Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(f\left(0\right)=\dfrac{b}{d}\Rightarrow f\left(f\left(0\right)\right)=0\Rightarrow f\left(\dfrac{b}{d}\right)=0\)
\(\Rightarrow\dfrac{\dfrac{ab}{d}+b}{\dfrac{cb}{d}+d}=0\Rightarrow b\left(a+d\right)=0\Rightarrow\left[{}\begin{matrix}b=0\\d=-a\end{matrix}\right.\)
TH1: \(b=0\)
\(f\left(1\right)=1\Rightarrow a=c+d\)
\(f\left(2\right)=2\Rightarrow2a=2\left(2c+d\right)\Rightarrow a=2c+d\)
\(\Rightarrow2c+d=c+d\Rightarrow c=0\) (ktm)
TH2: \(d=-a\)
\(f\left(1\right)=1\Rightarrow a+b=c+d=c-a\Rightarrow2a+b=c\) (1)
\(f\left(2\right)=2\Rightarrow2a+b=2\left(2c+d\right)=2\left(2c-a\right)\Rightarrow4a+b=4c\) (2)
Trừ (2) cho (1) \(\Rightarrow2a=3c\Rightarrow\dfrac{a}{c}=\dfrac{3}{2}\)
\(\Rightarrow\lim\limits_{x\rightarrow\infty}\dfrac{ax+b}{cx+d}=\dfrac{a}{c}=\dfrac{3}{2}\)
Hay \(y=\dfrac{3}{2}\) là tiệm cận ngang
a: \(f\left(1\right)=\dfrac{1-1}{1-2}=-1\)
\(f\left(-1\right)=\dfrac{-1-1}{-1-2}=-\dfrac{2}{-3}=\dfrac{2}{3}\)
\(f\left(0\right)=\dfrac{0-1}{0-2}=\dfrac{1}{2}\)
\(f\left(2\right)=\dfrac{2-1}{2-2}=\varnothing\)
b: f(x)=2 nên x-1=2x-4
=>2x-4=x-1
=>x=3
c: Để y là số ngyên thì \(x-2+1⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1\right\}\)
hay \(x\in\left\{3;1\right\}\)
\(y=f\left(x\right)=\dfrac{x-1}{x-2}\)
a)
\(y=f\left(1\right)=\dfrac{1-1}{1-2}=\dfrac{0}{-1}=0\)
\(y=f\left(-1\right)=\dfrac{\left(-1\right)-1}{\left(-1\right)-2}=\dfrac{-1-1}{-1-2}=\dfrac{-\left(1+1\right)}{-\left(1+2\right)}=\dfrac{-2}{-3}=\dfrac{2}{3}\)
\(y=f\left(0\right)=\dfrac{0-1}{0-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)
Đáp án D.