\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)

Ta có:

\(\dfrac{x}{3}=20\)

\(\Rightarrow\)\(x=60\)

Lại có:

\(x+y=20\)

\(\Rightarrow\)\(y=20-60\)

\(\Rightarrow\)\(y=-40\)

Vây x = 60 và y = - 40

\(\dfrac{x}{3}=x+y\)

\(\Leftrightarrow x-\dfrac{1}{3}x=-y\)

\(\Leftrightarrow y=-\dfrac{2}{3}x\)

Ta có: x+y=20

\(\Leftrightarrow x\cdot\dfrac{1}{3}=20\)

hay x=60

=> y=40

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=-\dfrac{54}{5}\)

\(\dfrac{x}{2}=-\dfrac{54}{5}\Rightarrow x=-\dfrac{54}{5}.2=-\dfrac{108}{5}\)

\(\dfrac{y}{3}=-\dfrac{54}{5}\Rightarrow y=-\dfrac{54}{5}.3=-\dfrac{162}{5}\)

Vậy \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)

Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}\)

nên \(\dfrac{2x}{4}=\dfrac{3y}{9}\)

mà 2x-3y=54

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y}{4-9}=\dfrac{-54}{5}\)

Do đó: \(x=-\dfrac{108}{5};y=-\dfrac{162}{5}\)

ta có : `x/2 = y/3 = z/4=> (2x)/4 =(3y)/9 = z/4`

`=> (2x)/4 =(3y)/9 = z/4` và `2x + 3y - z = 27`

Áp dụng t/c dãy tỉ số bằng nhau ta có:

`(2x)/4 =(3y)/9 = z/4 =(2x + 3y - z)/(4+9-4)=27/9=3`

`=>x/2=3=>x=3.2=6`

`=>y/3=3=>x=3.3=9`

`=>z/4=3=>z=3.4=12`

\(\Leftrightarrow\left(x+2\right)^3=343\)

=>x+2=7

hay x=5

\(x^3-3x^2+3x-1=-8\)

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

hay x=-1

\(\Leftrightarrow xy=63\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)

\(\dfrac{x^2}{20}=\dfrac{4}{5}\)

\(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?

\(\dfrac{x-1}{8}=\dfrac{4}{x-2}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)}{8.\left(x-2\right)}=\dfrac{4.8}{\left(x-2\right).8}\)

=>khử mẫu

=>(x-1)(x-2)=32

.......

Là \(\dfrac{x+1}{8}=\dfrac{4}{x-2}\) nhoa