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\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}-\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{6\left(2+\sqrt{3}\right)}-4+\sqrt{2\left(2+\sqrt{3}\right)}}{2\sqrt{6}}\right)\)
\(=\frac{\sqrt{2+\sqrt{3}}}{2}.\left(\frac{2\sqrt{6}}{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}\right)\)
\(=\frac{\sqrt{6\left(2+\sqrt{3}\right)}}{\left|\sqrt{3}+3\right|-4+\left|\sqrt{3}+1\right|}\)
\(=\frac{\left|\sqrt{3}+3\right|}{\sqrt{3}+3-4+\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+3}{2\sqrt{3}}\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7-2\sqrt{10}}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}\right)^2+6\sqrt{2}+9}-\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{6}+\left(\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}+1}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{10}+\left(\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{\left(\sqrt{2}+3\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+3-\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{5}+1-\sqrt{5}+\sqrt{2}}\)
\(=\frac{3}{2\sqrt{2}+1}\)
*****~~~~~~~~~~*****
\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+\sqrt{6}}{\sqrt{6}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{\sqrt{6}\left(\sqrt{6}+1\right)}{\sqrt{6}+1}\)
\(=\sqrt{3}+\sqrt{6}\)
\(=\sqrt{3}\left(1+\sqrt{2}\right)\)
*****~~~~~~~~~~*****
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(=\sqrt{3}+2+\sqrt{2}\)
(Chúc bạn học tốt nha!)
a) Đặt \(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(A^2=5-2\sqrt{6}+2\sqrt{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+5+2\sqrt{6}\)
\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)
\(\Rightarrow A=\sqrt{12}\)
b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
a) \(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{1}{2\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}=\frac{2\sqrt{3}}{12}+\frac{2\sqrt{3}}{6}-\frac{6-2\sqrt{3}}{6}\)
\(=\frac{2\sqrt{3}}{12}+\frac{4\sqrt{3}}{12}-\frac{12-4\sqrt{3}}{12}=\frac{-12+10\sqrt{3}}{12}=\frac{-6+5\sqrt{3}}{6}\)
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
=(căn 6-11)(căn 6+11)
=6-121=-115
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^2-1^2}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}\right)^2-2^2}-\dfrac{12\left(3+\sqrt{6}\right)}{3^2-\left(\sqrt{6}\right)^2}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}\right)^2-11^2\)
\(=6-121\)
\(=-115\)
a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)
\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)
\(=\dfrac{3+2\sqrt{2}}{\sqrt{2}+1}+\dfrac{3-2\sqrt{2}}{\sqrt{2}-1}\)
=căn 2+1+căn 2-1=2căn 2
b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)
bạn ơi cho mình hỏi câu b chi tiết hơn đước ko ạ
mình chưa hiểu lắm
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)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
các bạn giải giúp mình nha
\(\frac{\sqrt{6\sqrt{2}-1}-1}{\sqrt{6\sqrt{2}-1}}=1-\frac{1}{\sqrt{6\sqrt{2}-1}}\)??
chắc hết rút gọn đc rồi...