a: =>4/x=y/21=4/7

=>x=7; y=21*4/7=12

b: x/7=9/y

=>xy=63

mà x>y

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

c: x/15=3/y

=>xy=45

mà x<y<0

nên \(\left(x,y\right)\in\left\{\left(-45;-1\right);\left(-15;-3\right);\left(-9;-5\right)\right\}\)

d: x/y=21/28=3/4

=>x/3=y/4=k

=>x=3k; y=4k(k\(\in Z\))

\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}\\ \Rightarrow\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{4}{7}\)

\(\dfrac{4}{x}=\dfrac{4}{7}\Rightarrow x=7\)

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

1) Ta có: \(\dfrac{-4}{x}=\dfrac{y}{-21}=\dfrac{28}{49}\)

\(\Leftrightarrow\dfrac{-4}{x}=\dfrac{y}{-21}=\dfrac{4}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-4}{x}=\dfrac{4}{7}\\\dfrac{y}{-21}=\dfrac{4}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-4\cdot7}{4}=-7\\y=\dfrac{-21\cdot4}{7}=-12\end{matrix}\right.\)

Vậy: (x,y)=(-7;-12)

vì 4/x=28/49 nên x.28=4.49=196 ( dấu . là dấu nhân , vì sao x.28=4.49 cấp 2 học )

x=196:28=7

vì y/21=28/49 nên y.49=21.28=588

y=588:49=12

\(\frac{4}{x}=\frac{y}{21}=\frac{28}{49}\)

\(\Rightarrow\frac{4}{x}=\frac{12}{21}=\frac{28}{49}\)

\(\Rightarrow\frac{4}{7}=\frac{12}{21}=\frac{28}{49}\)

Vậy x = 7 , y = 12

Cách làm:

y = \(\frac{28\times21}{49}=12\)

x = \(\frac{21\times4}{12}=7\)

Công thức : nhân chéo

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\(\dfrac{4}{x}=\dfrac{y}{21}=\dfrac{28}{49}=\dfrac{28:7}{49:7}=\dfrac{4}{9}\\ Vậy:x=\dfrac{4.9}{4}=9\\ y=\dfrac{4.21}{9}=\dfrac{28}{3}\)

\(\dfrac{x}{2}=\dfrac{3}{y}\\ \Leftrightarrow x.y=2.3=6\\ Vậy:\left[{}\begin{matrix}\left(x;y\right)=\left(1;6\right)=\left(6;1\right)\\\left(x;y\right)=\left(2;3\right)=\left(3;2\right)\end{matrix}\right.\)

\(3,\\ \dfrac{42}{54}=\dfrac{42:6}{54:6}=\dfrac{7}{9}\\ \dfrac{42}{54}=\dfrac{7}{x}=\dfrac{7}{9}\\ Vậy:x=\dfrac{7.9}{7}=9\)

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

\(\frac{x}{15}\)\(=\frac{3}{y}\)

\(\Rightarrow x.y=15.3=45\)

Vói \(x,y\)nguyên ta có :

\(x=-15;-9;-45;9;15;45\)

\(y=-3;-5;-1;3;5;1\)

Giải thích các bước giải:

 a) x/15=3/y

=> x.y=45

Với x, y nguyên và x<y<0 , ta có :

   x   -15     -9     -45

   y    -3      -5       -1

Vậy có 3 cặp (x ; y)

\(\frac{X}{2}=\frac{Y}{3}=\frac{Z}{4}\)\(=\frac{X}{2}=\frac{2Y}{6}=\frac{3Z}{12}\)\(=\frac{X+2Y-3Z}{2+6-12}\)\(=5\)

\(=>X=2.5=10\)

\(=>y=3.5=15\)

\(=>z=4.5=20\)

vậy.....