Bạn xem lại đề chỗ : \(y=f\left(x\right)=\frac{2}{3}\)

TỰ TỬ ĐÊ CU AI THẤY HAY CHO

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

\(f\left(2\right)=-5.2^2+5=-5.4+5=-15\)

\(f\left(1\right)=-5.1^2+5=-5.1+5=0\)

\(f\left(3\right)=-5.3^2+5=-5.9+5=-40\)

\(f\left(-x\right)=100-\left(-x\right)^2=100-x^2=f\left(x\right)\)

\(f\left(-3\right)=\left(-3\right)^2+1=9+1=10\)

\(f\left(-1\right)=\left(-1\right)^2+1=1+1=2\)

\(f\left(3\right)=3^2+1=9+1=10\)

\(f\left(0\right)=0^2+1=1\)

\(f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2+1=\dfrac{1}{4}+\dfrac{4}{4}=\dfrac{5}{4}\)

\(f\left(\dfrac{1}{2}\right)=\left(\dfrac{1}{2}\right)^2+1=\dfrac{1}{4}+\dfrac{4}{4}=\dfrac{5}{4}\)

\(f\left(\dfrac{2}{3}\right)=\left(\dfrac{2}{3}\right)^2+1=\dfrac{4}{9}+\dfrac{9}{9}=\dfrac{13}{9}\)

\(f\left(-\dfrac{2}{3}\right)=\left(-\dfrac{2}{3}\right)^2+1=\dfrac{4}{9}+\dfrac{9}{9}=\dfrac{13}{9}\)