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NN
6 tháng 12 2017
a) \(A=\dfrac{1}{\sqrt{25}}+\dfrac{\sqrt{49}}{\sqrt{36}}-\dfrac{2}{\sqrt{100}}.\)
\(=\dfrac{1}{5}+\dfrac{7}{6}-\dfrac{1}{5}.\)
\(=\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\dfrac{7}{6}.\)
\(=0+\dfrac{7}{6}=\dfrac{7}{6}.\)
Vậy \(A=\dfrac{7}{6}.\)
b) \(B=\sqrt{\dfrac{0,01}{1,21}}+3.\dfrac{2}{\sqrt{10^2}+2^2+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{2}{10+4+40}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+3.\dfrac{1}{37}-\dfrac{3}{4}.\)
\(=\dfrac{1}{11}+\dfrac{1}{9}-\dfrac{3}{4}.\)
\(=\dfrac{36}{396}+\dfrac{44}{396}-\dfrac{297}{296}.\)
\(=-\dfrac{217}{396}.\)
Vậy \(B=-\dfrac{217}{396}.\)
KH
21 tháng 9 2017
a)\(\sqrt{1}\)+\(\sqrt{9}\)+\(\sqrt{25}\)+\(\sqrt{49}\)+\(\sqrt{81}\)
=1+3+5+7+9
=25
b)=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{4}\)
=\(\dfrac{6}{12}\)+\(\dfrac{4}{12}\)+\(\dfrac{2}{12}\)+\(\dfrac{3}{12}\)
=\(\dfrac{15}{12}\)
c) =0,2+0.3+0,4
= 0.9
d) =9-8+7
=8
j) =1,2-1,3+1.4
= (-0,1)+1,4
=1,4
g) \(\dfrac{2}{5}\)+\(\dfrac{5}{2}\)+\(\dfrac{9}{10}\)+\(\dfrac{3}{4}\)
= (\(\dfrac{4}{10}\)+\(\dfrac{15}{10}\)+\(\dfrac{9}{10}\))+\(\dfrac{3}{4}\)
= \(\dfrac{14}{5}\)+\(\dfrac{3}{4}\)
=\(\dfrac{56}{20}\)+\(\dfrac{15}{20}\)
= \(\dfrac{71}{20}\)
Nhớ tick cho mk nha~
14 tháng 10 2021
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
14 tháng 10 2021
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
24 tháng 12 2023
a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)
\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)
\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)
\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)
\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)
\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)
\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)
Y
0
LD
11 tháng 12 2017
\(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{\dfrac{8}{2}-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
O
8 tháng 1 2018
\(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{9}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)
\(A=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{\left(7.7\right)^2}}{\dfrac{8}{2}-\dfrac{4}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{2401}}{\dfrac{8}{2}-\dfrac{4}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{6}{7}+\dfrac{1}{49}-\dfrac{1}{2401}}{\dfrac{32}{9}+\dfrac{4}{49}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{43}{49}-\dfrac{1}{2401}}{\dfrac{1604}{441}-\dfrac{4}{343}}\)
\(A=\dfrac{\dfrac{2106}{2401}}{3,625526401}\)
\(A=\dfrac{2106}{2401}:3,625526401\)
\(A=\dfrac{9477}{39172}\)
8 tháng 6 2022
a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)
b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)
\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)
c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)
\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)
28 tháng 1 2019
\(\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\\ =\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\cdot\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)
Trả lời:
\(\sqrt{\dfrac{49}{100}}=\sqrt{\dfrac{7}{10}}\)
\(\sqrt{\dfrac{49}{100}}=\dfrac{7}{10}\)