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

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)

\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy x=100

Chúc bạn học tốt

`(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`

sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<

\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)

\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)

Ta có:\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)

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

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

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

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

(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 $x=-100$

\(\frac{x-2}{99}+\frac{x-3}{98}=\frac{x-4}{97}+\frac{x-5}{96}\Leftrightarrow\frac{x-2}{99}-1+\frac{x-3}{98}-1=\frac{x-4}{97}-1+\frac{x-5}{96}-1\)

<=>\(\frac{x-101}{99}+\frac{x-101}{98}=\frac{x-101}{97}+\frac{x-101}{96}\)

<=>\(\left(x-101\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

<=>x-101=0 \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\right)\)

<=>x=101

101

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