a) CMR: \(A=\left(36^3+41^3\right)⋮7\)
b) CMR: \(A=\left[75\cdot\left(4^{2018}+4^{2017}+...+4^2+4+1\right)+25\right]⋮10\)
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Lời giải:
\(A=2018^2-2017.2019=2018^2-(2018-1)(2018+1)\)
\(=2018^2-(2018^2-1^2)=1\)
\(B=9^8.2^8-(18^4-1)(18^4+1)\)
\(=(9.2)^8-[(18^4)^2-1^2]\)
\(=18^8-(18^8-1)=1\)
\(C=163^2+74.163+37^2=163^2+2.37.163+37^2\)
\(=(163+37)^2=200^2=40000\)
\(D=\frac{2018^3-1}{2018^2+2019}=\frac{(2018-1)(2018^2+2018+1)}{2018^2+2019}\)
\(=\frac{2017(2018^2+2019)}{2018^2+2019}=2017\)
Sử dụng công thức \((a-b)(a+b)=a^2-b^2\)
\(E=(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^8-1)(2^8+1)(2^{16}+1)-2^{32}\)
\(=(2^{16}-1)(2^{16}+1)-2^{32}\)
\(=(2^{32}-1)-2^{32}=-1\)
ta có 75(...) chia 100 dư 75
25 chia 100 dư 25
=> (75(....)+25)chia hết cho 100(tính chất chia hết của tổng)
học tốt
\(75\left(4^{2018}+4^{2017}+4^2+4+1\right)+25\)
\(=75\left(4^{2018}+4^{2017}+4^2+4\right)+75+25\)
\(=300\left(4^{2017}+4^{2016}+4+1\right)+100\)
Vì \(300\left(4^{2017}+4^{2016}+4+1\right)⋮100\)
và \(100⋮100\)nên
\(300\left(4^{2017}+4^{2016}+4+1\right)+100⋮100\)
Vậy \(75\left(4^{2018}+4^{2017}+4^2+4+1\right)+25⋮100\left(đpcm\right)\)
a: \(=\dfrac{13\left(3-18\right)}{40\left(15-2\right)}=\dfrac{13}{15-2}\cdot\dfrac{-15}{40}=\dfrac{-3}{8}\)
b: \(=\dfrac{18\left(34-124\right)}{36\left(-17-13\right)}=\dfrac{1}{2}\cdot\dfrac{-90}{-30}=\dfrac{3}{2}\)
c: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{\dfrac{-1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}\)
\(=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
Ta có: 75=25.3
mà các số trên đều là 4 mà 24.4=100 chia hết cho 10
Còn thừa số 40=1 khi nhân với 25=25 mà 25 +25( ở ngoài ngoặc)=50 chia hết cho 10
suy ra dãy tính trên chia hết cho 10
Rất đơn giản :)
Bài 2:
a: \(=44\cdot82-400+18\cdot44\)
\(=44\cdot100-400=4400-400=4000\)
b: \(=6^2:\left\{780:\left[390-125\cdot49+65\right]\right\}\)
\(=36:\left\{780:\left[-5670\right]\right\}\)
\(=36:\dfrac{-26}{189}=\dfrac{-3402}{13}\)
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)