CMR: \(\sqrt[3]{3+\sqrt[3]{3}}+\sqrt[3]{3-\sqrt[3]{3}}< 2\sqrt[3]{3}\)
các bn giúp mk vs. mk cần gấp!
tks
CMR:\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{2010\sqrt{2009}}\)< \(\frac{88}{45}\)
Các bn giúp mk vs! mk cần rất gấp
tks!
tính
\(\dfrac{2}{\sqrt{3}+1}-\dfrac{1}{\sqrt{3}-2}+\dfrac{12}{\sqrt{3}+3}\)
giúp mk vs ạ mk cần gấp
BT=\(\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{12\left(3-\sqrt{3}\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\dfrac{2\left(\sqrt{3}-1\right)}{2}+\dfrac{2+\sqrt{3}}{4-3}+\dfrac{12\left(3-\sqrt{3}\right)}{9-3}\)
\(=\sqrt{3}-1+2+\sqrt{3}+2\left(3-\sqrt{3}\right)\)
\(=\sqrt{3}-1+2+\sqrt{3}+6-2\sqrt{3}=7\)
1) rút gọn
A= \(3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
B= \(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
C= \(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
Giúp mk vs ạ mk cần gấp
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)
\(=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)
\(=\dfrac{3}{2}\)
\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)
\(A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)
\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+3+\sqrt{3}=5\)
1) so sánh
a) \(\sqrt{33}-\sqrt{17}\) và \(6-\sqrt{15}\)
b) \(4\sqrt{5}\) và \(5\sqrt{3}\)
c) \(\sqrt{3\sqrt{2}}\) và \(\sqrt{2\sqrt{3}}\)
d) \(\sqrt{10}+\sqrt{17}+1\) và \(\sqrt{61}\)
giúp mk vs ah mk cần gấp
b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)
\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)
mà 80>75
nên \(4\sqrt{5}>5\sqrt{3}\)
1) thực hiện phép tính
d)\(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
e) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ạ mk cần gấp
1) thực hiện phép tính:
a) \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
giúp mk vs ah mk đang cần gấp lắm
a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
=3
\(\sqrt{3}-3\sqrt{2}vs5\sqrt{2}-4\sqrt{3}\)
Đề bài: so sánh. Các bn giúp mk vs nha! Mk sẽ tick cho. Tks very much
Ta có: \(\sqrt{3}-3\sqrt{2}=\sqrt{3}-\sqrt{3^2.2}=\sqrt{3}-\sqrt{18}\)
Mà \(3< 18\Leftrightarrow\sqrt{3}< \sqrt{18}\Leftrightarrow\sqrt{3}-\sqrt{18}< 0\)(1)
Lại có: \(5\sqrt{2}-4\sqrt{3}=\sqrt{5^2.2}-\sqrt{4^2.3}=\sqrt{50}-\sqrt{48}\)
Mà \(50>48\Leftrightarrow\sqrt{50}>\sqrt{48}\Leftrightarrow\sqrt{50}-\sqrt{48}>0\)(2)
Từ (1) và (2), ta có: \(\sqrt{3}-3\sqrt{2}< 5\sqrt{2}-4\sqrt{3}\)
\(\frac{5\sqrt{6-1}}{\sqrt{6}+1}\) + \(\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}\)+ \(\sqrt{3-2\sqrt{2}}\)
Giúp mk vs mk cần gấp :) Ai nhanh mk tick :)
\(\frac{5\left(\sqrt{6}-1\right)\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}+\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}\)
\(=\frac{5\left(\sqrt{6}-1\right)^2}{5}-\frac{\left(\sqrt{2}-\sqrt{3}\right)^2}{1}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left(\sqrt{6}-1\right)^2-\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{2}-1\right)\)
\(=6-2\sqrt{6}+1-2+2\sqrt{6}-3+\sqrt{2}-1=\sqrt{2}\)
Tính:
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
giúp mk vs mk đang cần gấp//
Sủa lại đề:
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có \(a^2+b^2=6\), \(ab=2\), \(a+b=\sqrt{10}\), \(a-b=\sqrt{2}\), \(a^2-b^2=2\sqrt{5}\)
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)
\(=\frac{a^2.\left(\sqrt{10}+b\right)-b^2.\left(\sqrt{10}+a\right)}{\left(\sqrt{10}+a\right).\left(\sqrt{10}+b\right)}\)
\(=\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{\sqrt{10}.\left(a^2-b^2\right)+ab.\left(a-b\right)}{10+\sqrt{10}.\left(a+b\right)+ab}\)
\(=\frac{\sqrt{10}.2\sqrt{5}+\sqrt{10}.\sqrt{2}}{10+\sqrt{10}.\sqrt{10}+2}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}\)
\(=\frac{12\sqrt{2}}{22}\)
\(=\frac{6\sqrt{2}}{11}\)
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}
\)
\(=\frac{3+\sqrt{5}-3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=\frac{0}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=0\)
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