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\(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ A=\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ A=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
\(A=\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}=\dfrac{3\left(0,125-0,1+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(0,125-0,1+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{5}\left(2,5+\dfrac{5}{3}-1,25\right)}{2,5+\dfrac{5}{3}-1,25}=-\dfrac{3}{5}+\dfrac{3}{5}=0\)
Ta có : \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}\right).\frac{2005}{1890}+115\)
\(=\left(\frac{3}{5}-\frac{3}{5}\right).\frac{2005}{1890}+115=0+115=115\)
= ( \(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)+ \(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)) x \(\frac{2005}{1890}\)+ 115
= ( \(\frac{3(\frac{1}{2}+\frac{1}{3}-\frac{1}{4})}{5(\frac{1}{1}+\frac{1}{3}-\frac{1}{4})}\)+ \(\frac{3(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}{-5(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12})}\)) x \(\frac{2005}{1890}\)+ 115
=( \(\frac{3}{5}\)+\(\frac{3}{-5}\)) x \(\frac{2005}{1890}\)+115 = 0 +115 = 115
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
A=−85+105−115−12583−103+113+123+25+35−4523+33−43A=−5(81−101+111+121)3(81−101+111+121)+5(21+31−41)3(21+31−41)A=5−3+53=0
a, |0,75 - x| = 2,5
=> 0,75 - x = 2,5 hoặc 0,75 - x = -2,5
=> x = -1,75 hoặc x = 3,25
vậy_
\(|0,75-x|=2,5\)
\(\Rightarrow\hept{\begin{cases}0,75-x=2,5\\0,75-x=-2,5\end{cases}\Rightarrow\hept{\begin{cases}x=-1,75\\x=3,25\end{cases}}}\)
`0,75-43/80:(-4/5+2,5*3/4)`
`=0,75-43/80:(-4/5+15/8)`
`=0,75-43/80:43/40`
`=3/4-1/2`
`=1/4`
0,75 - \(\dfrac{43}{80}\) : ( - \(\dfrac{4}{5}\) + 2,5 \(\times\) \(\dfrac{3}{4}\))
= 0,75 - \(\dfrac{43}{80}\) : ( - \(\dfrac{4}{5}\) + \(\dfrac{15}{8}\))
= 0,75 - \(\dfrac{43}{80}\) : \(\dfrac{43}{40}\)
= 0,75 - \(\dfrac{1}{2}\)
= 0,75 - 0,5
= 0,25
a) − 5 24 + 0 , 75 + 7 12 : − 2 1 8 = − 5 24 + 3 4 + 7 12 : − 17 8 = 9 8 . − 8 17 = − 9 17 .
b) 75 % - 3 1 2 + 1,5 : 10 7 = 3 4 − 5 2 + 3 2 : 7 10 = 3 4 − 5 2 + 3 2 . 7 10 = − 14 20 = − 7 10 .
a, 1,5+1-0,75/2,5+5\3-1,25
=15\10+1-75\100/25\10+5\3-125\100
=7\4/35/12
Giúp mình với
2,5. 0,75 + 2,5.0,25 - 1,5
=2,5.(0,75 + 0,25) - 1,5
= 2,5.1 - 1,5
= 2,5 - 1,5
= 1