a)A=4(x+11/8)^2 -153/16
Min A=-153/16 khi x=-11/8
b)B=3(x-1/3)^2 -4/3
Min B=-4/3 khi x=1/3
Bài 1:
a) \(A=4x^2+11x-2=\left(4x^2+11x+\dfrac{121}{16}\right)-\dfrac{153}{16}=\left(2x+\dfrac{11}{4}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\)
\(minA=-\dfrac{153}{16}\Leftrightarrow x=-\dfrac{11}{8}\)
b) \(B=3x^2-2x-1=3\left(x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)-\dfrac{4}{3}=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minB=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{1}{3}\)
Bài 2:
a) \(A=-x^2+3x-1=-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{5}{4}=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)
\(maxA=\dfrac{5}{4}\Leftrightarrow x=\dfrac{3}{2}\)
b) \(B=-x^2-4x+7=-\left(x^2+4x+4\right)+11=-\left(x+2\right)^2+11\le11\)
\(maxB=11\Leftrightarrow x=-2\)
Bài 1:
a: Ta có: \(A=4x^2+11x-2\)
\(=4\left(x^2+\dfrac{11}{4}x-\dfrac{1}{2}\right)\)
\(=4\left(x^2+2\cdot x\cdot\dfrac{11}{8}+\dfrac{121}{64}-\dfrac{153}{64}\right)\)
\(=4\left(x+\dfrac{11}{8}\right)^2-\dfrac{153}{16}\ge-\dfrac{153}{16}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{11}{8}\)
b: Ta có: \(B=3x^2-2x-1\)
\(=3\left(x^2-\dfrac{2}{3}x-\dfrac{1}{3}\right)\)
\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{4}{9}\right)\)
\(=3\left(x-\dfrac{1}{3}\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)
Bài 2:
a: Ta có: \(A=-x^2+3x-1\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{5}{4}\right)\)
\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
b: ta có: \(B=-x^2-4x+7\)
\(=-\left(x^2+4x-7\right)\)
\(=-\left(x^2+4x+4-11\right)\)
\(=-\left(x+2\right)^2+11\le11\forall x\)
Dấu '=' xảy ra khi x=-2