`(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`

(x+1)/99 + (x+2)/98 + (x+3)/97 + (x+4)/96 = -4`

$\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}+4=0$

$\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0$

$\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0$

$(x+100)(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96})=0$

Vì $\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne 0$

$\Rightarrow x+100=0$

$\Rightarrow x=0-100$

$\Rightarrow x=-100$

Vậy $x=-100$

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

Ai tra loi nhanh nhat minh se cho

4x + (1/99+2/98+3/97 + 4/96)=-4

4x=-4 - (1/99+2/98+3/97 + 4/96)

4x=

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

Mn nhớ giải chi tiết ra nha

\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)

... rồi đặt x+100 làm nhân tử chung => x = -100

trả lời đầy đủ cho mình với

x=-100 chắn chắn luôn,cô giáo mình cho làm rồi

độ chính xác là 100%

(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

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Rightarrow\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\)

\(\Rightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\)

\(\Rightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

\(\Rightarrow x=-100\)(do \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\))