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![](https://rs.olm.vn/images/avt/0.png?1311)
(1)
a) x=\(\dfrac{-1}{12}-\dfrac{2}{3}\)=\(\dfrac{-3}{4}\)
b) 2x+1=3 => 2x=3-1=2 => x=1
(2)
f(2)=2.22+4=12
f(-1)=2.(-1)2+4=6
(1)
a) \(x+\dfrac{2}{3}=-\dfrac{1}{12}\\ \Rightarrow x=-\dfrac{1}{12}-\dfrac{2}{3}\\ \Rightarrow x=\dfrac{-1}{12}-\dfrac{8}{12}\\ \Rightarrow x=-\dfrac{9}{12}=-\dfrac{3}{4}\)
Vậy \(x=-\dfrac{3}{4}\)
b) \(\left(2x+1\right)^2=9\\ \Rightarrow\left(2x+1\right)^2=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{-2;1\right\}\)
(2)
\(y=f\left(x\right)=2x^2+4\\ f\left(2\right)=2\cdot2^2+4=8+4=12\\ f\left(-1\right)=2\cdot\left(-1\right)^2+4=2+4=6\)
Vậy \(f\left(2\right)=12\\ f\left(-1\right)=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho hàm số y=f(x)= −3x.
Ta có f(\(\dfrac{-3}{2}\)) = -3. (\(\dfrac{-3}{2}\))
= \(\dfrac{-3.\left(-3\right)}{2}\)
=\(\dfrac{9}{2}\)
Ta có f(-1) = -3. (-1)
= 3
Vậy f(\(\dfrac{-3}{2}\)) = \(\dfrac{9}{2}\) và f(-1) = 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Cho hàm số y = f(x) = -2x + 3.
Ta có: f(-2)= -2.(-2)+3
= 4+3=7
Ta có: f(0)= -2.0+3
= 0+3=3
Ta có: f(\(\dfrac{-1}{2}\))= -2.(-\(\dfrac{1}{2}\))+3
=\(\dfrac{-2.\left(-1\right)}{2}\)+3
=\(\dfrac{2}{2}\)+3
= 1+3= 4
Vậy f(-2)=7;f(0)=3;f( \(\dfrac{-1}{2}\))=4
b) Cho hàm số y = f(x) = -2x + 3
mà f(x)=5
Suy ra: f(x) = -2x + 3=5
hay -2x + 3=5
-2x=5-3
-2x=2
x=2:(-2)
x= -1
Cho hàm số y = f(x) = -2x + 3
mà f(x)=1
Suy ra: f(x) = -2x + 3=1
hay -2x + 3=1
-2x=1-3
-2x= -2
x= -2:(-2)
x=1
Vậy f(x)=5 thì x= -1 và f(x) = 1 thì x=1.
Lời giải:
a.
$f(-2)=(-2)(-2)+3=7$
$f(0)=(-2).0+3=3$
$f(\frac{-1}{2})=(-2).\frac{-1}{2}+3=4$
b.
$f(x)=-2x+3=5$
$\Rightarrow -2x=2$
$\Rightarrow x=-1$
$f(x)=-2x+3=1$
$\Rightarrow -2x=1-3=-2$
$\Rightarrow x=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: f(1) = 2.12 – 5 = 2 – 5 = -3
f(-2) = 2.(-2)2 – 5 = 8 – 5 = 3
f(0) = 2.02 – 5 = 0 – 5 = -5
f(2) = 2.22 – 5 = 2.4 – 5 = 3
\(•f\left(0\right)=2.0^2-3.0=0\\ •f\left(2\right)=2.2^2-3.2=2\\ •f\left(\dfrac{1}{2}\right)=2.\left(\dfrac{1}{2}\right)^2.3.\dfrac{1}{2}=-1\\ •f\left(\dfrac{-2}{3}\right)=2.\left(-\dfrac{2}{3}\right)^2-3.\left(-\dfrac{2}{3}\right)=\dfrac{26}{9}\\ •f\left(3\right)=2.3^2-3.3=9\)