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- 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 ^^
\(11^{12}< 11^{13}\)
\(7^4< 8^4\)
\(3^4>4^3\)
\(2^6>6^2\)
\(5^{15}>2^{30}\)
a)
C=1+3+32+33+34+35+...+311
C=(1+3+32)+(33+34+35)+...+(39+310+311)
C=13+(33.1+33.3+33.32)+...+(39.1+39.3+39.32)
C=13+33.(1+3+32)+...+39.(1+3+32)
C=13.1+33.13+...+39.13
C=13.(1+33+35+37+39)\(⋮\)3
\(\Rightarrow\)C\(⋮\)3
Câu b ghép 4 số lại với nhau rồi làm như trên
a)\(78-3\left(x-1\right)=15\)
\(3\left(x-1\right)=78-15\)
\(3\left(x-1\right)=63\)
\(x-1=63:3\)
\(x-1=21\)
\(x=21+1=22\)
b. \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\left(7x-11\right)^3=32.25+200\)
\(\left(7x-11\right)^3=800+200=1000\)
\(\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(7x=10+11\)
\(7x=21\)
\(x=21:7=3\)
c.\(5.\left(3x-13\right)=5^4\)
\(3x-13=5^4:5\)
\(3x-13=5^3=125\)
\(3x=125+13=138\)
\(x=138:3=46\)
d.\(3.\left(5x-13\right)=3^4\)
\(5x-13=3^4:3=3^3\)
\(5x-13=27\)
\(5x=27+13=40\)
\(x=40:5=8\)
(313.5+315.4):315
= [315(5+4)]:315
= (315.9):315
= (315.32):315
= (315+2):315
= 317:315 (317-15)
= 32 = 9
(313 . 5 + 313 . 4) : 315
=> [313 . (5+4)]: 315
=> (313.9) : 315
Mà 313.9=313.32=315
Vì 315 : 315 = 1 Suy ra : (313 . 5 + 313 . 4) : 315=1