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a) \(\left( { - 2,5} \right) + \left( { - 0,25} \right) = - \left( {2,5 + 0,25} \right)\)\( = - 2,75\)
b) \(\left( { - 1,4} \right) + 2,1 = 2,1 - 1,4 = 0,7\)
c) \(3,2-5,7 = -(5,7-3,2)=-2,5\)
a) \(\left( { - 31,5} \right):1,5 = - \left( {31,5:1,5} \right) = - 21\)
b) \(\left( { - 31,5} \right):\left( { - 1,5} \right) = 31,5:1,5 = 21\)
Ta có: \(\dfrac{5}{7} = \dfrac{{5.4}}{{7.4}} = \dfrac{{20}}{{28}}\) và \(\dfrac{{ - 3}}{4} = \dfrac{{ - 3.7}}{{4.7}} = \dfrac{{ - 21}}{{28}}\)
Như vậy, \(\dfrac{{20}}{{28}} + \dfrac{{ - 21}}{{28}} = \dfrac{{20 + \left( { - 21} \right)}}{{28}} = \dfrac{-1}{{28}}\)
a: \(\dfrac{1\cdot5\cdot6+2\cdot10\cdot12}{1\cdot3\cdot52\cdot6\cdot10}=\dfrac{5\cdot6\left(1+2\cdot2\cdot2\right)}{3\cdot52\cdot6\cdot10}\)
\(=\dfrac{1}{2}\cdot\dfrac{9}{156}\)
\(=\dfrac{9}{312}=\dfrac{3}{104}\)
\(a,12,3\cdot4,7+5,3\cdot12,3-1,4\cdot7,5+7,5\cdot5,4\)
\(=12,3\cdot\left(4,7+5,3\right)-7,5\cdot\left(1,4+5,4\right)\)
\(=12,3\cdot10-7,5\cdot6,8\)
\(=123-51\)
\(=72.\)
\(b,3\cdot x-1=\frac{3}{4}:\frac{15}{16}\)
\(\Leftrightarrow3\cdot x-1=\frac{3}{4}\cdot\frac{16}{15}\)
\(\Leftrightarrow3\cdot x-1=\frac{4}{5}\)
\(\Leftrightarrow3\cdot x=\frac{4}{5}+1\)
\(\Leftrightarrow3\cdot x=\frac{9}{5}\)
\(\Leftrightarrow x=\frac{9}{5}\div3\)
\(\Leftrightarrow x=\frac{9}{5}\cdot\frac{1}{3}\)
\(\Leftrightarrow x=\frac{3}{5}\)
a) \(12,3\times4,7+5,3\times12,3-1,4\times7,5+7,5\times5,4\)
\(=12,3\times\left(4,7+5,3\right)+7,5\times\left(5,4-1,4\right)\)
\(=12,3\times10+7,5\times4\)
\(=123+30\)
\(=153\)
b) \(3.x-1=\frac{3}{4}:\frac{15}{16}\)
\(\Rightarrow3.x-1=\frac{4}{5}\)
\(\Rightarrow3.x=\frac{9}{5}\Rightarrow x=\frac{3}{5}\)
a) 1,2.2,5 = 3;
125 : 0,25 = 500
b)
\(1,2.2,5 = \dfrac{6}{5}.\dfrac{5}{2} = \dfrac{{30}}{{10}} = 3\)
\(125:0,25 = 125:\dfrac{1}{4} = 125.4 = 500\)
a) 12,3 + 5,67 = 17,97
12,3 - 5,67 = 6,63
b) ( -12,3) + (-5,67) = -(12,3 + 5,67) = -17,97
5,67 - 12,3 = -(12,3 - 5,67)= - 6,63