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

<=> \(x+\dfrac{4}{3}=5,6\)

<=> \(x=\dfrac{64}{15}\)

x(x-3)-x(x+1)=12

<=>x²-3x-x²-x=12

<=>-4x=12

<=>x=-3

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...

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

x(x+1)-(x-1)(x+2)=8

\(\Leftrightarrow\)\(x^2+x-x^2-2x+x+2=8\)

\(\Leftrightarrow0x=6\left(ptvn\right)\)

\(\Rightarrow S=\varnothing\)

ta có :

\(12\times x=42:\frac{1}{10}\)

ha y \(12\times x=42\times10=420\) nên : \(x=420:12=35\)

Vậy x = 35

Ta có :\(\frac{x}{6}-\frac{7}{y}=\frac{1}{12}\)(y khác 0)

=> \(\frac{xy-42}{6y}=\frac{1}{12}\)

=> 12(xy - 42) = 6y

=> 12xy - 504 = 6y

=> 12xy - 6y = 504

=> 2xy - y = 84

=> y(2x - 1) = 84

Ta có 84 = 1.84 = (-1).(-84) = 42.2 = (-42).(-2) = 21.4 = (-21).(-4) = 7.12 = (-7).(-12) = (-3).(-28) = 28.3 = 14.6 = (-14).(-6)

Lập bảng xét 24 trường hợp

Vậy các cặp (y;x) thỏa mãn là : (84;1) ; (4 ; 11) ; (12 ; 4) ; (28 ; 2) ; (-4 ; - 10) ; (-12 ; -3) ; (-28 ; -1) ; (-84 ; 0)

\(...\Rightarrow x+x+\dfrac{x}{43}+\dfrac{x}{8}=14+148+\dfrac{10}{30}+\dfrac{5}{95}\)

\(\Rightarrow\left(1+1+\dfrac{1}{43}+\dfrac{1}{8}\right)x=162+\dfrac{1}{3}+\dfrac{1}{19}\)

\(\Rightarrow\left(\dfrac{2.43.8}{43.8}+\dfrac{1.8}{43.8}+\dfrac{1.43}{43.8}\right)x=\dfrac{162.3.19}{3.19}+\dfrac{1.19}{3.19}+\dfrac{1.3}{19.3}\)

\(\Rightarrow\left(\dfrac{688}{344}+\dfrac{8}{344}+\dfrac{43}{344}\right)x=\dfrac{9234}{57}+\dfrac{19}{57}+\dfrac{3}{57}\)

\(\Rightarrow\dfrac{739}{344}x=\dfrac{9256}{57}\)

\(\Rightarrow x=\dfrac{9256}{57}:\dfrac{739}{344}=\dfrac{9256}{57}.\dfrac{344}{739}=\dfrac{\text{3184064}}{\text{42123}}\)