\(\dfrac{3+\sqrt3}{\sqrt3+1}-\sqrt3\\=\dfrac{\sqrt3(\sqrt3+1)}{\sqrt3+1}-\sqrt3\\=\sqrt3-\sqrt3=0\)
\(\dfrac{3+\sqrt3}{\sqrt3+1}-\sqrt3\\=\dfrac{\sqrt3(\sqrt3+1)}{\sqrt3+1}-\sqrt3\\=\sqrt3-\sqrt3=0\)
1) \(\dfrac{2}{\sqrt{5}-2}+\dfrac{-2}{\sqrt{5}+2}\)
2) \(\dfrac{4}{1-\sqrt{3}}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3) \(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\)
4) \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}-3}{\sqrt{3}+1}\)
5) \(\dfrac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}+\dfrac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}\)
Tính:
1) \(\dfrac{3}{1-\sqrt{2}}+\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)
2) \(\dfrac{\sqrt{5}-1}{\sqrt{5}+1}+\dfrac{6}{1-\sqrt{5}}\)
3) \(\dfrac{\sqrt{2}+\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{6}+2}\)
4) \(\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}\)
5) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
Tính:
1) \(\dfrac{1}{1+\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}\)
2) \(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\)
3) \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}+\sqrt{2}}\)
4) \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{\sqrt{5}-3}\)
5) \(\dfrac{1}{\sqrt{2}-\sqrt{6}}-\dfrac{1}{\sqrt{6}+\sqrt{2}}\)
LM CHI TIẾT GIÚP MK NHÉ
Thực hiện phép tính
a) \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)
b) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)
c) \(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\)
e) \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}+7}\)
Rút gọn các biểu thức sau
a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\dfrac{\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\) b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\) c) \(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
d) \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}\) e) \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
chứng minh
\(\dfrac{3}{2}\)\(\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{6}\)
rút gọn
D=\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}\)\(-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
a)\(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\dfrac{1}{\sqrt{3}+5}\)
b)\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+3}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(\dfrac{\sqrt{6}-\sqrt{3}}{\sqrt{2}-1}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}+\dfrac{2}{\sqrt{2}+1}-\dfrac{4}{\sqrt{2}}\)
\(\dfrac{4}{\sqrt{5}+1}+\dfrac{5}{\sqrt{5}+2}+\dfrac{5}{\sqrt{5}+3}\)
có ai biết giải bài toán này k giúp mình với ?
1,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}}\)
2,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
3,\(\dfrac{3}{\sqrt{6}-\sqrt{3}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
4,\(\left(\sqrt{\dfrac{2}{3}-\sqrt{\dfrac{3}{2}}+\dfrac{5}{\sqrt{6}}}\right):\dfrac{6-\sqrt{6}}{1-\sqrt{6}}\)
5,\(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\times\left(\sqrt{3}+\sqrt{2}\right)\)
6,\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)
7, \(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
8,\(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
9,\(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
10,\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\)
11,\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
12,\(\dfrac{\sqrt{3}+2\sqrt{2}+\sqrt{3}-2\sqrt{2}}{\sqrt{3}+2\sqrt{2}-\sqrt{3}-2\sqrt{2}}\)
B1. ko sử dụng máy tính, rút gọn
\(D=\dfrac{1}{2}\sqrt{48}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}\)
\(E=\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}-\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}\)
\(F=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
B2.
\(G=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
so sánh G với 1
B3. giải pt
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)+2xy\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\)