Có: 3x + y = 3 => y = 3x - 3

a) M = 3x² + y² = 3x² + ( 3x - 3)² = 3x² + 9x² - 18x + 9 = 3(4x² - 6x + 3) = 3(4x² - 6x +9/4) + 9/4 = 3(2x - 3/2)² + 9/4 \(\ge\)9/4

Vậy min M là 9/4

b) N = 2xy = 2x(3x - 3) = 6x² - 6x = 6(x² - x + 1/4 - 1/4) = 6(x - 1/2)² - 3/2 \(\le\)-3/2

Vậy max N là -3/2

cảm ơn

Lời giải:

a)

Áp dụng BĐT Cauchy-Schwarz:

\(4M=(3x^2+y^2)(3+1)\geq (3x+y)^2\)

\(\Leftrightarrow 4M\geq 1\Leftrightarrow M\geq \frac{1}{4}\)

Vậy \(M_{\min}=\frac{1}{4}\Leftrightarrow x=y=\frac{1}{4}\)

b) Với mọi \(x,y\in\mathbb{R}\Rightarrow (3x-y)^2\geq 0\)

\(\Leftrightarrow 9x^2+y^2-6xy\geq 0\Leftrightarrow (3x+y)^2-12xy\geq 0\)

\(\Leftrightarrow xy\leq \frac{(3x+y)^2}{12}=\frac{1}{12}\)

Vậy \(K_{\max}=\frac{1}{12}\Leftrightarrow x=\frac{1}{6};y=\frac{1}{2}\)

\(M=x^2+y^2+xy-3x-3y+2018\)

\(=x^2+2x\frac{\left(y-3\right)}{2}+\left(\frac{y-3}{2}\right)^2+y^2-3y+2018-\left(\frac{y-3}{2}\right)^2\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3y^2-6y+8063}{4}\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y^2-2y+1\right)}{4}+2015\)

\(=\left(x+\frac{y-3}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}+2015\ge2015\)

\("="\Leftrightarrow x=y=1\)

Cảm ơn bạn nhiều nha

Ta có: \(y=1-3x\)

a/ \(M=3x^2+y^2=3x^2+\left(1-3x\right)^2\)

\(\Leftrightarrow12x^2-6x+1=\left(12x^2-\frac{2.2.3x}{2}+\frac{3}{4}\right)+\frac{1}{4}\)

\(=\left(2\sqrt{3}x-\frac{\sqrt{3}}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

Vậy GTNN là 0,25 đạt được khi x = 0,25

b/ \(N=xy=x\left(1-3x\right)=-3x^2+x\)

\(=\left(-3x^2+\frac{2.\sqrt{3}x}{2\sqrt{3}}-\frac{1}{12}\right)+\frac{1}{12}\)

\(=\frac{1}{12}-\left(\sqrt{3}x-\frac{1}{2\sqrt{3}}\right)^2\le\frac{1}{12}\)

Vậy max là \(\frac{1}{12}\) đạt được khi \(x=\frac{1}{6}\)

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