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1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)
\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)
2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)
\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)
4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)
\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)
\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)
a) \(x^2y+2xy+y=y\left(x^2+2x+1\right)=y\left(x+1\right)^2\)
b) \(4x^2-4xy-6y^2+6xy=4x\left(x-y\right)+6y\left(x-y\right)=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
c) \(18x^5y+18x^3y-2x^3y^5-2xy^5=18x^3y\left(x^2+1\right)-2xy^5\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(18x^3y-2xy^5\right)=2xy\left(x^2+1\right)\left(9x^2-y^4\right)=2xy\left(x^2+1\right)\left(3x-y^2\right)\left(3x+y^2\right)\)
d)
d) \(-12x^5-12x^3y-3xy^2+36x^4+36x^2y+9y^2=-3x\left(4x^4+4x^2y+y^2\right)+9y\left(4x^4+4x^2y+y^2\right)\)\(=\left(4x^4+4x^2y+y^2\right)\left(9-3x\right)\)
\(\left(12x-5\right)\left(3x-1\right)-36x^2=6\)
\(36x-12x-15x-5-36x^2=6\)
\(-12x-15x-5=6\)
\(-27x=5+6\)
\(x=11:\left(-27\right)\)
\(x=-\frac{11}{27}\)
chuk bn hok giỏi
\(\left(36x^2+12x+1\right)+\left(36x^2-12x+1\right)-2\left(1+6x\right)\left(1-6x\right)\)
\(=36x^2+12x+1+36x^2-12x+1-2\left(1-36x^2\right)\)
\(=72x^2+2-2+72x^2=144x^2\)
Chúc bạn học tốt!!!
36x2+5y2+12x-1
=36x2+12x+1+5y2-2
=(6x+1)2+5y2-2\(\ge\)-2
Dấu "=" xảy ra khi :
6x+1=0 và y=0
x =\(\frac{-1}{6}\) và y=0
Vậy GTNN của 36x2+5y2+12x-1 là -2 tại x=\(\frac{-1}{6}\)và y=0