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1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
a: \(\dfrac{18}{24}+\dfrac{35}{-21}-\dfrac{1}{2}\)
\(=\dfrac{3}{4}-\dfrac{5}{3}-\dfrac{1}{2}\)
\(=\dfrac{9-20-6}{12}=\dfrac{-17}{12}\)
b: \(\dfrac{-2}{3}-\dfrac{3}{-4}+\dfrac{5}{-6}\)
\(=-\dfrac{2}{3}+\dfrac{3}{4}-\dfrac{5}{6}\)
\(=\dfrac{-8+9-10}{12}=\dfrac{-9}{12}=\dfrac{-3}{4}\)
c: \(\left(\dfrac{3}{4}+\dfrac{-7}{2}\right)\cdot\left(\dfrac{10}{11}+\dfrac{2}{22}\right)\)
\(=\left(\dfrac{3}{4}-\dfrac{14}{4}\right)\cdot\left(\dfrac{10}{11}+\dfrac{1}{11}\right)\)
\(=-\dfrac{11}{4}\cdot1=-\dfrac{11}{4}\)
d: \(\left(\dfrac{7}{3}-\dfrac{-7}{2}\right):\left(-\dfrac{25}{6}\cdot\dfrac{1}{5}\right)\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\dfrac{-5}{6}\)
\(=\dfrac{35}{6}\cdot\dfrac{6}{-5}=\dfrac{35}{-5}=-7\)
a) \(\dfrac{18}{24}+\dfrac{35}{-21}-\dfrac{1}{2}\)
\(=\dfrac{3}{4}-\dfrac{5}{3}-\dfrac{1}{2}\)
\(=\dfrac{9}{12}-\dfrac{20}{12}-\dfrac{6}{12}\)
\(=\dfrac{-17}{12}\)
b) \(\dfrac{-2}{3}-\dfrac{3}{-4}+\dfrac{5}{-6}\)
\(=\dfrac{-8}{12}+\dfrac{9}{12}-\dfrac{10}{12}\)
\(=\dfrac{-9}{12}\)
\(=\dfrac{-3}{4}\)
c) \(\left(\dfrac{3}{4}+\dfrac{-7}{2}\right).\left(\dfrac{10}{11}+\dfrac{2}{22}\right)\)
\(=\dfrac{-11}{4}.1\)
\(=\dfrac{-11}{4}\)
d) \(\left(\dfrac{7}{3}-\dfrac{-7}{2}\right):\left(-\dfrac{25}{6}.\dfrac{1}{5}\right)\)
\(=\dfrac{35}{6}:\dfrac{-5}{6}\)
\(=-7\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\dfrac{3}{4}+\dfrac{2}{3}+\dfrac{3}{5}\)
`=`\(\dfrac{9}{12}+\dfrac{8}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{17}{12}+\dfrac{3}{5}\)
`=`\(\dfrac{85}{60}+\dfrac{36}{60}\)
`=`\(\dfrac{121}{60}\)
`b)`
\(\dfrac{1}{2}\cdot\dfrac{9}{13}\div\dfrac{27}{26}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{9}{13}\cdot\dfrac{26}{27}\)
`=`\(\dfrac{1}{2}\cdot\dfrac{2}{3}\)
`=`\(\dfrac{1}{3}\)
`c)`
\(\dfrac{2}{7}\cdot\dfrac{1}{9}+\dfrac{2}{7}\cdot\dfrac{2}{9}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\left(\dfrac{1}{9}+\dfrac{2}{9}\right)+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{2}{7}\cdot\dfrac{1}{3}+\dfrac{1}{3}\cdot\dfrac{5}{7}\)
`=`\(\dfrac{1}{3}\cdot\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
`=`\(\dfrac{1}{3}\cdot1=\dfrac{1}{3}\)
`d)`
\(11\div\dfrac{5}{2}+11\div\dfrac{7}{3}+11\div\dfrac{35}{6}\)
`=`\(11\cdot\dfrac{2}{5}+11\cdot\dfrac{3}{7}+11\cdot\dfrac{6}{35}\)
`=`\(11\cdot\left(\dfrac{2}{5}+\dfrac{3}{7}+\dfrac{6}{35}\right)\)
`=`\(11\cdot1=11\)
a) 3/4 + 2/3 + 3/5 = 45/60 + 40/60 + 36/60 = 121/60
b) 1/2 x 9/13 : 27/26 = 9/26 x 26/27 = 1/3
c) 2/7 x 1/9 + 2/7 x 2/9 + 1/3 x 5/7 = 2/7 x (1/9 + 2/9) + 5/21 = 2/7 x 1/3 + 5/21 = 2/21 + 5/21 = 1/3
d) 11 : 5/2 + 11 : 7:3 + 11 : 35/6 = 11 x (2/5 + 3/7 + 6/35) = 11 x 1 = 11
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
b. -6x=18
=> x = 18 : (-6)
=> x = -3
c. 35-3.|x|=5.(23-4)
=> 35-3.|x|=5.(8-4)
=> 35-3.|x|=5.4
=> 35-3.|x|=20
=> 3.|x|=35-20
=> 3.|x|=15
=> |x|=5
=> x \(\in\){-5; 5}
d. => 10+2.|x|=2.(9-1)
=> 10+2.|x|=2.8
=> 10+2.|x|=16
=> 2.|x|=16-10
=> 2.|x|=6
=> |x|=3
=> x \(\in\){-3; 3}