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\(A=\left(x+3\right)\left(x-4\right)+7=x^2-x-5=\left(x^2-x+\frac{1}{4}\right)-\frac{1}{4}-5\)
\(=\left(x-\frac{1}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)
"=" <=> x = 1/2
\(B=3-\left(x-1\right)\left(x-2\right)=3-\left(x^2-3x+2\right)\)
\(=3-\left(x-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}+2\right)\)
\(=3+\frac{1}{4}-\left(x-\frac{3}{2}\right)^2\le\frac{13}{4}\)
Xảy ra khi x = 3/2
\(B=3x^2-6xy+5y^2-y+3x+2016\)
\(3B=9x^2-18xy+15y^2-3y+9x+6048\)
\(3B=\left(9x^2-18xy+9y^2\right)+3\left(3x-3y\right)+\dfrac{9}{4}+\left(6y^2+6y+\dfrac{3}{2}\right)+6044,25\)
\(3B=\left(3x-3y\right)^2+3\left(3x-3y\right)+\dfrac{9}{4}+6\left(y^2+y+\dfrac{1}{4}\right)+6044,25\)
\(3B=\left(3x-3y+\dfrac{3}{2}\right)^2+6\left(y+\dfrac{1}{2}\right)^2+6044,25\ge6044,25\)
\(\Rightarrow B\ge2014,75\Leftrightarrow y=-\dfrac{1}{2};x=-1\)
Vậy MINB=2014,75<=>x=-1;y=-1/2
bài 1 : a. x^3 +27 -54-x^3 =-27
b. 8x^3 +y^3 -8x^3 +y^3 =2y^3
c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3
d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3
A=x2-4x+7
= x2-4x+4+3
= (x-2)2+3
Vì (x+2)2>/ 0
Nên (x-2)2+3>/3
Vậy MAX của A=3 khi x-2=0 => x=2
\(A=3\left(x+\frac{4}{3}\right)^2+\frac{146}{3}\ge\frac{146}{3}\)
\(A_{min}=\frac{146}{3}\) khi \(x=-\frac{4}{3}\)
\(B=-\left(x-2\right)^2-5\le-5\)
\(B_{max}=-5\) khi \(x=2\)