a) Ta có: \(f\left(0\right)=5\Rightarrow a.0^2+b.0+c=5\)
\(\Rightarrow c=5\)
\(f\left(1\right)=0\Rightarrow a.1^2+b.1+c=0\)
\(\Rightarrow a+b+c=0\left(1\right)\)
Thay \(c=5\) vào (1) được:
\(a+b+5=0\Rightarrow a+b=-5\left(2\right)\)
\(f\left(5\right)=0\Rightarrow a.5^2+5b+c=0\)
\(\Rightarrow25a+5b+c=0\)
\(\Rightarrow5\left(5a+b+1\right)=0\)
\(\Rightarrow5a+b+1=0\)
\(\Rightarrow5a+b=-1\)
\(\Rightarrow b=-1-5a\left(3\right)\)
Thay \(\left(3\right)\rightarrow\left(2\right):a+\left(-1-5a\right)=-5\)
\(\Rightarrow a-1-5a=-5\)
\(\Rightarrow-1-4a=-5\)
\(\Rightarrow4a=4\)
\(\Rightarrow a=1\)
Khi đó: \(1+b=-5\Rightarrow b=-6\)
Vậy \(\left\{{}\begin{matrix}a=1\\b=-6\\c=5\end{matrix}\right.\).
b) Kết hợp \(y=-3\) với câu a) ta có:
\(x^2-6x+5=-3\)
\(\Rightarrow x^2-3x-3x+5=-3\)
\(\Rightarrow x^2-3x-3x+ 9-4=-3\)
\(\Rightarrow x\left(x-3\right)-3\left(x-3\right)-4=-3\)
\(\Rightarrow\left(x-3\right)^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\).
a) thay f(0) = 5 vào hàm số ta có : \(5=a0^2+b0+c\) \(\Leftrightarrow\) \(c=5\)
thay f(1) = 0 và f(5) = 0 vào hàm số ta có hệ phương trình
\(\left\{{}\begin{matrix}a+b+5=0\\25a+5b+5=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=-5\\25a+5b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}5a+5b=-25\\25a+5b=-5\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}20a=20\\a+b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\1+b=-5\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=1\\b=-6\end{matrix}\right.\)
vậy \(a=1;b=-6;c=5\)
b) để y=-3
=> x^2- 6x+5=-3
=> -2x^2-6x+5+3=0
x^2-6x+8=0
x^2-2x-4x+4.2=0
(x^2-2x)-(4x-4.2)=0
x(x-2)-4(x-2)=0
(x-2).(x-4)=0
=>x-2=0 h x-4=0
x=2 x=4
