Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(f\left(0\right)=5=>c=5;f\left(2\right)=4.a+2.b+5=0;f\left(5\right)=25a+5b+5=0\Leftrightarrow5a+b+1=0\)
\(\hept{\begin{cases}4a+2b+5=0\\5a+b+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\10a+2b+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\6a-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=-\frac{7}{2}\\a=\frac{1}{2}\end{cases}}\)
\(f\left(x\right)=\frac{1}{2}x^2-\frac{7}{2}x+5\)
b)
\(f\left(-1\right)=\frac{1}{2}+\frac{7}{2}+5=9=>P\left(-1;3\right)kothuocHS\)
\(f\left(\frac{1}{2}\right)=\frac{1}{2}.\frac{1}{4}-\frac{7}{2}.\frac{1}{2}+5=\frac{\left(1-14+5.8\right)}{8}=\frac{27}{8}=>Qkothuoc\)
c)
\(\frac{1}{2}x^2-\frac{7}{2}x+5=-3\Rightarrow\frac{1}{2}x^2-\frac{7}{2}x+8=0\)
\(x^2-7x+16=0\Leftrightarrow\left(x^2-2.\frac{7}{2}x+\frac{49}{4}\right)+\frac{15}{4}\)vo nghiem
+f(0) = a.0+b.0 +c =5 => c =5
+f(1)= a.1 +b.1+ 5 = 0 => a+b =-5 (1)
+ f(5) =a.52 +b.5 +5 =0 => 5a +b =-1 (2)
(10(2) => 4a +(a+b) =-1 => 4a -5 =-1 => 4a =4 => a =1
=> b =-5-a = -5 -1 = -6
Vậy a =1; b =-6 ; c =5
f(x)=ax2 + bx+ c
f(0)=1, f(1)=2, f(2)=2
=>c=1;a+b+c=2;4a+2b+c=2
=>a+b=1;4a+2b=1
=>a+b=4a+2b
=>4a+2b-a-b=0
=>3a-b=0
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 \(a=1;b=-6;c=5\)