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Đặt \(A=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
\(A=\frac{\frac{44}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)=\(\frac{\frac{44.1}{132}+-\frac{33.1}{132}+\frac{60.1}{132}}{\frac{55.1}{132}+\frac{132.1}{132}-\frac{84.1}{132}}\)
\(A=\frac{\frac{1}{132}.\left(44-33+60\right)}{\frac{1}{132}.\left(55+132-84\right)}\)=\(\frac{\frac{1}{132}.71}{\frac{1}{132}.103}\)
\(A=\frac{71}{103}\)
Câu 1:
A = \(\dfrac{-7}{12}+\dfrac{11}{8}-\dfrac{5}{9}=\dfrac{-42}{72}+\dfrac{99}{72}-\dfrac{40}{72}=\dfrac{-42+99-40}{72}=\dfrac{17}{72}\)
\(B=\dfrac{1}{7}-\dfrac{8}{7}:8-3:\dfrac{3}{4}.2^2=\dfrac{1}{7}-\dfrac{8}{7}.\dfrac{1}{8}-3.\dfrac{4}{3}.4=\dfrac{1}{7}-\dfrac{1}{7}-16=0-16=-16\)
\(C=1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}=\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=-\dfrac{5}{21}\)
Vậy A=\(\dfrac{17}{72};B=-16;C=\dfrac{-5}{21}\)
Câu 2:
a. \(\dfrac{-11x}{12}+\dfrac{3}{4}=-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{-11x}{12}=-\dfrac{1}{6}-\dfrac{3}{4}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-2}{12}-\dfrac{9}{12}\)
\(\Rightarrow\dfrac{-11x}{12}=\dfrac{-11}{12}\)
\(\Rightarrow-11x=\dfrac{-11.12}{12}\)
\(\Rightarrow-11x=-11\Rightarrow x=1\)
Vậy x=1
b. \(3-(\dfrac{1}{6}-x).\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow3-\left(\dfrac{1}{6}-x\right)=1\)
\(\Rightarrow-(\dfrac{1}{6}-x)=1-3\Rightarrow\dfrac{1}{6}+x=-2\)
\(\Rightarrow x=2-\dfrac{1}{6}\Rightarrow x=\dfrac{11}{6}\)
Vậy x = \(\dfrac{11}{6}\)
Câu 4:
Ta có: \(\dfrac{1}{2.3}=\dfrac{1}{6}\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{3}{6}-\dfrac{2}{6}=\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
a)\(\dfrac{-6}{11}:\left(\dfrac{3}{5}.\dfrac{4}{11}\right)=\dfrac{-5}{2}\)
b)\(\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{14}{30}=\dfrac{67}{370}\)
c)\(\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)=-\dfrac{169}{50}\)
d)\(\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)=\dfrac{115}{103}\)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
=\(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
=\(\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
=\(\left(\dfrac{5}{10}+\dfrac{8}{10}\right)+0\)
=\(\dfrac{13}{10}\)
\(-\dfrac{7}{25}.\dfrac{11}{13}+\left(-\dfrac{7}{25}\right).\dfrac{2}{13}-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.\cdot\left(\dfrac{11}{13}+\dfrac{2}{13}\right)-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}.1-\dfrac{18}{25}\)
=\(-\dfrac{7}{25}-\dfrac{18}{25}\)
=\(-\dfrac{25}{25}\) = \(-1\)
a)9.x + 1=73
9x=73-1
9x=72
x=72:9
x=8
b)2.x - 5 = -17 - 12
2x-5=-29
2x=-29+5
2x=-24
x=-24:2
x=-12
c)10 - x - 5 = - 5 - 7 -11
10-x-5=-12-11
10-x-5=-23
10-x=-23+5
10-x=18
x=10-18
x=-8
d)(-9) . x + 3 = (-2) . (-7) +16
-9x+3=14+16
-9x+3=30
-9x=30-3
-9x=27
x=27:(-9)
x=-3
(-12) . x - 34 =2
-12x=2+34
-12x=36
x=36:(-12)
x=-3
(-11).x + 9 =130
-11x=130-9
-11x=121
x=121:(-11)
x=11
(-5) .x + 5 = (-15) .(-4) -12
-5x+5=60-12
-5x+5=48
-5x=48-5
-5x=43
x=43:(-5)
x=-8,6
IxI -3=0
|x|=3
=>x=+3
(7 - IxI).(2.x - 4) =0
*7-|x|=0 * 2x-4=0
|x|=7 2x=4
=>x=+7 x=4:2
x=2
280-(x-140):35 =270
(x-140):35=280-270
(x-140):35=10
x-140=10.35
x-140=350
x=350+140
x=490
(1900 - 2.x ) : 35- 32 =16
1900-2x=(16+32).35
1900-2x=1680
2x=1900-1680
2x=220
x=220:2
x=110
720 :[41-(2x -5 )] =23 .5
720:[41-(2x-5)]=40
41-(2x-5)=720:40
41-(2x-5)=18
2x-5=41-18
2x-5=23
2x=23+5
2x=28
x=28:2
x=14
(x - 5).(x2 - 4 ) =0
* x-5=0 * x2-4=0
x=0+5 x2=4
x=5 x2=22
=> x=+2
5.7.77-7.60+49.25-15.42
= 5.7.77-7.5.12+7.7.5.5-3.5-3.5.6.7
= 35.77-35.12+35.35-35.18
= 35.(77-12+35-18)
= 35.82
= 2870
2/3 - 1/4 + 5/11
= ( 2/3 - 1/4 ) + 5/11
= (1/3 - 1/2 ) + 5/11
= 5/6 + 5/11
= 85/66
\(\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
\(=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)
\(=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{103}{132}}\)
\(=\frac{115}{\frac{132}{\frac{103}{132}}}=\frac{115}{132}.\frac{132}{103}\)
\(=\frac{115}{103}\)