Đặt B= \(\sqrt{\sqrt{10}-3}-\sqrt{\sqrt{10}+3}\) với B<=0
Ta có B^2= ( \(\sqrt{\sqrt{10}-3}-\sqrt{\sqrt{10}+3}\))^2 =\(\sqrt{10}-3+\sqrt{10}+3-2.\sqrt{\left(\sqrt{10}-3\right).\left(\sqrt{10}+3\right)}\)
= \(2\sqrt{10}-2\sqrt{10-3^2}\) =
Đặt B= \(\sqrt{\sqrt{10}-3}-\sqrt{\sqrt{10}+3}\) với B<=0
Ta có B^2= ( \(\sqrt{\sqrt{10}-3}-\sqrt{\sqrt{10}+3}\))^2 =\(\sqrt{10}-3+\sqrt{10}+3-2.\sqrt{\left(\sqrt{10}-3\right).\left(\sqrt{10}+3\right)}\)
= \(2\sqrt{10}-2\sqrt{10-3^2}\) =
rút gọn biểu thức
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
rút gọn các biểu thức:
a) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}\)
c) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
d) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
e) \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
f) \(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\)
giải giúp mjk vs m.n :]] arigatou <3
rút gọn biểu thức
\(\dfrac{\sqrt{2}}{\sqrt{5+1}}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
Rút gọn các biểu thức sau
a,\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
b,\(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
c,\(C=\left(1-\dfrac{x+3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
d,\(D=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
e,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Hãy so sánh hai biểu thức sau:
\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
( \(\dfrac{3\sqrt{x}+6}{x-4}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{x-9}{\sqrt{x}-3}\)
rút gọn biểu thức
cho biểu thức A=\(\left(\dfrac{4x-9}{2\sqrt{x}-3}+\sqrt{x}\right)\cdot\dfrac{1}{x+2\sqrt{x}+1}\)
a)rút gọn
rút gọn biểu thức
a, \(\dfrac{1}{\sqrt{7-\sqrt{24}+1}}-\dfrac{1}{\sqrt{7+\sqrt{24}+1}}\)
b,\(\sqrt{\dfrac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
c,\(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4}+\sqrt{7}}+\dfrac{4-\sqrt{7}}{3\sqrt{7}-\sqrt{4}-\sqrt{7}}\)
1/Tính
A=\(\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}\)
B=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2}-\sqrt{3}}\)
C=\(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
D=\(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}\)
2/So sánh
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}\) và \(\dfrac{2.1016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)