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Bài 1:
a: \(\dfrac{25}{42}-\dfrac{20}{63}=\dfrac{75-40}{126}=\dfrac{35}{126}=\dfrac{5}{18}\)
b: \(\dfrac{9}{20}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{135}{300}-\dfrac{52}{300}-\dfrac{50}{300}=\dfrac{33}{300}=\dfrac{11}{100}\)
a: \(A=\dfrac{5}{7}-\dfrac{2}{7}+\dfrac{8}{11}+\dfrac{3}{11}+\dfrac{1}{2}=\dfrac{3}{7}+\dfrac{1}{2}+1=\dfrac{6+7+14}{14}=\dfrac{27}{14}\)
b: \(B=\dfrac{11}{17}+\dfrac{6}{17}-\dfrac{8}{19}-\dfrac{30}{19}+\dfrac{-3}{4}=1-2-\dfrac{3}{4}=-1-\dfrac{3}{4}=-\dfrac{7}{4}\)
c: \(C=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{49}{50}\)
Bài 3:
a: \(=-8\left(72+19-1\right)=-8\cdot90-720\)
b: \(=-27\left(1011-12-1\right)=-27\cdot998=-26946\)
c: \(=17\cdot\left[29+111\right]+29\cdot\left(-17\right)\)
\(=17\left(29+111-29\right)=17\cdot111=1887\)
d: \(=43\cdot\left(-1\right)+40=-43+40=-3\)
1) 125.(-24) + 24.225.
\(=24\left(-125+225\right)=24.100=2400.\)
2) 512.(2 – 128) – 128. (-512).
\(=512\left(2-128+128\right)=512.2=1024.\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
a)= \(2^3\left(17-14\right)\)
\(=8.3\)
\(=24\)
b)\(=17\left(85+15\right)-120\)
\(=17.100-120\)
\(=1700-120\)
\(=1580\)
c)\(=\left(25.4\right).\left(125.8\right).\left(2.5\right)\)
\(=100.1000.10\)
\(=1000000\)
d)\(=24.53+24.87-24.40\)
\(=24\left(53+87-40\right)\)
\(=24.100\)
\(=2400\)
e)\(=5.7.77-7.60-7.7.25-15.6.7\)
\(=7\left(5.77-60-7.25-15.6\right)\)
\(=7\left(385-60-175-90\right)\)
\(=7.60\)
\(=420\)
a)-11+(-83)+183+(-89)=[(-11)+(-89)]+[(-83)+183]=-100+100=0
b)-42-(-215)-(-85)-58=-42+215+85-58=-(42+58)+(215+85)=-100+300=200
c)1-2+3-4+...+2015-2016=(1-2)+(3-4)+...+(2015-2016)=(-1)+(-1)+...+(-1)=(-1)*1008=-1008
d)25*(-17)*(-2)^2*(-5)=25*(-17)*4*(-5)=(25*4)*[(-17)*(-5)]=100*85=8500