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10 tháng 2 2020
\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)
\(=\frac{31}{25}.\frac{-25}{29}\)
\(=-\frac{31}{29}\)
\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)
\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)
\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)
\(=\frac{253}{15}.\frac{3}{19}\)
\(=\frac{253}{95}\)
Số to :v
4 tháng 3 2018
\(\left(\dfrac{1}{\sqrt{625}}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{\sqrt{25}}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
\(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\left(-\dfrac{29}{25}\right)\)
\(=\dfrac{31}{25}.\left(-\dfrac{25}{29}\right)\)
\(=-\dfrac{31}{29}\)
2 tháng 6 2022
a: \(C=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=\dfrac{-1}{5}+\dfrac{1}{5}-\dfrac{1}{2}-\dfrac{1}{2}\)
=-1
b: \(E=5\cdot4-4\cdot3+5-0.3\cdot20\)
\(=20-12+5-6=7\)
KV
29 tháng 10 2020
a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)
\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)
\(=\frac{4}{5}+11-2\)
\(=\frac{4}{5}+9\)
\(=\frac{49}{9}\)
b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+4-5+64\)
= 55
c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)
\(=\frac{\sqrt{48}}{91-7}\)
\(=\frac{4\sqrt{3}}{84}\)
\(=\frac{\sqrt{3}}{41}\)
d) Xem lại đề nhé em!
e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
\(=5-3.\frac{2}{3}\)
= 5 - 2
= 3
h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)
\(=-9.\frac{1}{3}-7+125:5\)
\(=-3-7+25\)
= 15
VM
4 tháng 8 2019
a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=-1.\)
b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=6.\frac{5}{4}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}.\)
c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)
\(=\frac{6}{7}.\)
d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)
\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)
\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)
\(=\frac{107}{100}+\frac{1}{5}\)
\(=\frac{127}{100}.\)
Chúc bạn học tốt!
4 tháng 8 2019
a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{-59}{45}\)
b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)
\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)
\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)
\(\Rightarrow\frac{31}{4}\)
c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)
\(\Rightarrow\frac{6}{7}\)
d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)
\(\Rightarrow\frac{93}{100}\)
LD
29 tháng 12 2017
1.
a. \(0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}=5-\dfrac{2}{5}=\dfrac{23}{5}>1\)
\(\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}=\dfrac{\dfrac{\sqrt{10}}{3}-\dfrac{3}{4}}{5}=\dfrac{-9+4\sqrt{10}}{60}\approx0,06< 1\)
\(\Rightarrow0,5\sqrt{100}-\sqrt{\dfrac{4}{25}}>\dfrac{\left(\sqrt{1\dfrac{1}{9}}-\sqrt{\dfrac{9}{16}}\right)}{5}\)
2.
Ta có:
\(\left(\sqrt{a+b}\right)^2=a+b\)
\(\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2+2\sqrt{ab}+\left(\sqrt{b}\right)^2=a+2\sqrt{ab}+b\)
=> \(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)
1b.
Áp dụng công thức trên
=> \(\sqrt{25+9}< \sqrt{25}+\sqrt{9}\)
NT
29 tháng 12 2017
2.
\(\sqrt{a+b}< \sqrt{a}+\sqrt{b}\\ \Rightarrow a+b< a+2\sqrt{ab}+b\\ \Rightarrow2\sqrt{ab}>0\\ \Rightarrow\sqrt{ab}>0\)
Luôn đúng với mọi a;b dươn g
=> đpcm
30 tháng 10 2017
1.
0,2 . \(\sqrt{100}\) - \(\sqrt{\dfrac{16}{25}}\)
= 0,2 . 10 - \(\dfrac{4}{5}\)
= 2 - \(\dfrac{4}{5}\)
= \(\dfrac{6}{5}\)
PT
30 tháng 10 2017
1/ \(0,2.\sqrt{100}-\sqrt{\dfrac{16}{25}}\)
\(=0,2.10-0,8\)
\(=2-0,8=1,2\)
2/ \(\dfrac{2^7.9^3}{6^5.8^2}\)
\(=\dfrac{93312}{497664}=\dfrac{3}{16}=0,1875\)
3/ \(\sqrt{0,01}-\sqrt{0,25}\)
\(=0,1-0,5\)
\(=-0,4\)
4/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{4}}\)
\(=0,5.10-0,5\)
\(=5-0,5=4,5\)
5/ \(7.\sqrt{0,01}+2.\sqrt{0,25}\)
\(=7.0,1+2.0,5\)
\(=0,7+1=1,7\)
6/ \(0,5.\sqrt{100}-\sqrt{\dfrac{1}{25}}\)
\(=0,5.10-0,2\)
\(=5-0,2=4,8\)
27 tháng 11 2022
a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
16 tháng 11 2022
\(\sqrt{484}-\dfrac{1}{\sqrt{5}}< \sqrt{529}-\dfrac{1}{19}< \sqrt{576}-\dfrac{1}{\sqrt{7}}< \sqrt{625}-\dfrac{1}{\sqrt{8}}\)
\(\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)
\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)
\(=\frac{31}{25}:\frac{-29}{25}\)
\(=\frac{-31}{29}\)