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\(=-\dfrac{3}{4}+3-\dfrac{1}{4}-\dfrac{9}{4}+\dfrac{9}{2}=-\dfrac{13}{4}+3+\dfrac{9}{2}=\dfrac{17}{4}\)
\(-\dfrac{3}{4}+\left(3-\dfrac{1}{4}\right)-\left(2,25-\dfrac{9}{2}\right)\)
= \(-1+3-\left(\dfrac{9}{4}-\dfrac{9}{2}\right)\)
= \(2+\dfrac{9}{4}=\dfrac{17}{4}\)
\(B=\left(8-2,25+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+1\dfrac{1}{4}\right)-\left(3+0,5-1\dfrac{2}{7}\right)\\ B=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{1}{2}+\dfrac{9}{7}\\ B=\left(8+6-3\right)+\left(-\dfrac{9}{4}-\dfrac{5}{4}\right)+\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\dfrac{1}{2}\)
\(B=11-\dfrac{7}{2}+2-\dfrac{1}{2}\\ B=\left(11+2\right)+\left(-\dfrac{7}{2}-\dfrac{1}{2}\right)\\ B=13-4\\ B=9.\)
a.=> \(\frac{\left(\frac{1}{3}\right).x}{\frac{2}{3}}=\frac{\frac{7}{4}}{\frac{2}{5}}\)
=> \(\frac{1}{3}.x=\frac{7}{4}.\frac{2}{3}:\frac{2}{5}\)
=>\(\frac{1}{3}.x=\frac{35}{12}\)
=> x\(=\frac{35}{12}:\frac{1}{3}\)
Vậy x=\(\frac{35}{4}\).
b. => \(\frac{4,5}{0,3}=\frac{2,25}{0,1.x}\)
=>\(0,1.x=\frac{2,25.0,3}{4,5}\)
=>\(0,1.x=0,15\)
=>\(x=0,15:0,1\)
Vậy x=1,5
c. =>\(\frac{8}{\frac{1}{4}.x}=\frac{2}{0,02}\)
=>\(\frac{1}{4}.x=\frac{8.0,02}{2}\)
=>\(\frac{1}{4}.x=0,08\)
=>\(x=0,08:\frac{1}{4}\)
Vậy x=0,32.
d. =>\(\frac{3}{\frac{9}{4}}=\frac{\frac{3}{4}}{6.x}\)
=>\(3.6x=\frac{9}{4}.\frac{3}{4}\)
=>\(18x=\frac{27}{16}\)
=>\(x=\frac{27}{16}:18\)
Vậy x=\(\frac{3}{32}\)
a: \(x=\left(-\dfrac{2}{3}\right)^5:\left(-\dfrac{2}{3}\right)^2=\left(-\dfrac{2}{3}\right)^3=-\dfrac{8}{27}\)
b: =>x-1/2=1/3
=>x=5/6
c: =>2/3x-1=0 hoặc 3/4x+1/2=0
=>x=3/2 hoặc x=-1/2:3/4=-1/2*4/3=-4/6=-2/3
d =>4/9:x=10/3:9/4=10/3*4/9=40/27
=>x=4/9:40/27=4/9*27/40=108/360=3/10
a: =5*1,4-4*1,5+3*1,3
=7-6+3,9=4,9
b: =1/3*7+3/4*18-2/3*20
=7/3+54/4-40/3
=-11+54/4
=2,5
c: =5/6*17-1/2*16+2/5*15
=85/6-8+6
=85/6-2
=73/6
d: =15*4/5+12*3/4-18*4/9
=12+9-8
=12+1=13
\(2^3.\left(2,25-\dfrac{1}{4}\right):\left(\dfrac{10}{3}-1\dfrac{1}{3}\right)^2\\ =8.\left(\dfrac{9}{4}-\dfrac{1}{4}\right):\left(\dfrac{10}{3}-\dfrac{4}{3}\right)^2\\ =8.2:2^2\\ =16:4\\ =4\)