Bài 5: Hàm số

\(\left\{{}\begin{matrix}y=2x-7=3\Rightarrow2x=10\Rightarrow x=5\\y=2x-7=5\Rightarrow2x=12\Rightarrow x=6\\y=2x-7=-6\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\end{matrix}\right.\)

\(f\left(1\right)=-\dfrac{1}{2}\\ f\left(2\right)=-\dfrac{1}{2}\left(-5\right)=\dfrac{5}{2}\\ f\left(-\dfrac{3}{2}\right)=-\dfrac{1}{2}\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\)

\(\left\{{}\begin{matrix}f\left(1\right)=-\dfrac{1}{2}.1=-\dfrac{1}{2}\\f\left(-5\right)=-\dfrac{1}{2}.\left(-5\right)=\dfrac{5}{2}\\f\left(-\dfrac{3}{2}\right)=\left(-\dfrac{1}{2}\right).\left(-\dfrac{3}{2}\right)=\dfrac{3}{4}\end{matrix}\right.\)

f(1)=−12+1-12+1
=-1+1
=0
f(2)=−22+1-22+1
=-4+1
= -3
f(3/2)=−(32)2+1-(32)2+1
= -2,25+1
= -1,25
​f(-2)=−(−2)2+1-(-2)2+1
=-4 +1 
=-3

biểu thức đâu

thiếu đề

\(f\left(-2\right)=-4+a-3=5\Leftrightarrow a=12\)

Ta có hàm số $2x+a-3$ 

Có $f(-2)=5$

Có $:f(-2)=2.(-2)+a-3=5$

$\Rightarrow a=-4+a-3=5$

$\Rightarrow a=-4+a=8$

$\Rightarrow a=12$

\(f\left(0\right)=1\Leftrightarrow c=1\left(1\right)\\ f\left(1\right)=2\Leftrightarrow a+b+c=2\left(2\right)\\ f\left(2\right)=4\Leftrightarrow4a+2b+c=4\left(3\right)\\ \left(1\right)\left(2\right)\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}c=1\\a+b+1=2\\4a+2b+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\4a+2b=3\\c=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{2}\\c=1\end{matrix}\right.\)

Vậy \(y=f\left(x\right)=\dfrac{1}{2}x^2+\dfrac{1}{2}x+1\)

\(\left\{{}\begin{matrix}f\left(0\right)=c=1\\f\left(1\right)=a+b+c=2\\f\left(2\right)=4a+2b+c=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=1\\4a+2b=3\end{matrix}\right.\)\(\Rightarrow a=b=\dfrac{1}{2}\)

Vậy (a;b;c)=\(\left(\dfrac{1}{2};\dfrac{1}{2};1\right)\)

\(f\left(-x\right)=-x-\dfrac{1}{x}=-\left(x+\dfrac{1}{x}\right)=-f\left(x\right)\)