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1) \(A=20,11\times2,4+1,4\times20,11\times2-10,05\times10,4\) (tách 40,22 = 20,11 x 2)
\(=20,11\times\left(2,4+1,4\times2\right)-10,05\times10,4\)
\(=20,11\times5,2-10,05\times5,2\times2\) (tách 10,4 = 5,2 x 2)
\(=5,2\times\left(20,11-20,1\right)\)
\(=5,2\times0,01\)
\(=0,052\)
(11/1×12 + 11/12×23 + 11/23×34 + ... + 11/89×100) + x = 2/3
(1 - 1/12 + 1/12 - 1/23 + 1/23 - 1/34 + ... + 1/89 - 1/100) + x = 2/3
(1 - 1/100) + x = 2/3
99/100 + x = 2/3
x = 2/3 - 99/100
x = 200/300 - 297/300
x = -97/300
Vậy x =-97/300
^_^ ^_^ ☆☆☆
A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100
A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100
A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100
A/100=1-1/2 + 1/2-1/3+...+1/99-1/100
A/100=1-1/100
A/100=99/100
A=99/100x100=99
Vậy A=99.
Ta có:
\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)
\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)
A. \(\left(x+1\right)+\left(x+2\right)+......+\left(x+100\right)=5750\)
\(x+1+x+2+....+x+100=5750\)
\(100x+\left(1+2+3+.......+100\right)=5750\)
\(100x+5050=5750\)
\(100x=700\)
\(x=700:100=7\)
B. x+(1+2+......+100) = 2000
x + 5050 = 2000
x = 2000 - 5050
x= -3050
C. ( x-1 )+(x-2)+......+( x - 100 ) = 50
x-1+x-2+.........+x-100 = 50
100x + ( -1-2-........-100 ) = 50
100x + ( - 5050 ) = 50
100x = 50 + 5050
100 x = 5100
x = 5100 : 100
x = 51
A . \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(100x+5050=5750\)
\(100x=5750-5050\)
\(100x=700\)
\(\Rightarrow x=\frac{700}{100}=7\)
B. \(x+\left(1+2+3+4+5+....+100\right)=2000\)
\(x+\frac{\left(100+1\right).100}{2}=2000\)
\(x+5050=2000\)
\(\Rightarrow x=2000-5050=-3050\)
C. \(\left(x-1\right)+\left(x-2\right)+\left(x-3\right)+....+\left(x-100\right)=50\)
\(\left(x+x+x+...+x\right)-\left(1+2+3+...+100\right)=50\)
\(100x-5050=50\)
\(100x=5100\)
\(\Rightarrow x=\frac{5100}{100}=51\)
Biểu thức trên sẽ bằng
(12+13-25)x(1+2+3+..+100)
=0.(1+2+...+100)=0
vÌ 0 NHÂN VS BẤT KÌ SỐ NÀO CX =0
----------------Hok tốt-----------------
[1 . 12 + 1 . 13 - 1 . 25 ] . [ 1 + 2 + 3 + .... + 100 ]
= [1 ( 12 + 13 - 25 ) ] . [ 1 + 2 + 3 + ... + 100 ]
= 0 . [ 1 + 2 + 3 + ... + 100 ]
= 0