=> f(-x) = ( -x ) ^2 + 5 = x^2 +5 

mà f(x) = x^2 + 5 nên f(x) = f( -x) ( đpcm)

\(a,\)

\(y=f\left(3\right)=4.3^2-5=31\)

\(y=f\left(-\frac{1}{2}\right)=4.\left(-\frac{1}{2}\right)^2-5=-4\)

\(b,\)

\(y=f\left(x\right)=4x^2-5\)

\(\Leftrightarrow4.x^2-5=-1\)

\(\Leftrightarrow4.x^2=4\)

\(\Leftrightarrow x^2=1\)

\(\Leftrightarrow x=1\)

 y=ƒ (3)=4.3²−5=31

y=ƒ (−1/2 )=4.(−1/2 )2−5=−4

b,

y=ƒ (x)=4x2−5

⇔4.x2−5=−1

⇔4.x²=4

⇔x²=1

⇔x=1

chúc bn học tốt 

f(1) = \(\frac{1^2-2}{3.1+1}=\frac{-1}{4}\)

f(-1) = \(\frac{\left(-1\right)^2-2}{3.\left(-1\right)+1}=\frac{-1}{-2}=\frac{1}{2}\)

f(0) = \(\frac{0^2-2}{3.0+1}=-2\)

f(-3) = \(\frac{\left(-3\right)^2-2}{3.\left(-3\right)+1}=\frac{-7}{8}\)

f(5) = \(\frac{5^2-2}{3.5+1}=\frac{23}{16}\)

a) Ta có: \(f\left(\frac{1}{3}\right)=\frac{1}{3}+\frac{1^2}{3^2}+\frac{1^3}{3^3}+....+\frac{1^{2016}}{3^{2016}}\)

\(\Rightarrow3.f\left(\frac{1}{3}\right)=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2015}}\)

\(\Rightarrow3.f\left(\frac{1}{3}\right)-f\left(\frac{1}{3}\right)=\left(1+\frac{1}{3}+...+\frac{1}{3^{2015}}\right)\)\(-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2016}}\right)\)

\(\Rightarrow2.f\left(\frac{1}{3}\right)=1-\frac{1}{3^{2016}}\)

\(\Rightarrow f\left(\frac{1}{3}\right)=\frac{1-\frac{1}{3^{2016}}}{2}\)

\(f\left(-2\right)+f\left(-4\right)=-2\left(3m-2\right)+\left(-4\right)\left(3m-2\right)=-6\left(3m-2\right)\)

\(6\cdot f\left(-x\right)=6\cdot\left(-1\right)\cdot\left(3m-2\right)=-6\left(3m-2\right)\)

Do đó: f(-2)+f(-4)=6f(-x)

f(-1) = \(\dfrac{3}{2}.-1=-\dfrac{3}{2}\)

f(-2) = \(\dfrac{3}{2}.2=3\)

f(-4) = \(\dfrac{3}{2}.-4=-6\)

a, Ta có: f(-2) = |-2 - 1| + 2 = |-3| + 2 = 5

              f(1/2) = |1/2 - 1| + 2 = |-0,5| + 2 =2,5

b, Ta có: f(x) = 3 =>|x - 1| + 2 = 3 => |x - 1| = 3 - 2 => |x - 1| = 1

=> x - 1 = 1 hoặc x - 1 = -1

=> x = 2 hoặc x = 0