3x+y=1

y^2=1-6x+9x^2

a) M=12(x^2-2.1/4x+1/16)+1-12/16

GTNN=1-3/4=1/4 khi x=1/4=>y=1/4

b) N=xy=x(1-3x)=-3x^2+x=-3(x^2-2.1/6x+1/36)+3/36

GTLN =1/12 khi x=1/6 ;y=1/2

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

1) \(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)\(\Leftrightarrow\)\(x+y\ge8\)

\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\)\(\Leftrightarrow\)\(xy=2\left(x+y\right)\ge16\)

\(A=\sqrt{x}+\sqrt{y}\ge2\sqrt[4]{xy}\ge2\sqrt[4]{16}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=4\)

2) \(B=\sqrt{3x-5}+\sqrt{7-3x}\ge\sqrt{3x-5+7-3x}=\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{7}{3}\end{cases}}\)

\(B=\sqrt{3x-5}+\sqrt{7-3x}\le\frac{3x-5+1+7-3x+1}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=2\)

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