Chứng minh rằng:
a) \(\overline{aaa}\) chia hết cho \(3\) ?
b) \(\overline{ab}+\overline{ba}\) chia hết cho \(11\) ?
c) \(\overline{ab}-\overline{ba}\) chia hết cho \(9\) ?
d) Số \(\overline{abcabc}\) chia hết cho \(11\) ?
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\(250:\left\{5\cdot\left[1997-1869\right]-78\right\}\)
\(=250:\left\{5\cdot128-78\right\}\)
\(=250:\left\{640-78\right\}=\dfrac{250}{562}=\dfrac{125}{281}\)
\(6\cdot1235\cdot20-5\cdot235\cdot24\)
\(=120\cdot1235-120\cdot235=120\cdot1000\)
=120000
\(\dfrac{9}{4}-\dfrac{1}{4}:y-\dfrac{1}{2}=\dfrac{2}{3}\\ =>\dfrac{1}{4}:y=\dfrac{9}{4}-\dfrac{1}{2}-\dfrac{2}{3}\\ =>\dfrac{1}{4}:y=\dfrac{27}{12}-\dfrac{6}{12}-\dfrac{8}{12}\\ =>\dfrac{1}{4}:y=\dfrac{13}{12}\\ =>y=\dfrac{1}{4}:\dfrac{13}{12}\\ =>y=\dfrac{3}{13}\)
`9/4- 1/4:y - 1/2 = 2/3`
`<=> 9/4 - 1/4:y - 2/4 = 2/3`
`<=> 7/4 - 1/4:y= 2/3`
`<=> 1/4:y = 7/4 - 2/3`
`<=> 1/4:y = 21/12 - 8/12`
`<=> 1/4:y = 13/12`
`<=> y = 1/4 : 13/12`
`<=> y = 1/4 . 12/13`
`<=> y = 3/13`
Vậy ..
\(5,1\times y-y:10=36,25\times4\)
\(5,1\times y-y\times0,1=145\)
\(\left(5,1-0,1\right)\times y=145\)
\(5\times y=145\)
\(y=145:5\)
\(y=29\)
`5,1 xx y - y : 10 = 36,25 xx 4`
`=>5,1 xx y - y xx 0,1 = 36 x 4 + 0,25 x 4`
`=>y xx (5,1 - 0,1) = 144+1`
`=>y xx 5 = 145`
`=>y=145:5`
`=>y=29`
Vậy: ...
`x(x-5)=0`
TH1: `x=0`
TH2: `x-5=0`
`=>x=0+5`
`=>x=5`
Vậy: `x∈{0;5}`
Bổ sung cho @HT.Phong
Kết luận: \(x\) \(\in\) {0; 5}
\(124\times\left(56+67\right)-24\times\left(67+56\right)\\ =124\times123-24\times123\\ =\left(124-24\right)\times123\\ =100\times123\\ =12300\)
124 x (56 + 67) - 24 x (67 + 56)
= (56 + 67) x (125 - 24)
= 123 x 100
= 12300
a) \(\overline{aaa}=\overline{a00}+\overline{a0}+a=a.100+a.10+a.1\\ =a.\left(100+10+1\right)=a.111=a.37.3⋮3\) (dpcm)
b) \(\overline{ab}+\overline{ba}=\overline{a0}+b+\overline{b0}+a\\ =a.10+b+b.10+a\\ =a.\left(10+1\right)+b.\left(1+10\right)\\ =a.11+b.11\\ =11\left(a+b\right)⋮11\) (dpcm)
c) \(\overline{ab}-\overline{ba}=\overline{a0}+b-\left(\overline{b0}+a\right)\\ =a.10+b-b.10-a\\ =a.\left(10-1\right)+b.\left(1-10\right)\\ =a.9+b.\left(-9\right)\\ =9.\left(a-b\right)⋮9\) (dpcm)
d) \(\overline{abcabc}=\overline{abc000}+\overline{abc}\\ =\overline{abc}.1000+\overline{abc}.1\\ =\overline{abc}.1001=\overline{abc}.11.91⋮11\) (dpcm)