Tính
1/2 - 1/6 - 1/9 =
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1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
\(1,=\dfrac{1}{12}-\dfrac{4}{12}=\dfrac{3}{12}\\ 2,=\dfrac{6}{36}-\dfrac{1}{36}=\dfrac{5}{36}\\ 3,\dfrac{-1}{6}-\dfrac{2}{5}=\dfrac{-5}{30}-\dfrac{12}{30}=\dfrac{-17}{30}\\ 4,=\dfrac{15}{20}-\dfrac{1}{20}=\dfrac{14}{20}\)
1,=112−412=3122,=636−136=5363,−16−25=−530−1230=−17304,=1520−120=1420
Đáp án :
`-1/2`
Giải thích các bước giải :
`1/2` - (`1/3` + `2/3`)
= `1/2` - 1
= `-1/2`
________________________
Đáp án :
`7/9`
Giải thích các bước giải :
`4/9` + `1/2` . `2/3`
= `4/9` + `1/3`
= `12/27` + `9/27`
= `21/27` = `7/9`
\(1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{4}{12}-\dfrac{3}{12}=\dfrac{1}{12}\)
\(\dfrac{1}{4}-\dfrac{1}{5}=\dfrac{5}{20}-\dfrac{4}{20}=\dfrac{1}{20}\)
\(\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{6}{30}-\dfrac{5}{30}=\dfrac{1}{30}\)
\(\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{7}{42}-\dfrac{6}{42}=\dfrac{1}{42}\)
`@mt`
1: \(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
\(=7\cdot\sqrt3+3\sqrt6-4\cdot3\sqrt3\)
\(=3\sqrt6+7\sqrt3-12\sqrt3=3\sqrt6-5\sqrt3\)
2: \(\sqrt{28}-4\cdot\sqrt{63}+7\cdot\sqrt{112}\)
\(=2\sqrt7-4\cdot3\sqrt7+7\cdot4\sqrt7\)
\(=2\sqrt7-12\sqrt7+28\sqrt7=18\sqrt7\)
3: \(\sqrt{49}-5\cdot\sqrt{28}+\frac12\cdot\sqrt{63}\)
\(=7-5\cdot2\sqrt7+\frac12\cdot3\sqrt7=7-10\sqrt7+1,5\sqrt7=7-8,5\cdot\sqrt7\)
4: \(\left(2\sqrt6-4\sqrt3-\frac14\sqrt8\right)\cdot3\sqrt6\)
\(=6\cdot\sqrt{36}-12\sqrt{18}-\frac14\cdot2\cdot\sqrt2\cdot3\sqrt6\)
\(=36-36\sqrt2-\frac32\sqrt{12}=36-36\sqrt2-3\sqrt3\)
6: \(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt3=\sqrt{\frac{48}{3}}-3\cdot\sqrt{\frac{27}{3}}-\sqrt{\frac{147}{3}}\)
\(=\sqrt{16}-3\cdot\sqrt9-\sqrt{49}=4-3\cdot3-7\)
=-3-9
=-12
5: \(\left(2\cdot\sqrt{1\frac{9}{16}}-5\cdot\sqrt{5\frac{1}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\sqrt{\frac{25}{16}}-5\cdot\sqrt{\frac{81}{16}}\right):\sqrt{16}\)
\(=\left(2\cdot\frac54-5\cdot\frac94\right):4=\frac{10-45}{4\cdot4}=\frac{-35}{16}\)
7: \(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt2-\sqrt{162}:\sqrt2\)
\(=\left(5\sqrt2-21\right):\sqrt2-9\sqrt2:\sqrt2\)
\(=5-\frac{21}{\sqrt2}-9=-4-\frac{21\sqrt2}{2}=\frac{-8-21\sqrt2}{2}\)
8: \(\left(2\cdot\sqrt{1\frac{9}{10}}-\sqrt{5\frac{1}{10}}\right):\sqrt{10}=\left(2\cdot\sqrt{\frac{19}{10}}-\sqrt{\frac{51}{10}}:\sqrt{10}\right)\)
\(=2\cdot\frac{\sqrt{19}}{10}-\frac{\sqrt{51}}{10}=\frac{2\sqrt{19}-\sqrt{51}}{10}\)
9: \(2\cdot\sqrt{\frac{16}{3}}-3\cdot\sqrt{\frac{1}{27}}-6\cdot\sqrt{\frac{4}{75}}\)
\(=2\cdot\frac{4}{\sqrt3}-3\cdot\frac{1}{3\sqrt3}-6\cdot\frac{2}{5\sqrt3}\)
\(=\frac{8}{\sqrt3}-\frac{1}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7}{\sqrt3}-\frac{12}{5\sqrt3}=\frac{7\sqrt3}{3}-\frac{12\sqrt3}{15}=\frac{35\sqrt3-12\sqrt3}{15}=\frac{23\sqrt3}{15}\)
10: \(2\sqrt{27}-6\cdot\sqrt{\frac43}+\frac35\cdot\sqrt{75}\)
\(=2\cdot3\sqrt3-6\cdot\frac{2}{\sqrt3}+\frac35\cdot5\sqrt3\)
\(=6\sqrt3-4\sqrt3+3\sqrt3=5\sqrt3\)
11: \(\frac{\sqrt{18}}{\sqrt2}-\frac{\sqrt{12}}{\sqrt3}=\sqrt9-\sqrt4=3-2=1\)
1: \(=\dfrac{1}{3}\cdot\dfrac{3}{7}\cdot\dfrac{1}{2}=\dfrac{1}{7\cdot2}=\dfrac{1}{14}\)
2: =5/12+1/7
=35/84+12/84=47/84
3: =8(8/9-6/9)
=8*2/9=16/9
4: \(=\dfrac{5}{12}\cdot\dfrac{12}{5}=1\)
5: =16/5+6
=16/5+30/5=46/5
6: =10*1/2-10*1/5
=5-2=3
1) \(\sqrt{1\dfrac{9}{16}}=\sqrt{\dfrac{25}{16}}=\dfrac{5}{4}\)
2) \(\dfrac{\sqrt{12.5}}{0.5}=\sqrt{\dfrac{12.5}{0.25}}=5\sqrt{2}\)
3) \(\sqrt{\dfrac{25}{64}}=\dfrac{5}{8}\)
4) \(\dfrac{\sqrt{230}}{\sqrt{2.3}}=\sqrt{\dfrac{230}{2.3}}=\sqrt{100}=10\)
5) \(\left(\sqrt{\dfrac{2}{3}}+\sqrt{\dfrac{50}{3}}-\sqrt{24}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{5\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=\left(\dfrac{6\sqrt{2}}{\sqrt{3}}-2\sqrt{6}\right)\cdot\sqrt{6}\)
\(=0\cdot\sqrt{6}=0\)
`1/2 -1/6- 1/9`
`= 3/6 -1/6 -1/9`
`= 2/6 -1/9`
`= 1/3-1/9`
`=3/9-1/9`
`=2/9`
\(\dfrac{1}{2}-\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{9}{18}-\dfrac{3}{18}-\dfrac{2}{18}=\dfrac{4}{18}=\dfrac{2}{9}\)