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\(A=x^2-20x+101\)
\(=x^2-20x+100+1\)
\(=\left(x-10\right)^2+1\)
\(\Rightarrow A_{min}=1\Leftrightarrow\left(x-10\right)^2=0\)
\(\Rightarrow x-10=0\)
\(\Rightarrow x=10\)
a) \(x^2+6x-3\)
\(=x^2+6x+9-12\)
\(=\left(x+3\right)^2-12\ge-12\)
Vậy GTNN của bt là -12\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
a) \(A=25x^2-10x+9\)
\(A=\left(5x\right)^2-2\cdot5x\cdot1+1^2+9\)
\(A=\left(5x-1\right)^2+9\ge9\)
Dấu "=" xảy ra \(\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)
a) \(A=\left(x^2-10x+25\right)\)\(-28\)
\(A=\left(x-5\right)^2-28\)\(>=\)-28
MinA = -28 <=> x-5=0 <=> x=5
b)\(B=-\left(x^2+2x+1\right)+6\)
\(B=-\left(x+1\right)^2+6\)\(< =\)6
MaxB = 6 <=> x+1=0 <=> x=-1
c)\(C=-5\left(x^2-\frac{6}{5}x+\frac{9}{25}\right)-\frac{26}{5}\)
\(C=-5\left(x-\frac{3}{5}\right)^2-\frac{26}{5}\)\(< =-\frac{26}{5}\)
MaxC = \(-\frac{26}{5}\)<=> \(x-\frac{3}{5}=0\)<=> x=\(\frac{3}{5}\)
d)\(D=-3\left(x^2+\frac{1}{3}x+\frac{1}{36}\right)+\frac{61}{12}\)
\(D=-3\left(x+\frac{1}{6}\right)^2+\frac{61}{12}\)\(< =\frac{61}{12}\)
MacD = \(\frac{61}{12}\)<=> \(x+\frac{1}{6}=0\)<=> \(x=\frac{-1}{6}\)
Đúng thì nhớ tích cho minh nha
a) \( A=5x-x^2\)
\(\Leftrightarrow A=-x^2+2.x\dfrac{5}{2}-\left(\dfrac{5}{2}\right)^2+\left(\dfrac{5}{2}\right)^2\)
\(\Leftrightarrow A=-\left[x^2-2.x\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]+\left(\dfrac{5}{2}\right)^2\)
\(\Leftrightarrow A=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\)
Vậy GTLN của \(A=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
b) \(B=x-x^2\)
\(\Leftrightarrow B=-x^2+2.x\dfrac{1}{2}-\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow B=-\left[x^2-2.x\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]+\left(\dfrac{1}{2}\right)^2\)
\(\Leftrightarrow B=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\)
Vậy GTLN của \(B=\dfrac{1}{4}\) khi \(x=\dfrac{1}{2}\)
c) \(C=4x-x^2+3\)
\(\Leftrightarrow C=-x^2+4x-4+7\)
\(\Leftrightarrow C=-\left(x^2-2.x.2+2^2\right)+7\)
\(\Leftrightarrow C=-\left(x-2\right)^2+7\)
Vậy GTLN của \(C=7\) khi \(x=2\)