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PA
17 tháng 7 2017
\(\dfrac{1+3\sqrt{2}-2\sqrt{3}}{\sqrt{6}+\sqrt{3}+\sqrt{2}}\)
\(=\dfrac{\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}{\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]}\)
Tử:
\(\left[1+\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)
\(=\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)+6\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{6}\) (nhân phân phối)
\(=5\sqrt{3}-7\sqrt{2}\)
Mẫu:
\(\left[\sqrt{6}+\left(\sqrt{3}+\sqrt{2}\right)\right]\left[\sqrt{6}-\left(\sqrt{3}+\sqrt{2}\right)\right]\)
\(=6-\left(5+2\sqrt{6}\right)\)
\(=1-2\sqrt{6}\)
Ta có:
\(\dfrac{5\sqrt{3}-7\sqrt{2}}{1-2\sqrt{6}}\)
\(=\dfrac{\left(5\sqrt{3}-7\sqrt{2}\right)\left(1+2\sqrt{6}\right)}{1-24}\)
\(=\dfrac{5\sqrt{3}+30\sqrt{2}-7\sqrt{2}-28\sqrt{3}}{-23}\)
\(=\dfrac{-23\left(\sqrt{3}-\sqrt{2}\right)}{-23}\)
\(=\sqrt{3}-\sqrt{2}\)
G
please help me
0
vs
please help me!!!!
GD
20 tháng 7 2017
Bài 1:
a)
\(A=\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)\left(\dfrac{x-\sqrt{x}}{\sqrt{x}+1}-\dfrac{x+\sqrt{x}}{\sqrt{x}-1}\right)\) ĐKXĐ: x >1
\(=\left(\dfrac{2\sqrt{x}.\sqrt{x}}{2.2\sqrt{x}}-\dfrac{2}{2.2\sqrt{x}}\right)\left(\dfrac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)^2}-\dfrac{\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)^2}\right)\\ =\left(\dfrac{2x-2}{4\sqrt{x}}\right)\left(\dfrac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{\left(x-1\right)^2}\right)\\ =\left(\dfrac{x-1}{2\sqrt{x}}\right)\left(\dfrac{-4x}{\left(x-1\right)^2}\right)\\ =\dfrac{\left(x-1\right).\left(-4x\right)}{2\sqrt{x}.\left(x-1\right)^2}=\dfrac{-2\sqrt{x}}{x-1}\)
b)
Với x >1, ta có:
A > -6 \(\Leftrightarrow\dfrac{-2\sqrt{x}}{x-1}>-6\Rightarrow-2\sqrt{x}>-6\left(x-1\right)\)
\(\Leftrightarrow-2\sqrt{x}+6x-6>0\\ \Leftrightarrow x-\dfrac{2}{6}\sqrt{x}-1>0\\ \Leftrightarrow x-2.\dfrac{1}{6}\sqrt{x}+\left(\dfrac{1}{6}\right)^2>1+\dfrac{1}{36}\\ \Leftrightarrow\left(\sqrt{x}-\dfrac{1}{6}\right)^2>\dfrac{37}{36}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{6}-\sqrt{x}>\dfrac{\sqrt{37}}{6}\\\sqrt{x}-\dfrac{1}{6}>\dfrac{\sqrt{37}}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-\sqrt{x}>\dfrac{\sqrt{37}-1}{6}\\\sqrt{x}>\dfrac{\sqrt{37}+1}{6}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}-x>\dfrac{19-\sqrt{37}}{18}\\x>\dfrac{19+\sqrt{37}}{18}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x< \dfrac{\sqrt{37}-19}{18}\\x>\dfrac{19+\sqrt{37}}{18}\end{matrix}\right.\)
Vậy không có x để A >-6
QP
21 tháng 7 2017
1)\(\sqrt{x-2+2\sqrt{x-3}}\)=\(\sqrt{(x-3)+2\sqrt{x-3}+1}=\sqrt{(\sqrt{x-3}+1)^2}=\sqrt{x-3}+1 \)
2)\(\sqrt{x-1-2\sqrt{x-2}}=\sqrt{x-2-2\sqrt{x-2}+1}=\sqrt{(\sqrt{x-2}-1)^2}=\sqrt{x-2}-1\)
....
1. ĐK: x>3\(\sqrt{x-2+2\sqrt{x-3}}=\sqrt{\left(x-3\right)+2\sqrt{x-3}+1}=\sqrt{\left(\sqrt{x-3}+1\right)^2}=\sqrt{x-3}+1\\ \)2( Tương tự) Bớt 1 thêm 1