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

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

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

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

#Hok tốt !

awioufWQCB

x=9,16515139?

4 /x=y/21=28/49

vì 21.4/12=7 nên x=7

    28.21/49 =12 nên y=12 (công thức nhân chéo)

vậy ,x=7 ;y=12

x =7

y = 12

KHONG HIEU

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\(\in Z\))

LAM THE NAY KHONG HIEU

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

\(\Rightarrow y=\frac{28}{49}\times21\)

\(\Rightarrow y=12\)

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

\(\Rightarrow x=7\)

Vậy x = 7 và y = 12

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

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

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

a: =>2xy+y=7

=>(2x+1)*y=7

=>(2x+1;y) thuộc {(1;7); (7;1); (-1;-7); (-7;-1)}

=>(x,y) thuộc {(0;7); (3;1); (-1;-7); (-4;-1)}

b: =>(2x+1)^2+(y+1)^2=179-169=10

=>((2x+1)^2;(y+1)^2) thuộc {(1;9); (9;1)}

TH1: (2x+1)^2=1 và (y+1)^2=9

=>\(\left\{{}\begin{matrix}2x+1\in\left\{1;-1\right\}\\y+1\in\left\{3;-3\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{0;-1\right\}\\y\in\left\{2;-4\right\}\end{matrix}\right.\)

TH2: (2x+1)^2=9 và (y+1)^2=1

=>\(\left\{{}\begin{matrix}2x+1\in\left\{3;-3\right\}\\y+1\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{1;-2\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

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