![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(f\left(1\right)=2.1^2+5.1-3=2+5-3=4\)
\(f\left(0\right)=0+0-3=-3\)
\(f\left(1,5\right)=2.\left(1,5\right)^2-5.1,5-3=4,5-7,5-3=-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
f(-1)=b-a=10
f(1)=b+a=6
=>b=(10+6):2=8=> a=6-8=-2
Vậy a=-2,b=8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(f\left(3\right)=3a-3=9\)
\(3a=12\Rightarrow a=4\)
\(f\left(5\right)=5a-3=11\)
\(5a=14\Rightarrow a=\dfrac{14}{5}\)
\(f\left(-1\right)=-a-3=6\)
\(-a=9\Rightarrow a=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)`
`@f(1)=2.1^2+5.1-3=2.1+5-3=2+5-3=4`
`@f(0)=2.0^2+5.0-3=-3`
`@f(1,5)=2.(1,5)^2+5.1,5-3=4,5+7,5-3=9`
_____________________________________________________
`b)`
`***f(3)=9`
`=>3a-3=9`
`=>3a=12=>a=4`
`***f(5)=11`
`=>5a-3=11`
`=>5a=14=>a=14/5`
`***f(-1)=6`
`=>-a-3=6`
`=>-a=9=>a=-9`
a: f(1)=2+5-3=4
f(0)=-3
f(1,5)=4,5+7,5-3=9
b: f(3)=9 nên 3a-3=9
hay a=4
f(5)=11 nên 5a-3=11
hay a=14/5
f(-1)=6 nên -a-3=6
=>-a=9
hay a=-9
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(f\left(1\right)=a-7=-4\)\(\Leftrightarrow a=3\)
\(f\left(2\right)=2a-7=5\Leftrightarrow a=6\)
\(f\left(3\right)=3a-7=6\Leftrightarrow a=\frac{13}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) Ta có : f(2) = 5
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(2\right)\\\text{ax}-3=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\a.2-3=5\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2\\a=4\end{cases}}\)
Vậy a = 4
b ) Ta có : f(0) = 3
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(0\right)\\\text{ax}+b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\a.0+b=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\b=3\end{cases}}\) ( 1 )
Ta có : f ( 1 ) = 4
\(\Leftrightarrow\hept{\begin{cases}f\left(x\right)=f\left(1\right)\\\text{ax}+b=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\a.1+b=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\a+b=4\end{cases}}\) ( 2 )
Thay b = 3 ở ( 1 ) vào a+b=4 ở ( 2 ) ta được : a + 3 = 4
a = 1
Vậy a = 1 ; b = 3
\(f\left(x\right)=ax-3\)
a) \(f\left(3\right)=a.3-3=9\)
\(\Leftrightarrow f\left(3\right)=a.3=9+3\)
\(\Leftrightarrow f\left(3\right)=a.3=12\)
\(\Leftrightarrow f\left(3\right)=a=12:3\)
\(\Leftrightarrow f\left(3\right)=a=4\)
Vậy \(f\left(3\right)=9\) thì \(a=4\)
b) \(f\left(5\right)=a.5-3=11\)
\(\Leftrightarrow f\left(5\right)=a.5=11+3\)
\(\Leftrightarrow f\left(5\right)=a.5=14\)
\(\Leftrightarrow f\left(5\right)=a=14:5\)
\(\Leftrightarrow f\left(5\right)=a=\dfrac{14}{5}\)
Vậy \(f\left(5\right)=11\) thì \(a=\dfrac{14}{5}\)
c) \(f\left(-1\right)=a.\left(-1\right)-3=6\)
\(\Leftrightarrow f\left(-1\right)=a.\left(-1\right)=6+3\)
\(\Leftrightarrow f\left(-1\right)=a.\left(-1\right)=9\)
\(\Leftrightarrow f\left(-1\right)=a=9:\left(-1\right)\)
\(\Leftrightarrow f\left(-1\right)=a=-9\)
Vậy \(f\left(-1\right)=6\) thì \(a=-9\)