\(f\left(0\right)=5=>c=5;f\left(2\right)=4.a+2.b+5=0;f\left(5\right)=25a+5b+5=0\Leftrightarrow5a+b+1=0\)

\(\hept{\begin{cases}4a+2b+5=0\\5a+b+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\10a+2b+2=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}4a+2b+5=0\\6a-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}b=-\frac{7}{2}\\a=\frac{1}{2}\end{cases}}\)

\(f\left(x\right)=\frac{1}{2}x^2-\frac{7}{2}x+5\)

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

Theo de ta co:

f(0) = a.0²+b.0+c = c =1

f(1)=a.1²+b.1+c = a+b+1 = 2  => a+b = 1

f(2)=a.2²+b.2+c = 4a+2b+1=2(2a+b)+1 = 4  => 2(2a+b) = 3  => 2a+b = 3/2 => b = 3/2 - 2a

Thay b=3/2 - 2a vao bieu thuc: a+b=1  ta duoc:

a+3/2-2a = 1

3/2-a= 1

=> a = 3/2 - 1 = 1/2

Suy ra: b = 3/2 - 2.1/2  = 1/2

Vay: a = 1/2   ;    b=1/2       ;      c=1

f(0) = 1

\(\Rightarrow\) a.0² + b.0 + c = 1 

\(\Rightarrow\) c = 1

Vậy hệ số a = 0; b = 0; c = 1

f(1) = 2

\(\Rightarrow\) a.1² + b.1 + c = 2

\(\Rightarrow\) a + b + c = 2

Vậy hệ số a = 1; b = 1; c = 1

f(2) = 4

\(\Rightarrow\) a.2² + b.2 + c = 4

\(\Rightarrow\) 4a + 2b + c = 4

Vậy hệ số a = 4; b = 2; c = 1

Chúc bn học tốt! (chắc vậy :D)

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}a\cdot0+b\cdot0+c=-2\\4a-2b+c=0\\a+b+c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-2\\4a-2b=2\\a+b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=-2\\a=1\\b=1\end{matrix}\right.\)

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 \(a=1;b=-6;c=5\)

câu b thay a;b;c vào tìm x là được