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Ta có hàm số sau :
\(f\left(1\right)=3.1^2-1=2\)
\(f\left(\frac{-2}{3}\right)=3.\frac{-2}{3}-1=-2-1=-3\)
Vậy hàm số f(1) = 2
Hàm số :\(f\left(\frac{-2}{3}\right)=-3\)
Ta có y = f(x) = 3x2 + 1. Do đó
f(\(\dfrac{1}{2}\)) = 3.\(\left(\dfrac{1}{2}\right)^2\) + 1 = \(\dfrac{3}{4}\)+ 1 = \(\dfrac{7}{4}\)
f(1) = 3.12 + 1 = 3.1 + 1 = 3 + 1 = 4
f(3) = 3.32 + 1 = 3.9 + 1 = 27 + 1 = 28.
\(f\left(x\right)=2x+\dfrac{1}{2}\)
a) \(f\left(0\right)=2.0+\dfrac{1}{2}=0+\dfrac{1}{2}=\dfrac{1}{2}\)
b) \(f\left(\dfrac{1}{2}\right)=2.\dfrac{1}{2}+\dfrac{1}{2}=1+\dfrac{1}{2}=\dfrac{3}{2}\)
c) \(f\left(-2\right)=2.\left(-2\right)+\dfrac{1}{2}=-4+\dfrac{1}{2}=\dfrac{-7}{2}\)
f(x)=2x+12f(x)=2x+12
a) f(0)=2.0+12=0+12=12f(0)=2.0+12=0+12=12
b) f(12)=2.12+12=1+12=32f(12)=2.12+12=1+12=32
c) f(−2)=2.(−2)+12=−4+12=−72
a) \(y=f\left(x\right)=1-5x\)
\(y=f\left(1\right)=1-5.1=1-5=-4\)
\(y=f\left(-2\right)=1-5.\left(-2\right)=1-\left(-10\right)=1+10=11\)
\(y=f\left(\dfrac{1}{5}\right)=1-5.\dfrac{1}{5}=1-1=0\)
\(y=f\left(\dfrac{-3}{5}\right)=1-5.\left(\dfrac{-3}{5}\right)=1-\left(-3\right)=1+3=4\)
b) \(y=f\left(x\right)=1-5x=-4\)
\(\Rightarrow5x=1-\left(-4\right)\)
\(\Rightarrow5x=1+4\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=\dfrac{5}{5}=1\)
Vậy \(f\left(x\right)=-4\) thì \(x=1\)
a: \(f\left(-1\right)=3-7=-4\)
\(f\left(\dfrac{1}{5}\right)=\dfrac{3}{25}-7=\dfrac{-172}{25}\)
b: f(x)=-20/3
\(\Leftrightarrow3x^2-7=-\dfrac{20}{3}\)
\(\Leftrightarrow3x^2=\dfrac{1}{3}\)
\(\Leftrightarrow x^2=\dfrac{1}{9}\)
=>x=1/3 hoặc x=-1/3
a) \(y=f\left(x\right)=\dfrac{6}{x}\)
*) \(f\left(1\right)=\dfrac{6}{1}=6\Rightarrow y=f\left(1\right)=6\)
*) \(f\left(1.5\right)=\dfrac{6}{1.5}=1,2\Rightarrow y=f\left(1.5\right)=1,2\)
*) \(f\left(3\right)=\dfrac{6}{3}=2\Rightarrow y=f\left(3\right)=2\)
*) \(f\left(-\dfrac{2}{3}\right)=\dfrac{6}{-\dfrac{2}{3}}=-9\Rightarrow y=f\left(-\dfrac{2}{3}\right)=-9\)
b) \(x:y=3\)
Tại \(y=-2\)
\(\Rightarrow x:\left(-2\right)=3\)
\(\Rightarrow x=3.\left(-2\right)\)
\(\Rightarrow x=-6\)
Vậy \(x=-6\)
- Xin lỗi ☹ làm lại cậu b cho ~ tại đề bài không rõ
b) \(y=f\left(x\right)=\dfrac{6}{x}\)
*)Tại y=3 \(\Rightarrow3=\dfrac{6}{x}\) \(\Rightarrow x=2\)
Vậy tại y = 3 thì x = 2
*) Tại y = -2 \(\Rightarrow-2=\dfrac{6}{x}\Rightarrow x=-3\)
Vậy tại y = -2 thì x = -3
Phần này khó chú ý nè bạn
Giải
Ta có f(x1+x2) = f(x1) + f(x2)
nên f(7) = f(3)+f(4)= f(2)+f(1) + f(2)+f(2) = f(1)+f(1)+f(1)+f(1)+f(1)+f(1)+f(1)=7
\(f\left(\dfrac{1}{7}\right)=\dfrac{1}{49}.f\left(7\right)=\dfrac{1}{49}.7=\dfrac{1}{7}\)
Ta có :\(f\left(\dfrac{5}{7}\right)=f\left(\dfrac{2}{7}\right)+f\left(\dfrac{3}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{2}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)=\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{5}{7}\)
Lời giải:
Ta có:
\(f(x)=x^2+x\Rightarrow \frac{1}{f(x)}=\frac{1}{x^2+x}=\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}\)
Do đó:
\(\frac{1}{f(1)}=1-\frac{1}{2}\)
\(\frac{1}{f(2)}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{f(3)}=\frac{1}{3}-\frac{1}{4}\)
......
\(\frac{1}{f(2014)}=\frac{1}{2014}-\frac{1}{2015}\)
\(\frac{1}{f(2015)}=\frac{1}{2015}-\frac{1}{2016}\)
Cộng theo vế:
\(\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+...+\frac{1}{f(2014)}+\frac{1}{f(2015)}=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
\(f\left(1\right)=-\dfrac{3}{2}.1=-\dfrac{3}{2}\)
\(f\left(-1\right)=-\dfrac{3}{2}.\left(-1\right)=\dfrac{3}{2}\)
\(f\left(2\right)=-\dfrac{3}{2}.2=-3\)
\(f\left(-2\right)=-\dfrac{3}{2}.\left(-2\right)=3\)
\(f\left(\dfrac{1}{2}\right)=-\dfrac{3}{2}.\dfrac{1}{2}=\dfrac{-3}{4}\)
\(f\left(-\dfrac{1}{2}\right)=-\dfrac{3}{2}.\left(-\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(f\left(a\right)< f\left(-a\right)\)