x + 1 : 0,75 = 1,4 : 0,25

<=> \(x+\dfrac{4}{3}=5,6\)

<=> \(x=\dfrac{64}{15}\)

`x-(5/6 -x) =x-2/3`

`x-5/6 +x -x+2/3 =0`

`x = 5/6-2/3 = 5/6 -4/6 = 1/6`

bạn có thể giải chi tiết hơn dc ko TV Cuber

Ta có :

 \(x-\dfrac{8}{5}< -6\\ \Rightarrow x< -6+\dfrac{8}{5}\\ \Rightarrow x< -\dfrac{22}{5}=-4\dfrac{2}{5}\\ \Rightarrow-6< x< -1\dfrac{2}{5}\\ \Rightarrow x=-5\)

    Vậy...

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{8}{5}< -6\\-6< x\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< -6+\dfrac{8}{5}\\x>-6\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x< -\dfrac{22}{5}\\x>-6\end{matrix}\right.\\ \Rightarrow-6< x< -\dfrac{22}{5}\)

\(\frac{x-3}{x+5}=\frac{5}{7}\Rightarrow7\cdot\left(x-3\right)=5\cdot\left(x+5\right)\Rightarrow7x-21=5x+25\Rightarrow7x-5x=25+21\Rightarrow2x=46\Rightarrow x=\frac{46}{2}=23\)

cảm ơn nghe

\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Leftrightarrow x^3-25x-x^3-8=42\)

\(\Leftrightarrow-25x-8=42\)

\(\Leftrightarrow-25x=42+8\)

\(\Leftrightarrow-25x=50\)

\(\Leftrightarrow x=-\dfrac{50}{25}=-2\)

x.(x-5)+x-5=o

=> x.(x-5)+(x-5)=0

=>(x-5)(x+1)=0

=> x-5=0    =>x=5

     x+1=0        x=-1

x ( x - 5 ) + x -  5 = 0

x ( x - 5 ) + ( x - 5 ) = 0

(x - 5 ) ( x + 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)

Vậy...

\(x.\left(x-5\right)+x-5=0\)

\(\Leftrightarrow x.\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right).\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x.2-5.x=0\)

\(\Rightarrow-3x=0\)

\(\Rightarrow x=0\)

Vậy \(x=0\)

\(2x-5x=0\)

\(=>2x=5x\)

\(=>x=0\)

5/3

5/3

\(x\div\dfrac{2}{3}=\dfrac{5}{2}\)

\(x=\dfrac{2}{3}\times\dfrac{5}{2}\)

\(x=\dfrac{5}{3}\)

Đề còn thiếu gì ko bn nhỉ?