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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: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)
\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)
\(=\dfrac{\left(x-2y\right)^3}{\left(x-2y\right)^2}\)
=x-2y
c: \(\dfrac{x^3+y^3}{x+y}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)
\(=x^2-xy+y^2\)
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
a: \(N=\left(5x\right)^3-\left(2y\right)^3=1^3-1^3=0\)
b: \(Q=x^3+27y^3=\dfrac{1}{8}+\dfrac{27}{8}=\dfrac{28}{8}=\dfrac{7}{2}\)
\(x^2+4y^2-4\left(x-y\right)+2=x^2-4x+4+4y^2+4y+1-3\)
\(=\left(x-2\right)^2+\left(2y+1\right)^2-3\ge-3\)
Dấu ''='' xảy ra khi \(x=2;y=-\frac{1}{2}\)
Vậy GTNN biểu thức trên là -3 khi \(x=2;y=-\frac{1}{2}\)
Ta có
\(x^2+4y^2-4\left(x-y\right)+2=\left(x^2-4x+4\right)+\left(4y^2+4y+1\right)-3=\left(x-2\right)^2+\left(2y+1\right)^2-3\)
Vì \(\hept{\begin{cases}\left(x-2\right)^2\ge0\\\left(2y+1^2\right)\ge0\end{cases}}\)
\(\Rightarrow\left(x-2\right)^2+\left(2y+1\right)^2-3\ge-3\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{1}{2}\end{cases}}\)