{x}}{x-1}\)
\(=\dfrac{4\sqrt{x}\left(1+x-1\right)}{x-1}\cdot\dfrac{\sqrt{x}}{x-1}=\dfrac{4x^2}{\left(x-1\right)^2}\)
b: Để E=2 thì \(4x^2=2\left(x-1\right)^2\)
\(\Leftrightarrow4x^2-2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow4x^2-2x^2+4x-2=0\)
\(\Leftrightarrow2x^2+4x-2=0\)
\(\Leftrightarrow x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)^2=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{2}\\x+1=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}-1\left(nhận\right)\\x=-\sqrt{2}-1\left(loại\right)\end{matrix}\right.\)
c: \(x=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
Thay x=2 vào E, ta được:
\(E=\dfrac{4\cdot2^2}{\left(2-1\right)^2}=16\)
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
a: \(P=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Khi \(x=4+2\sqrt{3}\) thì \(P=\dfrac{-\left(4+2\sqrt{3}-1\right)}{\sqrt{3}+1}=\dfrac{-\left(3+2\sqrt{3}\right)}{\sqrt{3}+1}=\dfrac{-3-\sqrt{3}}{2}\)
c: Để P<0 thì -(x-1)<0
=>x-1>0
=>x>1
\(D=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{x+2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(D=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
\(E=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right)\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\)
\(E=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
ĐK : a >= 0 , a khác 1
\(C=\left[\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\div\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\times\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\frac{a}{\sqrt{a}+1}\)
a: \(=\dfrac{\sqrt{m}\left(m+4n-4\sqrt{mn}\right)}{\sqrt{mn}\left(\sqrt{m}-2\sqrt{n}\right)}\)
\(=\dfrac{1}{\sqrt{n}}\cdot\left(\sqrt{m}-2\sqrt{n}\right)\)
b: \(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
c: \(=\sqrt{5^2\cdot2\cdot x^2y^4\cdot xy}-\dfrac{2y^2}{x^2}\cdot4\sqrt{2}\cdot x^3\sqrt{xy}+\dfrac{3}{2}xy\cdot\sqrt{2}\cdot y\cdot\sqrt{xy}\)
\(=5xy^2\sqrt{2xy}-8\sqrt{2xy}xy^2+\dfrac{3}{2}xy^2\cdot\sqrt{2xy}\)
\(=-\dfrac{3}{2}\sqrt{2xy}\)
d: \(=\left(x+2\right)\cdot\dfrac{\sqrt{2x-3}}{\sqrt{x+2}}=\sqrt{\left(2x-3\right)\left(x+2\right)}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
2+ 6/ căn x -1