\(a,\dfrac{1}{3x-3y}=\dfrac{x-y}{3\left(x-y\right)^2};\dfrac{1}{x^2-2xy+y^2}=\dfrac{3}{3\left(x-y\right)^2}\\ b,\dfrac{3}{x^2-3x}=\dfrac{6}{2x\left(x-3\right)};\dfrac{5}{2x-6}=\dfrac{5x}{2x\left(x-3\right)}\\ c,\dfrac{x}{x+3}=\dfrac{x^2-3x}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{3-x}=\dfrac{-x-3}{\left(x-3\right)\left(x+3\right)};\dfrac{1}{x^2-9}=\dfrac{1}{\left(x-3\right)\left(x+3\right)}\)

\(d,\dfrac{1}{x^2+xy}=\dfrac{xy-y^2}{xy\left(x+y\right)\left(x-y\right)};\dfrac{1}{xy-y^2}=\dfrac{x^2+xy}{xy\left(x-y\right)\left(x+y\right)};\dfrac{2}{y^2-x^2}=\dfrac{-2xy}{xy\left(x-y\right)\left(x+y\right)}\)

Áp dụng t/c dtsbn:

\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6}=\dfrac{3x-3}{6}=\dfrac{4y+12}{16}=\dfrac{5z-25}{30}=\dfrac{-3x+3-4y-12+5z-25}{-6-16+30}=\dfrac{50+3-12-25}{8}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x-1=4\\y+3=8\\z-5=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=5\\y=5\\z=17\end{matrix}\right.\)

a) Để y nguyên thì \(6x-4⋮2x+3\)

\(\Leftrightarrow-13⋮2x+3\)

\(\Leftrightarrow2x+3\in\left\{1;-1;13;-13\right\}\)

\(\Leftrightarrow2x\in\left\{-2;-4;10;-16\right\}\)

hay \(x\in\left\{-1;-2;5;-8\right\}\)

b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)

Đặt \(x=15k;y=20k;z=24k\)

Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)

lk

a, \(\dfrac{x}{7}-\dfrac{1}{2}=\dfrac{y}{y+1}\Leftrightarrow\dfrac{2x-7}{14}=\dfrac{y}{y+1}\Rightarrow\left(2x-7\right)\left(y+1\right)=14y\)

\(\Leftrightarrow2xy+2x-7y-7=14y\Leftrightarrow2xy+2x-21y-7=0\)

\(\Leftrightarrow2x\left(y+1\right)-21\left(y+1\right)+14=0\Leftrightarrow\left(2x-21\right)\left(y+1\right)=-14\)

\(\Rightarrow2x-21;y+1\inƯ\left(-14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

7) vì \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)và x-y+z=36

Nên theo tính chất của dãy tỉ số bằng nhau ta có:

 \(\dfrac{x}{5}\)=\(\dfrac{y}{6}\)=\(\dfrac{z}{7}\)=\(\dfrac{x-y+z}{5-6+7}\)=\(\dfrac{36}{6}\)=6

 \(\Rightarrow\)x=6.5=30

     y=6.6=36

     z=6.7=42

vậy x=30,y=36,z=42

a: 2x-3y-4z=24

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{1}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{2x-3y-4z}{2\cdot1-3\cdot6-4\cdot3}=\dfrac{24}{-28}=\dfrac{-6}{7}\)

=>x=-6/7; y=-36/7; z=-18/7

b: 6x=10y=15z

=>x/10=y/6=z/4=k

=>x=10k; y=6k; z=4k

x+y-z=90

=>10k+6k-4k=90

=>12k=90

=>k=7,5

=>x=75; y=45; z=30

d: x/4=y/3

=>x/20=y/15

y/5=z/3

=>y/15=z/9

=>x/20=y/15=z/9

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{9}=\dfrac{x-y-z}{20-15-9}=\dfrac{-100}{-4}=25\)

=>x=500; y=375; z=225

1) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{x+y}{5+7}=\dfrac{48}{12}=4\)

\(\dfrac{x}{5}=4\Rightarrow x=20\\ \dfrac{y}{7}=4\Rightarrow y=28\)

2) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{4}=\dfrac{y}{-7}=\dfrac{x-y}{4+7}=\dfrac{33}{11}=3\)

\(\dfrac{x}{4}=3\Rightarrow x=12\\ \dfrac{y}{-7}=3\Rightarrow y=-21\)