\(\sqrt{900}\div\sqrt{800}\)
Những câu hỏi liên quan
Tính tổng S = \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{900\sqrt{899}+899\sqrt{900}}\)
Xét \(a_n=\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n}}{n}-\frac{\sqrt{n+1}}{n+1}\)
\(\Rightarrow S=\frac{\sqrt{1}}{1}-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{3}+...+\frac{\sqrt{899}}{899}-\frac{\sqrt{900}}{900}\)
\(S=1-\frac{\sqrt{900}}{900}=1-\frac{1}{30}=\frac{29}{30}\)
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B1: Tính:a, sqrt{72}divsqrt{8}b, (sqrt{28}-sqrt{7}+sqrt{112})divsqrt{7}B2: Tính:a, sqrt{dfrac{49}{8}}divsqrt{3dfrac{1}{8}}b, sqrt{54x}divsqrt{6x}c, sqrt{dfrac{1}{125}}timessqrt{dfrac{32}{35}}divsqrt{dfrac{56}{225}}giúp em với ạ , em cảm mơn
Đọc tiếp
B1: Tính:
a, \(\sqrt{72}\div\sqrt{8}\)
b, \((\sqrt{28}-\sqrt{7}+\sqrt{112})\div\sqrt{7}\)
B2: Tính:
a, \(\sqrt{\dfrac{49}{8}}\div\sqrt{3\dfrac{1}{8}}\)
b, \(\sqrt{54x}\div\sqrt{6x}\)
c, \(\sqrt{\dfrac{1}{125}}\times\sqrt{\dfrac{32}{35}}\div\sqrt{\dfrac{56}{225}}\)
giúp em với ạ , em cảm mơn 
Bài 1:
a) \(\sqrt{72}:\sqrt{8}=\sqrt{72:8}=3\)
b) \(\left(\sqrt{28}-\sqrt{7}+\sqrt{112}\right):\sqrt{7}=5\sqrt{7}:\sqrt{7}=5\)
Bài 2:
a) \(\sqrt{\dfrac{49}{8}}:\sqrt{3\dfrac{1}{8}}=\sqrt{\dfrac{49}{8}:\dfrac{25}{8}}=\sqrt{\dfrac{49}{25}}=\dfrac{7}{5}\)
b) \(\sqrt{54x}:\sqrt{6x}=\sqrt{54x:6x}=\sqrt{9}=3\)
c) \(\sqrt{\dfrac{1}{125}}\cdot\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}\)
\(=\dfrac{\sqrt{5}}{25}\cdot\dfrac{4\sqrt{2}}{\sqrt{35}}:\dfrac{2\sqrt{14}}{15}\)
\(=\dfrac{\sqrt{5}\cdot4\sqrt{2}\cdot15}{25\cdot\sqrt{35}\cdot\sqrt{14}\cdot2}\)
\(=\dfrac{6}{35}\)
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So sánh: a,\(\sqrt{2}+\sqrt{3}\) và \(\sqrt{10}\) b,\(\sqrt{2003}+\sqrt{2005}\) và \(2\sqrt{2004}\) c,\(\sqrt{5\sqrt{3}}\) và \(\sqrt{3\sqrt{5}}\)
Xem chi tiết
\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)
\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)
\(=\sqrt{14+\sqrt{130^2}}-\sqrt{19+\sqrt{30^2}}+\sqrt{45+\sqrt{55^2}}\)
\(=\sqrt{14+130}-\sqrt{19+30}+\sqrt{45+55}\)
\(=\sqrt{144}-\sqrt{49}+\sqrt{100}\)
\(=\sqrt{12^2}-\sqrt{7^2}+\sqrt{10^2}\)
\(=12-7+10\)
\(=5+10\)
\(=15\)
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Mk nghĩ bạn cx lm đc mak cần J đăng :)
\(\sqrt{14+\sqrt{16900}}-\sqrt{19+\sqrt{900}}+\sqrt{45+\sqrt{3025}}\)
\(=\sqrt{14+130}-\sqrt{19+30}+\sqrt{45+55}=12-7+10=15\)
Mk đang rảnh nên thik lm mấy câu dễ dễ "___"
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\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{899}+\sqrt{900}}\)
Rút gọn biểu thức
\(=\dfrac{\left(\sqrt{1}-\sqrt{2}\right)}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+....+\dfrac{\sqrt{899}-\sqrt{900}}{899-900}\\ =\sqrt{900}-\sqrt{899}+\sqrt{899}-.....+\sqrt{3}-\sqrt{2}+\sqrt{2}-\sqrt{1}=\\ =\sqrt{900}-\sqrt{1}\\ =30-1=29\)
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Rút gon:
\(\frac{\sqrt{1+2\sqrt{5\sqrt{\text{ 7 }}-13}}-\sqrt{\sqrt{\text{ 7 }}-2}}{\sqrt{3-\sqrt{\text{ 7 }}}}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
Tính
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{899}+\sqrt{900}}\)
\(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{899}+\sqrt{900}}\)
=\(\frac{\sqrt{1}-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+....+\frac{\sqrt{899}-\sqrt{900}}{899-900}\)
\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+....+\frac{\sqrt{899}-\sqrt{900}}{-1}\)
\(=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+....+\sqrt{899}-\sqrt{900}}{-1}\)
\(=\frac{\sqrt{1}-\sqrt{900}}{-1}\)
\(=\frac{1-30}{-1}=\frac{-29}{-1}=29\)
=
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\(\frac{15}{x}-\frac{11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
so sanh 900^800 voi 800^900
