*\(A=\left(x-1\right)\left(x-3\right)+11\)
\(A=x^2-4x+4+11\)
\(A=\left(x-2\right)^2+11\ge11\)
\(\Rightarrow Min_A=11\Leftrightarrow x=2\)
*\(B=5-4x^2+4x\)
\(B=-\left(4x^2-4x-5\right)\)
\(B=-\left(4x^2-4x+1\right)+6\)
\(B=-\left(2x-1\right)^2-4\le6\)
\(\Rightarrow Max_B=6\Leftrightarrow x=\dfrac{1}{2}\)
Giải:
a) \(A=\left(x-1\right)\left(x-3\right)+11\)
\(\Leftrightarrow A=x^2-4x+3+11\)
\(\Leftrightarrow A=x^2-4x+4+10\)
\(\Leftrightarrow A=\left(x-2\right)^2+10\ge10\)
Dấu "=" xảy ra:
\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy ...
b) \(B=5-4x^2+4x\)
\(\Leftrightarrow B=6-4x^2+4x-1\)
\(\Leftrightarrow B=6-\left(4x^2-4x+1\right)\)
\(\Leftrightarrow B=6-\left(2x-1\right)^2\le6\)
Dấu "=" xảy ra:
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
Câu a còn cách khác ko bik đúng ko
A = \(x^2-4x+3+11\)
=> A = \(\left(x^2-2x.2+2^2\right)-2^2+3+11\)
<=> A = \(\left(x-2\right)^2+10\ge10\)
Vậy A min : 10 khi x-2 = 0 hay x = 2
Câu b tương tự 2 bạn kia
