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b)Ghi đầu baì
=(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.555)
=(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.5.111)
=(1+2+3+...+100).(12+22+32+....+1002).(111.(65-65))
=(1+2+3+...+100).(12+22+32+....+1002).111.0
=(1+2+3+...+100).(12+22+32+....+1002).0
=0
(1+2+3+...+100)*(12+22+32+...+102)*(65*111--13*15*17)
(1+2+3+...+100)*(12+22+32+...+102)*(7215-72150)
(1+2+3+...+100)*(12+22+32+...+102)*0
sUY ra =0
a,3 . ( 5 - 3 + 1 ) + ( 2 - 13 ) - ( 10 - 13 )=3.3+(-11)-(-3)=9+(-11)+3=1
b,28 . 76 - 13 . 28 + 11 x 28=28.(76-13+11)=28.74=2072
c) 100 - [ 75 - ( 7 - 2 ) 2 ] =100-(75-5^2)=100-(75-25)=100-50=50
CÂU 1
\(5^{n+1}+5^n=750\)
\(=>5^n\cdot5+5^n=750\)
\(=>5^n\cdot\left(5+1\right)=750\)
\(=>5^n\cdot6=750\)
\(=>5^n=750:6\)
\(=>5^n=125\)
\(=>5^n=5^3\)
\(=>n=3\)
- Bài 1:
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.13+2^{10}.13.5}{2^8.2^2.13.2}\)
\(=\frac{2^{10}.13\left(1+5\right)}{2^{10}.13.2}=\frac{2^{10}.13.6}{2^{10}.13.2}=\frac{6}{2}=3\)
\(B=\left(1+2+3+...+100\right)\left(1^2+2^2+3^2+...+100^2\right)\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right)\left(1^2+2^2+...+100^2\right)\left(65.111-13.5.3.37\right)\)
\(=\left(1+2+...+100\right)\left(1^2+2^2+...+100^2\right)\left(65.111-65.111\right)\)
\(=\left(1+2+...+100\right)\left(1^2+2^2+...+100^2\right).0\)
\(=0\)
- Bài 2:
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(x+1+x+2+x+3+...+x+100=5750\)
\(x+x+x+...+x+1+2+3+...+100=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(x=700:100\)
\(x=7\)
t_i_c_k cho mình nha ^^
a) (105 + 155 - 55) : 55
= 105 : 55 + 155 : 55 - 55 : 55
= 25 + 35 - 1
= 32 + 243 - 1
= 274
Bn giải hộ mik câu b với