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Lời giải:
a. $=5.9+112-31=45+112-31=157-31=126$
b. $=-31-82-17=-(31+82+17)=-130$
c. $=6.(-8)+20+12=-48+32=-(48-32)=-16$
d. Trùng với câu b.
a: =-3/24-40/24
=-43/24
b: \(=\dfrac{6}{54}\cdot\dfrac{49}{35}=\dfrac{1}{9}\cdot\dfrac{7}{5}=\dfrac{7}{45}\)
c: \(=\dfrac{6}{5}+\dfrac{4}{3}=\dfrac{18+20}{15}=\dfrac{38}{15}\)
d: \(=\dfrac{31}{17}-\dfrac{14}{17}-\dfrac{5}{13}-\dfrac{8}{13}=1-1=0\)
a: =35/17-18/17-9/5+4/5
=1-1=0
b: =-7/19(3/17+8/11-1)
=7/19*18/187=126/3553
c: =26/15-11/15-17/3-6/13
=1-6/13-17/3
=7/13-17/3=-200/39
a) 49 - 9 . [ 34 : 31 - ( 12 - 4 ) : 22 ]
= 49 - 9 . [ 33 - 8 : 22 ]
= 49 - 9 . [ 33 -23 : 22 ]
= 49 - 9 . [ 33 - 2 ]
= 49 - 9 . [ 27 - 2 ]
= 49 - 9 . 25
= 49 - 225 = -176
b) 4.52 + 16 : 22
= 4. 25 + 24 : 22
= 100 + 22 = 100+ 4 =104
--tho--là tớ💋
`@` `\text {Ans}`
`\downarrow`
`a.`
\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)
`=`\(0,3-0,4+1\)
`= -0,1 + 1`
`= 0,9`
`b.`
\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)
`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)
`=`\(1+4\cdot\left(-2,25\right)\)
`= 1+ (-9) = -8`
`c.`
\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)
`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)
`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)
`=`\(\dfrac{13}{6}\div2\)
`=`\(\dfrac{13}{12}\)
`d.`
\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)
`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)
`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)
`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)
`e.`
\(\left(\dfrac{4}{5}-1\right)\div\dfrac{3}{5}-\dfrac{2}{3}\cdot0,5\)
`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)
`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)
`f.`
\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)
`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)
`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)
`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)
`= 0,8 \div (-2/15)`
`=-6`
`@` `yHGiangg.`