tìm các số nguyên
7\x=y\21=-42\54
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\(\frac{7}{x}=\frac{y}{21}=\frac{-42}{54}\)
Ta có : \(\frac{-42}{54}=\frac{-42:6}{54:6}=\frac{-7}{9}\)
+) \(\frac{7}{x}=\frac{-7}{9}\)=> 7.9 = (-7).x => (-7).x = 63 => x = -9
+) \(\frac{y}{21}=\frac{-7}{9}\)=> y.9 = 21.(-7) => y.9 = -147 => y = -49/3
Vậy x = 9,y = -49/3
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ta có:
\(\frac{7}{x}=\frac{y}{21}=\frac{-42}{54}\)
=> \(\frac{7}{x}=\frac{-42}{54}\Rightarrow x=\frac{7\cdot54}{\left(-42\right)}=-9\)
=> \(\frac{y}{21}=\frac{-42}{54}\Rightarrow y=\frac{21\cdot\left(-42\right)}{54}=-16\frac{1}{3}\)
`A)2/3=x/60`
`=>40/60=x/60`
`=>x=40`
`B)-1/2=y/18`
`=>-9/18=y/18`
`=>y=-9`
`C)3/x=y/35=-36/84`
Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`
`=>3/x=-3/7`
`=>x=-7`
`y/35=-3/7=-15/35`
`=>y=-15`
`D)7/x=y/27=-42/54`
Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`
`=>7/x=-7/9`
`=>x=-9`
`y/27=-7/9=-21/27`
`=>y=-21`
7/x=y/27=-42/54=-7/9
7/x=-7/9 nên x=-9
y/27=-7/9=-21/27 nên y=-21
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\(\frac{7}{x}=\frac{y}{27}=\frac{-42}{54}\)
+) => \(\frac{y\cdot2}{27\cdot2}=\frac{-42}{54}\)
=> \(y\cdot2=-42\)
\(\Rightarrow y=-21\)
+) \(\frac{7}{x}=\frac{-21}{27}\)
=> \(\frac{7\cdot\left(-3\right)}{x\cdot\left(-3\right)}=\frac{-21}{27}\)
=> \(x\cdot\left(-3\right)=27\)
=> \(x=-9\)
* Lớp 6 chưa học đến tỉ lệ thức nên đây là cách đơn giản nhất r *
\(\frac{y}{27}\)=\(\frac{-42}{54}\)=>\(\frac{2y}{54}\)=\(\frac{-42}{54}\)=>2y= -42 y= \(\frac{-42}{2}\)= -21 \(\frac{7}{x}\)=\(\frac{-21}{27}\)=> -21x=7*27=189 x=\(\frac{189}{-21}\)= -9 Vậy x= -9, y= -21
a: =>x/-3=3
hay x=-9
b: =>x/9=-1/9
hay x=-1
c: =>x+1/5=-1/3
hay x=-8/15
d: =>-7/x=-7/9
hay x=9
a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)
b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)
c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)
d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)
x=-9, y=-49/3
Ta có
7/x = -42/54 => x= 7.54/-42 =>x=-9 (tmđk)
y/21 = -42/54 => y= -42.21/54 => y= -49/3