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H9
HT.Phong (9A5)
CTVHS
23 tháng 8 2023
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
H
23 tháng 8 2023
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
9 tháng 7 2019
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
PK
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}}\)
0
TH
rút gọn biểu thức
\(D=\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)
1
2 tháng 10 2016
\(D=\sqrt{\frac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\frac{\left(5-2\sqrt{6}\right)^2}{25-24}}=5+2\sqrt{6}+5-2\sqrt{6}=10\)
15 tháng 7 2019
a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\)
\(=\frac{-2\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
15 tháng 7 2019
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)
\(=\frac{3-\sqrt{3}}{3}\)
\(=1-\frac{\sqrt{3}}{3}\)
\(\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{6}+\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}+\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{6}-\sqrt{2}\right)}\)
\(=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\frac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}\)
\(=\sqrt{5}+\sqrt{6}\)