Tính
\(\sqrt{\frac{1}{25}}.\sqrt{\frac{32}{35}}:\sqrt{\frac{56}{225}}\)
Những câu hỏi liên quan
Chứng minh rằng:
\(\sqrt[3]{\sqrt[5]{\frac{32}{5}}-\sqrt[5]{\frac{27}{5}}}=\sqrt[5]{\frac{1}{25}}+\sqrt[5]{\frac{3}{25}}-\sqrt[5]{\frac{9}{25}}\)
TÍNH GIÁ TRỊ CÁC BIỂU THỨC SAU
A,\(\sqrt{0,09}-\sqrt{0,64}\)
B,\(0,1\times\sqrt{225}-\sqrt{\frac{1}{4}}\)
C,\(\sqrt{0,36}\times\sqrt{\frac{25}{16}+\frac{1}{4}}\)
D,\(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
a) \(\sqrt{0,09}-\sqrt{0,64}=\frac{-1}{2}=-0,5\)
b) \(0,1\cdot\sqrt{225}-\sqrt{\frac{1}{4}}=0,1\cdot15-\frac{1}{2}=1\)
c) \(\sqrt{0,36}\cdot\sqrt{\frac{25}{16}+\frac{1}{4}}=\frac{3\sqrt{29}}{20}\)
d) đề baì có sai ko ban?
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Tính Nhanh:\(\frac{\left(\frac{1}{10}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
Tính nhanh: \(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\left(-\frac{4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}\left(\frac{1}{2}-\sqrt{2}+\frac{3\sqrt{2}}{5}\right).\frac{-4}{15}}{\frac{1}{5}\left(\frac{1}{2}+\frac{3\sqrt{2}}{5}-\sqrt{2}\right).\frac{5}{7}}\)
\(=\frac{\frac{1}{7}.\frac{-4}{15}}{\frac{1}{5}.\frac{5}{7}}=\frac{\frac{-4}{105}}{\frac{1}{7}}=\frac{-4}{15}\)
\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\left(\frac{1}{10}+\frac{3.\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{10}.\frac{5}{7}+\frac{3.\sqrt{2}}{25}.\frac{5}{7}-\frac{\sqrt{2}}{5}.\frac{5}{7}}\)
\(=\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3.\sqrt{2}}{35}\right).\frac{-4}{15}}{\frac{1}{14}+\frac{3.\sqrt{2}}{35}-\frac{\sqrt{2}}{7}}\)
\(=\frac{-4}{15}\)
Học tốt
tính :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(B=\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{10}}+....+\frac{1}{\sqrt{220}+\sqrt{225}}\)
\(\sqrt[]{\dfrac{1}{125}}.\sqrt[]{\dfrac{32}{35}}:\sqrt[]{\dfrac{56}{225}}\)
\(\sqrt{\dfrac{1}{125}}.\sqrt{\dfrac{32}{35}}:\sqrt{\dfrac{56}{225}}=\sqrt{\dfrac{1}{125}.\dfrac{32}{35}:\dfrac{56}{225}}=\sqrt{\dfrac{36}{1225}}=\dfrac{6}{35}\)
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Tính:
B=\(\frac{\left(\frac{1}{14}-\frac{\sqrt{2}}{7}+\frac{3\sqrt{2}}{35}\right).\left(\frac{-4}{15}\right)}{\left(\frac{1}{10}+\frac{3\sqrt{2}}{25}-\frac{\sqrt{2}}{5}\right).\frac{5}{7}}\)
Tính :\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
Sorry mới lớp 6 chưa học
thông cảm
no chửi
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Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
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cmr\(\frac{1}{\left(\sqrt{2}+\sqrt{5}\right)^3}+\frac{1}{\left(\sqrt{5}+\sqrt{8}\right)^3}+...+\frac{1}{\left(\sqrt{32+\sqrt{35}}\right)^3}<\frac{5}{72}\)