\(\dfrac{x}{30}+\dfrac{x}{40}=\dfrac{3}{4}\)\(\Leftrightarrow x\left(\dfrac{1}{40}+\dfrac{1}{30}\right)=\dfrac{3}{4}\)\(\Leftrightarrow x.\dfrac{7}{120}=\dfrac{3}{4}\)\(\Leftrightarrow x=\dfrac{90}{7}\)

Vậy tập nghiệm của pt là S=\(\left\{x=\dfrac{90}{7}\right\}\)

\(\dfrac{x}{40}+\dfrac{x}{30}=\dfrac{3}{4}\\ \dfrac{30x}{120}+\dfrac{40x}{120}=\dfrac{90}{120}\\ \dfrac{70x}{120}=\dfrac{90}{120}\\ \Rightarrow70x=90\\ \Rightarrow x=\dfrac{90}{70}\\ x=\dfrac{9}{7}\)

Mẫu chung là 120
<=> 3x + 4x = 90 ( Quy Đồng Khử Mẫu ) 
<=> 7x = 90
<=> x = 90/7 
Vậy Phương trình có tập nghiệm S = {90/7}
Bạn đổi ra số thập phân cũng được , nó xấp xỉ 12.86

\(x=126\)

x/60 + x/45 + 1,5 = 32/5

3x + 4x + 270 = 1152

7x + 270 = 1152

7x = 1152 − 270

7x = 882

x = 882/7

x = 126

\(\dfrac{x}{60}+\dfrac{x}{45}+1,5=\dfrac{32}{5}\)

\(\Leftrightarrow\)​​​\(\dfrac{x}{60}+\dfrac{x}{45}+\dfrac{3}{2}=\dfrac{32}{5}\)​

\(\Leftrightarrow\dfrac{3x}{180}+\dfrac{4x}{180}+\dfrac{270}{180}=\dfrac{1152}{180}\)

\(\Leftrightarrow7x+270=1152\)

\(\Leftrightarrow7x=882\)

\(\Leftrightarrow x=126\)

\(\Leftrightarrow x^2-x-x-1=-1\)

=>x(x-2)=0

=>x=2

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne0\end{matrix}\right.\)

\(\dfrac{x-1}{x+1}-\dfrac{1}{x}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x\left(x-1\right)}{x\left(x+1\right)}-\dfrac{\left(x+1\right)}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2-x-x-1}{x\left(x+1\right)}=\dfrac{-1}{x\left(x+1\right)}\\ \Leftrightarrow x^2-2x-1=-1\\ \Leftrightarrow x^2-2x=0\\ \Leftrightarrow x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(\dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow2x^2+2x-x^2+x=x^2-1\)

\(\Leftrightarrow2x^2+2x-x^2+x-x^2+1=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow x=\dfrac{-1}{3}\left(nhận\right)\)

-Vậy \(S=\left\{\dfrac{-1}{3}\right\}\)

`[x+35]/1984-[x+30]/1989+[x+19]/2000+[x+23]/[1996=-2`

`<=>[x+35]/1984+1-[x+30]/1989-1+[x+19]/2000+1+[x+23]/1996+1=0`

`<=>[x+2019]/1984-[x+2019]/1989+[x+2019]/2000+[x+2019]/1996=0`

`<=>(x+2019)(1/1984-1/1989+1/2000+1/1996)=0`

  `=>x+2019=0`

`<=>x=-2019`

\(\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}\text{=}-2\)

\(\Leftrightarrow\dfrac{x+35}{1984}-\dfrac{x+30}{1989}+\dfrac{x+19}{2000}+\dfrac{x+23}{1996}+3-1\text{=}0\)

\(\Leftrightarrow\left(\dfrac{x+35}{1984}+1\right)-\left(\dfrac{x+30}{1989}+1\right)+\left(\dfrac{x+19}{2000}+1\right)+\left(\dfrac{x+23}{1996}+1\right)\text{=}0\)

\(\Leftrightarrow\dfrac{x+2019}{1984}-\dfrac{x+2019}{1989}+\dfrac{x+2019}{2000}+\dfrac{x+2019}{1996}\text{=}0\)

\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{1984}-\dfrac{1}{1989}+\dfrac{1}{2000}+\dfrac{1}{1996}\right)\text{=}0\)

\(\Leftrightarrow\left(x+2019\right)\text{=}0\)

\(\Leftrightarrow x\text{=}-2019\)

*Sửa: Lỗi ở đề

`[x+35]/1984-[x+30]/1989+[x+19]/2000+[x+23]/1996=-2`

\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\) ( ĐK : \(x\ne-2\) )

\(\Leftrightarrow\dfrac{12}{x^3+2^3}=1+\dfrac{1}{x+2}\)

\(\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow12=\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4\)

\(\Leftrightarrow x^3+8+x^2-2x+4=12\)

\(\Leftrightarrow x^3+x^2-2x=0\)

\(\Leftrightarrow x\left(x^2+x-2\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-2\left(L\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;1\right\}\)