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\(1,\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{3x+y}{9+5}=\dfrac{28}{14}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ 2,\\ a,a=2\Rightarrow y=f\left(x\right)=2x\\ b,f\left(-0,5\right)=2\left(-0,5\right)=-1\\ f\left(\dfrac{3}{4}\right)=2\cdot\dfrac{3}{4}=\dfrac{3}{2}\\ c,\text{Thay }x=-4;y=2\Rightarrow-4a=2\Rightarrow a=-\dfrac{1}{2}\)
Ta có: x/y=3/5 ⇒ x/3=y/5
Theo tính chất của dãy tỉ số bằng nhau ta có:x/3=y/5=3x/3.3=y/5= 3x+y9/y9+5=28/14=2
Do đó:
x/3=2 ⇒x=2.3=6
y/5=2 ⇒y=2.5=10
Vậy x=6 và y=10.
\(f\left(-3\right)=\left(-3\right)^2+1=9+1=10\)
\(f\left(-1\right)=\left(-1\right)^2+1=1+1=2\)
\(f\left(3\right)=3^2+1=9+1=10\)
\(f\left(0\right)=0^2+1=1\)
\(f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2+1=\dfrac{1}{4}+\dfrac{4}{4}=\dfrac{5}{4}\)
\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^2+1=\dfrac{1}{4}+\dfrac{4}{4}=\dfrac{5}{4}\)
\(f\left(\dfrac{2}{3}\right)=\left(\dfrac{2}{3}\right)^2+1=\dfrac{4}{9}+\dfrac{9}{9}=\dfrac{13}{9}\)
\(f\left(-\dfrac{2}{3}\right)=\left(-\dfrac{2}{3}\right)^2+1=\dfrac{4}{9}+\dfrac{9}{9}=\dfrac{13}{9}\)
a: f(0)=1
\(f\left(-\dfrac{1}{3}\right)=1-3\cdot\left(-\dfrac{1}{3}\right)^2=1-3\cdot\dfrac{1}{9}=1-\dfrac{1}{3}=\dfrac{2}{3}\)
a: f(0)=1
\(f\left(-\dfrac{1}{3}\right)=1-3\cdot\left(-\dfrac{1}{3}\right)^2=1-3\cdot\dfrac{1}{9}=1-\dfrac{1}{3}=\dfrac{2}{3}\)
a)Ta có:
f(3)= 3.3-8=1
f(2)=3.2-8=-2
b) y=1
=>3x-8=1
=>3x=9
=>x=3
a) y = f(x) = 3x - 8
=> f(3) = 3 . 3 - 8 = 9 - 8 = 1
f(-2) = 3 . (-2) - 8 = -6 - 8 = -14
b) y = 1 => 3x - 8 = 1 => 3x = 9 => x = 3
Vậy ..............
1.
y=f(-1)=3*(-1)-2=-5
y=f(0)=3*0-2=-2
y=f(-2)=3*(-2)-2=-8
y=f(3)=3*3-2=7
Câu 2,3a làm tương tự,chỉ việc thay f(x) thôi.
3b
Khi y=5 =>5=5-2*x=>2*x=0=> x=0
Khi y=3=>3=5-2*x=>2*x=2=>x=1
Khi y=-1=>-1=5-2*x=>2*x=6=>x=3
f(-1)=3.1-2=3-2=1
f(0)=3.0-2=0-2=-2
f(-2)=3.(-2)-2=-6-2=-8
f(3)=3.3-2=9-2=7
a.\(f\left(0\right)=3.0=0\)
\(f\left(-1\right)=3.\left(-1\right)=\left(-3\right)\)
\(f\left(2\right)=3.2=6\)