CMR\(\sqrt{1^3+2^3}=1+2\)
ta thấy :
\(\sqrt{1^3+2^3}\)
=\(\sqrt{\left(1+2\right)^2}\)
= \(1+2\)
Vậy => \(\sqrt{1^3+2^3}=1+2\)
CMR\(\sqrt{1^3+2^3}=1+2\)
ta thấy :
\(\sqrt{1^3+2^3}\)
=\(\sqrt{\left(1+2\right)^2}\)
= \(1+2\)
Vậy => \(\sqrt{1^3+2^3}=1+2\)
\(A=\frac{3+\sqrt{3+\sqrt{3+\sqrt{3+....+\sqrt{3}}}}}{6.\sqrt{3+\sqrt{3+\sqrt{3+...+\sqrt{3}}}}}có2013\sqrt{3}\)
CM :A< \(\frac{1}{2}\)
P=\(\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
a,Rút gọn P
b,Tính P khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
P=\(\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)
a,Rút gọn P
b,Tính P khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
thực hiện phép tính:
1, \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2,\(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}+\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\)
3,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}+\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}\)
4,\(\dfrac{3-\sqrt{3}}{2\sqrt{3}-1}+\dfrac{3+\sqrt{3}}{2\sqrt{3}-1}\)
5,\(\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+\sqrt{3}}\)
Tìm giá trị của biểu thức;
a,\(\sqrt{5}\left(\sqrt{6}+1\right):\frac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}-\sqrt{2}}}\)
b,\(\frac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}+\frac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
c,\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
d,\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\dfrac{\sqrt{2}-1}{\sqrt{2}+1}-\dfrac{3-\sqrt{2}}{3+\sqrt{2}}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
B=\((\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}):\frac{\sqrt{x}}{\sqrt{x}-3}\)) (vs x>0; x \(\ne\)9)
a,Rut gon B
b,Tinh gt cua B khi x=\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
c,CM: B>\(\frac{1}{3}\)
(giup mk vs..)
rút gọn biểu thức
K=\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{25\sqrt{24}+24\sqrt{25}}\)
M=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
N=\(\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)