a: =>-2x=90/91

hay x=-45/91

b: =>2x=-7

hay x=-7/2

c: ->-3x=-12

hay x=4

a. x=2

b, y=-8

c,x=-1

nhớ k hộ mình nhé

a) \(\frac{x}{7}=\frac{6}{28}\Rightarrow x=\frac{6.7}{28}\Rightarrow x=\frac{3}{2}\)

Vậy \(x=\frac{3}{2}\)

b) \(\frac{-5}{y}=\frac{15}{24}\Rightarrow y=\frac{\left(-5\right).24}{15}\Rightarrow y=-8\)

Vậy \(y=-8\)

c) \(\frac{x}{-3}=\frac{4}{12}\Rightarrow x=\frac{\left(-3.\right).4}{12}\Rightarrow x=-1\)

Vậy \(x=-1\)

49/156 

49/156

:)))) :>>> :+ := :- :

185/741

a)\(\dfrac{x}{7}\)= \(\dfrac{6}{28}\)

x.28 = 6.7

x.28 = 42

x = 42 : 28

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

Vậy x = \(\dfrac{3}{2}\)

b)\(\dfrac{-5}{y}=\dfrac{15}{24}\)

-5.24 = y.15

-120 = y.15

-120:15 = y

-8 = y

Vậy y = -8

c)\(\dfrac{x}{-3}=\dfrac{1}{3}\)

x = 1

Vậy x = 1

a, \(\frac{X}{3}=\frac{5}{Y}\Rightarrow XY=3.5=15\Rightarrow X.Y=15\)

\(\Rightarrow\)X = 1 thì Y = 15

\(\Rightarrow\)X = 3 thì Y = 5

\(\Rightarrow\)X = 5 thì Y = 3

\(\Rightarrow\)X = 15 thì Y = 1

\(\Rightarrow\)X = -1 thì Y = -15

\(\Rightarrow\)X = -3 thì Y = -5

\(\Rightarrow\)X = -5 thì Y = -3

\(\Rightarrow\)X = -15 thì Y = -1

\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)

\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)

\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)

\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)

\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)

\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)

\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)

\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)

\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)

\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)

\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)

\(< =>5x-35=32x-8< =>32x-5x=-35+8\)

\(< =>27x=-27< =>x=-1\)

\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)

\(< =>12=6x-3< =>6x=12+3\)

\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)

\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)

\(< =>8x-4=9x+3< =>9x-8x=-4-3\)

\(< =>9x-8x=-7< =>x=-7\)

\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)

\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)

\(< =>12x-7x=14-4< =>5x=10\)

\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)

\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)

\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)

\(< =>6x-4x=4+6< =>2x=10\)

\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)

\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)

\(< =>\left(x+1\right)\left(x+1\right)=3.3\)

\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)

\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)

\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)

\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)

có thể khẳng định ngay vì trong các tích a.d và b.c luôn có một tích dương và một tích âm