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

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

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

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

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

Lời giải:

a. $\frac{x}{7}=\frac{6}{21}$

$x=\frac{6}{21}.7$

$x=2$

b.

$\frac{-5}{y}=\frac{20}{28}$

$y=-5:\frac{20}{28}$

$y=-7$

c.

$\frac{-4}{8}=\frac{-7}{y}$

$y=-7:\frac{-4}{8}$

$y=14$

a, \(\dfrac{x}{7}=\dfrac{6}{21}\Leftrightarrow\dfrac{3x}{21}=\dfrac{6}{21}\Rightarrow x=2\)

b, \(\dfrac{-5}{y}=\dfrac{20}{28}\Leftrightarrow\dfrac{20}{-4y}=\dfrac{20}{28}\Leftrightarrow y=-7\)

c, \(\dfrac{-4}{8}=-\dfrac{7}{y}\Rightarrow-4y=-56\Leftrightarrow y=14\)

a) Ta có: \(\dfrac{x}{7}=\dfrac{6}{21}\)

nên \(x=\dfrac{6\cdot7}{21}=\dfrac{42}{21}=2\)

b) Ta có: \(\dfrac{-5}{y}=\dfrac{20}{28}\)

nên \(y=\dfrac{-5\cdot28}{20}=\dfrac{-140}{20}=-7\)

c) Ta có: \(\dfrac{-4}{8}=\dfrac{-7}{y}\)

nên \(y=\dfrac{-7\cdot8}{-4}=\dfrac{-56}{-4}=14\)

a) \(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{4}{3}+\dfrac{3}{4}\)

\(\dfrac{1}{2}:y\times\dfrac{3}{5}=\dfrac{25}{12}\)

\(\dfrac{1}{2}:y=\dfrac{25}{12}:\dfrac{3}{5}\)

\(\dfrac{1}{2}:y=\dfrac{125}{36}\)

\(y=\dfrac{1}{2}:\dfrac{125}{36}\)

\(y=\dfrac{18}{125}\)

b) \(\dfrac{4}{3}-\dfrac{1}{2}\times y=1\)

\(\dfrac{1}{2}\times y=\dfrac{4}{3}-1\)

\(\dfrac{1}{2}\times y=\dfrac{1}{3}\)

\(y=\dfrac{1}{3}:\dfrac{1}{2}\)

\(y=\dfrac{2}{3}\)

c) \(\dfrac{1}{4}+y:\dfrac{1}{3}=\dfrac{5}{6}\)

\(y:\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{1}{4}\)

\(y:\dfrac{1}{3}=\dfrac{7}{12}\)

\(y=\dfrac{7}{12}\cdot\dfrac{1}{3}\)

\(y=\dfrac{7}{36}\)

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

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

a) \(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)

\(\dfrac{4}{y}\) = \(\dfrac{x}{3}-\dfrac{1}{5}\)

\(\dfrac{4}{y}\) = \(\dfrac{5x-3}{15}\)

=> 4.15 = y.(5x-3)

60 = y.(5x-3)

Ta có bảng

Vậy y=30 và x=1 ; y=5 và x=3

a.

\(\dfrac{x}{3}-\dfrac{4}{y}=\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{x}{3}-\dfrac{1}{5}=\dfrac{5x-3}{15}\)

\(\Rightarrow y\left(5x-3\right)=60\)

Lập bảng.................

b,c tương tự