Ta có :
\(B\left(x\right)=ax^2+bx+c\)
+) \(B\left(1\right)=3\Leftrightarrow a.1^2+b.1+c=3\)
\(\Leftrightarrow a+b+c=3\)\(\left(1\right)\)
+) \(B\left(-1\right)=1\Leftrightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=1\)
\(\Leftrightarrow a-b+c=1\)\(\left(2\right)\)
+) \(B\left(2\right)=7\Leftrightarrow a.2^2+b.2+c=7\)
\(\Leftrightarrow4a+2b+c=7\)
\(\Leftrightarrow3a+a+2b+c=7\) \(\left(5\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\left(a+b+c\right)+\left(a-b+c\right)=3+1\)
\(\Leftrightarrow2a+2c=4\)
\(\Leftrightarrow2\left(a+c\right)=4\)
\(\Leftrightarrow a+c=2\)
Từ \(\left(3\right)+\left(4\right)\Leftrightarrow\left(a+b+c\right)-\left(a-b+c\right)=3-1\)
\(\Leftrightarrow2b=2\)
\(\Leftrightarrow b=1\)
Từ \(\left(5\right)\Leftrightarrow3a+2.1+\left(a+c\right)=7\)
\(\Leftrightarrow3a+2+2=7\)
\(\Leftrightarrow3a=3\)
\(\Leftrightarrow a=1\)
Lại có :
\(a+c=2\)
\(\Leftrightarrow c=1\)
Vậy \(a=b=c=1\)
Ta có:
\(B\left(1\right)=3\Rightarrow a.1^2+b.1+c=3\Rightarrow a+b+c=3\) (1)
\(B\left(-1\right)=1\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=1\Rightarrow a-b+c=1\) (2)
\(B\left(2\right)=7\Rightarrow a.2^2+b.2+c=7\Rightarrow4a+2b+c=7\) (3)
Từ \(\left(1\right),\left(2\right)\Rightarrow a+b+c+a-b+c=3+1=4\)
\(\Rightarrow2a+2b=4\Rightarrow2\left(a+b\right)=4\Rightarrow a+b=4\)
Mà a + b + c = 3 \(\Rightarrow b=1\)
Từ (2) và (3) \(\Rightarrow4a+2b+c-\left(a-b+c\right)=7-1=6\)
\(\Rightarrow3a+3b=6\Rightarrow3\left(a+b\right)=6\Rightarrow a+b=2\)
mà \(b=1\Rightarrow a=1\)
Lại có a + b + c = 3 \(\Rightarrow c=1\) ( vì a = b = 1 )
Vậy a = b = c = 1
B(x) = ax2 + bx + c
Ta có :
B(1) = a + b + c = 3
B(-1) = a - b + c = 1
B(2) = 4a + 2b + c = 7
=> B(1) - B(-1) = a + b + c - a + b - c = 2b = 3-1 = 2
=> b = 1
=> 4a + c = 5
Mà a+c=2
=> 3a=3
Hay a = 1
=> c = 1
Vậy a=b=c=1