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a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
Đặt VT là T
Áp dụng AM-GM cho 3 số dương, ta có:
\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)
\(T\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)(đpcm)
\(P=\dfrac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{2}{x+2\sqrt{x}}+\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+2\sqrt{x}\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{.....}+\dfrac{x+2}{....}\)
\(=\dfrac{\sqrt{x^3}+2x+2\sqrt{x}-2+x+2}{.....}=\dfrac{\sqrt{x^3}+3x+2\sqrt{x}}{....}\)
\(=\dfrac{\sqrt{x}\left(x+3\sqrt{x}+2\right)}{....}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{....}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P/S: Chú ý điều kiện khi rút gọn, tự tìm.
a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
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=\(\dfrac{3}{4}\)
2 = 1.2 => \(\dfrac{1}{2}\) = \(\dfrac{1}{1.2}\) = 1 - \(\dfrac{1}{2}\)
TT \(\dfrac{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\)
.................
=> VT = 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{\sqrt{2012-x}+2012}{\sqrt{2012-x}+2013}\)
Đặt \(\sqrt{2012-x}+2012=y\)
=> 1 - \(\dfrac{1}{x+1}\) = \(\dfrac{y}{y+1}\)
=> \(\dfrac{x}{x+1}\) = \(\dfrac{y}{y+1}\)
=> x = y
<=> x = \(\sqrt{2012-x}+2012\)
<=> 2012 - x + \(\sqrt{2012-x}\) = 0
<=> \(\sqrt{2012-x}=0\)
<=> x = 2012