Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Trục căn thức ở mẫu và rút gọn
a, (\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\))(\(\sqrt{6} +11\))
b,(\(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\))(\(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\))
c,\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)- (\(\sqrt{2}+\sqrt{3}\))
d,(\(\frac{5-2\sqrt{5}}{2-\sqrt{5}}-2\))(\(\frac{5+3\sqrt{5}}{3+\sqrt{5}}-2\))
Trục căn thức ở mẫu
a)\(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
b)\(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
a/ \(\frac{1}{2+\sqrt{3}}-\frac{1}{2-\sqrt{3}}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}-\frac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+5\sqrt{3}\)
\(=\frac{2-\sqrt{3}}{4-3}-\frac{2+\sqrt{3}}{4-3}+5\sqrt{3}\)
\(=2-\sqrt{3}-2-\sqrt{3}+5\sqrt{3}\)
\(=3\sqrt{3}\)
Vậy..
b/ \(\frac{1}{\sqrt{5}+2}-\sqrt{9+4\sqrt{5}}\)
\(=\frac{1}{\sqrt{5}+2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\frac{1}{\sqrt{5}+2}-\left|\sqrt{5}+2\right|\)
\(=\frac{\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}-\sqrt{5}-2\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
Vậy..
1. Tìm trục căn ở mẫu
a, \(\frac{3}{\sqrt{5}-\sqrt{2}}\) ; b,\(\frac{\sqrt{2}+1}{\sqrt{2}-1}\) ; c, \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\) ; d, \(\frac{1-a\sqrt{a}}{1-\sqrt{a}}\)
2. Rút gọn biểu thức:
A=\(\frac{1}{7+\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
B=\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
Trục căn thức ở mẫu và rút gọn:
a) \(\frac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
b) \(\frac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2.\sqrt{3+2\sqrt{5}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
giúp mik vs
trục căn thức ở mẫu
\(\frac{1\sqrt{2}}{\sqrt{2}.\sqrt{2}}\) \(\frac{1\sqrt[]{3}}{\sqrt{3}.\sqrt{3}}\)\(\frac{2\sqrt{5}-5}{\sqrt{5}-2}\)\(\frac{6-2\sqrt{6}}{2-\sqrt{6}}\)\(\frac{5\sqrt{6}-6\sqrt{5}}{\sqrt{5}-\sqrt{6}}\)\(\frac{1}{\sqrt{5}-1}\)\(\frac{1}{4-2\sqrt{a}+a}\)\(\frac{1}{b-2\sqrt{b}+4}\)\(\frac{1}{b-3\sqrt{b}+9}\)
a) Rút gọn biểu thức:\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}\)
b) Tìm giá trị nhỏ nhất của biểu thức B=\(x^2-x\sqrt{3}+1\)
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 \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
TRỤC CĂN THỨC Ở MẪU
a)\(\frac{1}{\sqrt{3}+\sqrt{2}+\sqrt{1}}\)
b)\(\frac{2\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\)
c)\(\frac{12\sqrt{5}}{\sqrt{5}+\sqrt{2}-\sqrt{7}}\)
mọi người giúp em với ạ, em cảm ơn <3
Trục căn ở mẫu:
a, \(\frac{9}{\sqrt{3}}\)
b, \(\frac{3}{\sqrt{5}-\sqrt{2}}\)
c, \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
d, \(\frac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}\)
a) \(\frac{9}{\sqrt{3}}=\frac{9\sqrt{3}}{3}=3\sqrt{3}\)
b) \(\frac{3}{\sqrt{5}-\sqrt{2}}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{\left(\sqrt{5}-\sqrt{2}\right)\left(\sqrt{5}+\sqrt{2}\right)}=\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}=\sqrt{5}+\sqrt{2}\)
c) \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\left(\sqrt{5}-\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\frac{5-2\sqrt{15}+3}{5-3}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}\)
d) \(\frac{1}{\sqrt{18}+\sqrt{8}-2\sqrt{2}}=\frac{1}{3\sqrt{2}+2\sqrt{2}-2\sqrt{2}}=\frac{1}{3\sqrt{2}}=\frac{\sqrt{2}}{3\sqrt{2}\cdot\sqrt{2}}=\frac{\sqrt{2}}{6}\)
