help me
\(\dfrac{3x-74}{y+15}=k\) biết y=3 khi x=2 Tìm x khi y=12
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\(k=\frac{3x-4}{y+15}\)(1)
Thế x = 2 ; y = 3 vào (1) ta được : \(k=\frac{3\cdot2-4}{3+15}=\frac{2}{18}=\frac{1}{9}\)
=> k = 1/9
Khi y = 12, thế y vào (1) ta được :
\(k=\frac{3x-4}{12+15}\)
<=> \(\frac{1}{9}=\frac{3x-4}{27}\)
<=> 1.27 = 9( 3x - 4 )
<=> 27 = 27x - 36
<=> 27x = 63
<=> x = 63/27 = 7/3
Vậy khi y = 12 thì x = 7/3
a: f(5)=75/2
=>\(a\cdot5^2=\dfrac{75}{2}\)
=>\(a=\dfrac{75}{2}:25=\dfrac{3}{2}\)
Vậy: \(y=f\left(x\right)=\dfrac{3}{2}x^2\)
Khi x=-3 thì \(y=\dfrac{3}{2}\left(-3\right)^2=\dfrac{3}{2}\cdot9=\dfrac{27}{2}\)
b: y=15
=>\(\dfrac{3}{2}x^2=15\)
=>\(x^2=10\)
=>\(x=\pm\sqrt{10}\)
\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\)
\(\dfrac{x-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\)
\(\dfrac{x-1+3}{9}=\dfrac{1}{y+2}\)
\(\dfrac{x-\left(1-3\right)}{9}=\dfrac{1}{y+2}\)
\(\dfrac{x-\left(-2\right)}{9}=\dfrac{1}{y+2}\)
\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\)
\(\left(x+2\right)\left(y+2\right)=9\)
=> (X+2) ; (y+2) ϵ Ư(9)
TH1: x+2 = 1 => x = -1
y+2=9 => y = 7
TH2: x+2 = 9 => x = 7
=> y +2 = 1 => y =-1
TH3:x+2 = -9 => x = -11
y+2 = -1 => y=-3
TH4: x+2 = -1 => x =-3
y+2 = -9 => x=-11
TH5: x+2 = -3 => x =-5
y+2 = -3 => y=-5
TH6: x+2 =3 => x = 1
y+2=3 => y=1
Vì khi x=2 thì y=3 ,nên:
=>\(\dfrac{3.2-74}{3+15}\)=k
=>\(\dfrac{6-74}{18}\)=k
=>\(\dfrac{-34}{9}\)=k
Vì k=\(\dfrac{-34}{9}\)và y=12,suy ra:
\(\dfrac{3x-74}{12+15}\)=\(\dfrac{-34}{9}\)
=>\(\dfrac{3x-74}{27}=\dfrac{-34}{9}\)
=>\(\dfrac{3x-74}{27}=\dfrac{-102}{27}\)
=>3x-74=-102
=>3x= -102+74
=>3x= - 28
=>x=\(\dfrac{-28}{3}\)
Vậy x=\(\dfrac{-28}{3}\)khi y=12