Lần sau bạn chú ý ghi đầy đủ yêu cầu đề.
Coi đây là bài toán rút gọn.
Lời giải:
$\frac{3}{\sqrt{3}-2}+\frac{2}{2+\sqrt{3}}=\frac{3(\sqrt{3}+2)}{(\sqrt{3}-2)(\sqrt{3}+2)}+\frac{2(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}$
$=\frac{3\sqrt{3}+6}{3-4}+\frac{4-2\sqrt{3}}{2^2-3}$
$=-3\sqrt{3}-6+4-2\sqrt{3}=-2-5\sqrt{3}$
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
\(\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{2}+\sqrt{3}}-\frac{5}{1+\sqrt{6}}\)
\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
A=\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
B=\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}-\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
rút gọn biểu thức
Rút gọn: \(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
J = \(\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)\) . \(\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
Q = \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !
\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}\)
chứng minh rằng
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}=\frac{2}{\sqrt[]{x}}\)
1. Rút gọn
D = \(\frac{\sqrt{1+\frac{2\sqrt{2}}{3}}+\sqrt{1-\frac{2\sqrt{2}}{3}}}{\sqrt{1+\frac{2\sqrt{2}}{3}}-\sqrt{1-\frac{2\sqrt{2}}{3}}}\)
2. Chứng minh rằng:
\(\frac{a\sqrt{b}+b}{a-b}.\sqrt{\frac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\left(\sqrt{a}+\sqrt{b}\right)=b\) với ( a > b > 0 )
rút gọn biểu thức:
1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2/ \(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3/ \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
4/ \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
giúp minh với cần gấp lắm
\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
