4 | + | 4 | + … + | 4 | = | |
2 x 3 | 3 x 4 | 97 x 98 |
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x+1)/99+(x+2)/98+(x+3)/97+(x+4)/96=-4`
`=>(x+1)/99+1+(x+2)/98+1+(x+3)/97+1+(x+4)/96+1=-4+4`
`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`
`=>(x+100)(1/99+1/98+1/97+1/96)=0`
`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)
`=>x=-100`
Vậy ...
`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙
\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)
mà `1/99+1/98+1/97+1/96 \ne 0`
nên `x+100=0`
`x=-100`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )
Vậy x = 1
b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)
=> x + 100 = 0
=> x = -100
c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)
\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)
\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)
=> x - 100 = 0
=> x = 100
Chúc bạn học tốt
có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)
hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)
b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)
\(\Leftrightarrow x+100=0\)
hay x=-100
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
4x + (1/99+2/98+3/97 + 4/96)=-4
4x=-4 - (1/99+2/98+3/97 + 4/96)
4x=
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
... rồi đặt x+100 làm nhân tử chung => x = -100
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+1)/99 + (x+2)/98 + (x+3)/97 + (x+4)/96 = -4
=> [(x+1)/99 +1] +[(x+2)/98+1]+[(x+3)/97+1]+[(x+4)/96+1] = 0
=>[(x+100)/99] + [(x+100)/98] +[(x+100)/97] + [(x+100)/96]=0
=>(x+100)(1/99+1/98+1/97+1/96)=0
=>x+100=0
=>x= -100
\(\dfrac{4}{2.3}+\dfrac{4}{3.4}+\dfrac{4}{5.6}+...+\dfrac{4}{97.98}\)
=\(4.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{97.98}\right)\)
=\(4.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{97}-\dfrac{1}{98}\right)\)
=\(4.\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
=\(4.\) \(\dfrac{24}{49}\)
=\(\dfrac{96}{49}\)