tính giá trị của biểu thức
15/4:3/2-1/4
4-5/3:2+2/3
365,04:23,4x0,01
(547-5,65):15:0,1
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Đề bài như thế này phải ko
a)4- 5/3 : 2 +2/3
b)365,04 : 13,4 × 0,01
c)(547 - 5,65) : 15: 0,1
a: \(17628+3547\cdot6\)
\(=17628+21282\)
\(=38910\)
b: \(57924-15760:5\)
\(=57924-3152\)
=54772
\(\dfrac{8}{15}\times\dfrac{11}{2}+\dfrac{7}{15}\times\dfrac{11}{2}=\dfrac{11}{2}\left(\dfrac{8}{15}+\dfrac{7}{15}\right)=\dfrac{11}{2}\times\dfrac{15}{15}=\dfrac{11}{2}\times1=\dfrac{11}{2}\)
\(\dfrac{8}{12}+\dfrac{30}{12}-\dfrac{9}{12}=\dfrac{19}{12}\)
\(\dfrac{24}{30}-\dfrac{15}{30}+\dfrac{10}{30}=\dfrac{19}{30}\)
`8 / 15 : 2 / 11 + 7 / 15 : 2 / 11`
`= ( 8 / 15 + 7 / 15 ) : 2 / 11`
`= 15 / 15 xx 11 / 2`
`= 1 xx 11 / 2 = 11 / 2`
_________________________________
`2 / 3 + 5 / 2 - 3 / 4`
`= 8 / 12 + 30 / 12 - 9 / 12`
`= 29 / 12`
_________________________________
`4 / 5 - 1 / 2 + 1 / 3`
`= 24 / 30 - 15 / 30 + 10 / 30`
`= 19 / 30`
a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)
b)
\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)
a: \(M=\dfrac{631}{315}\cdot\dfrac{1}{651}-\dfrac{1}{105}\cdot\dfrac{2603}{651}-\dfrac{4}{315\cdot651}+\dfrac{4}{105}\)
\(=\dfrac{1}{315\cdot651}\cdot\left(631-4\right)-\dfrac{1}{105}\left(\dfrac{2603}{651}-4\right)\)
\(=\dfrac{1}{105}\cdot\dfrac{1}{1953}\cdot627+\dfrac{1}{105\cdot651}\)
\(=\dfrac{1}{105\cdot651}\left(\dfrac{1}{3}\cdot627+1\right)=\dfrac{1}{105\cdot651}\cdot210=\dfrac{2}{651}\)
b: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)
\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)