giúp mình bài này nữa đi
1/1.2+1/2.3+1/3.4+...+1/98.99+1/99.100
gấp lắm !!!
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1/1.2 + 1/2.3 + .................+ 1/99.100 =
1/1 - 1/2 + 1/2 - 1/3 +....................+ 1/99 - 1/100 =
1/1 - 1/100 = 99/100
`1/( 1.2 ) + 1/( 2.3 ) + .......+1/(99.100)`
`= 1-1/2+1/2-1/3+.....+1/99-1/100`
`=1-1/100`
`=99/100`
Ta có : 1.98 + 2.97 + 3.96 + ...+ 98.1 = 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + .....+ ( 1 + 2 + 3 + ...+ 97 + 98 ) = \(\frac{1.2}{2}\)+ \(\frac{2.3}{2}\)+ \(\frac{3.4}{2}\)+ ...+ \(\frac{98.99}{2}\)= \(\frac{1}{2}\)( 1 . 2 + 2 . 3 + 3 . 4 +...+ 98 . 99).
Vậy A = \(\frac{1}{2}\)
Nè bạn giải cụ thể chi tiết cho mình đk k thì mình mới k cho đk
2.
a) (2x + 1)3 = 125
(2x + 1)3 = 53
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4:2
x = 2
Vậy x = 2
b) 5x+1 = 54
x + 1 = 4
x = 4 - 1
x = 3
Vây x = 3
a) \(A=1.2+2.3+3.4+...+98.99+99.100\)
\(3A=1.2.3+2.3.4+3.4.3+...+99.100.3\)
\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)....99.100.\left(101-98\right)\)
\(3A=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(0.1.2+1.2.3+2.3.4+...+98.99.100\right)\)
\(3A=99.100.101-0.1.2\)
\(3A=999900-0\)
\(3A=999900\)
\(A=999900:3\)
\(\Rightarrow A=333300\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
vì \(\frac{99}{100}< 1\)
nên \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}< 1\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}< 1\)
Vậy A<1
Đặt \(A=1.2+2.3+...+98.99\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+....+98.99.\left(100-97\right)\)
\(3A=1.2.3+2.3.4-1.2.3+...+98.99.100-97.98.99\)
\(3A=98.99.100\Rightarrow A=\frac{98.99.100}{3}\)
\(\Rightarrow\frac{\frac{98.99.100}{3}.x}{26950}=\frac{-60}{7}\)\(\Rightarrow98.99.100.x=-\frac{60}{7}.80850\)
\(\Rightarrow98.99.100.x=-693000\)
Đến đây bạn tự tính nhé
cảm ơn bạn nhé, nếu không có bạn chắc mai mình bị cô mắng chết
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)