Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(A=x^2-6x+11\)
\(=x^2-2.3.x+9+2\)
\(=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)
Vậy \(MinA=3\Leftrightarrow x=3\)
b, \(B=2x^2+10x-1\)
\(=2\left(x^2+5x\right)-1\)
\(=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{21}{4}\)
\(=2\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(MinB=-\frac{21}{4}\Leftrightarrow x=-\frac{5}{2}\)
c, \(C=5x-x^2\)
\(=-x^2+5x\)
\(=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{25}{4}\)
\(=-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\)
Ta có: \(-\left(x+\frac{5}{2}\right)^2\le0\Leftrightarrow-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)
Vậy \(MaxB=\frac{25}{4}\Leftrightarrow x=-\frac{5}{2}\)
a)72 -272 =(7-27)(7+27)
=-20.34
=-680
b)372 -132=(37-13)(37+13)=24.50
=1200
c)20022-22=(2002-2)(2002+2)
=2000.2004
=4008000
A= (x^2 - 2.x.1/2 + 1/4) -1/4
=(x-1/2)^2 -1/4 >= -1/4
Dấu"=" xảy ra <=> x-1/2 = 0 <=>x=1/2
Vậy Min A= -1/4 <=> x=1/2
\(A=-x^2+6x-15\)
\(A=-x^2+2.3x-9-6\)
\(\Rightarrow-A=x^2-2.3x+9+6\)
\(-A=\left(x^2-2.3.x+3^2\right)+6\)
\(-A=\left(x-3\right)^2+6\)
\(\Rightarrow A=-\left(x-3\right)^2-6\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Rightarrow-\left(x-3\right)^2-6\le-6\forall x\)
\(A=-6\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy Amax =-6\(\Leftrightarrow\)x=3
\(B=-2x^2+8x-15\)
\(-2B=4x^2-16x+30\)
\(-2B=\left[\left(2x\right)^2-2.2x.4+4^2\right]+14\)
\(-2B=\left(2x-4\right)^2+14\)
\(\Rightarrow B=-\frac{\left(2x-4\right)^2}{2}-7\)
Ta có: \(-\frac{\left(2x-4\right)^2}{2}\le0\forall x\)
Đến đây b làm tương tự như trên nhé.
Chúc b học tốt
a) \(A=-x^2+6x-15\)
\(-A=x^2-6x+15\)
\(-A=\left(x^2-6x+9\right)+6\)
\(-A=\left(x-3\right)^2+6\)
Mà \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow-A\ge6\)
\(\Leftrightarrow A\le-6\)
Dấu "=" xảy ra khi :
\(x-3=0\Leftrightarrow x=3\)
Vậy \(A_{Max}=-6\Leftrightarrow x=3\)