\(=\frac{4y^2+2y}{4}=\frac{\left(2y\right)^2+2.2y.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}}{4}\)

\(=\frac{\left(2y+\frac{1}{2}\right)^2-\frac{1}{4}}{4}\ge\frac{0-\frac{1}{4}}{4}=\frac{-1}{16}\)

Dấu "=" xảy ra khi \(2y=\frac{-1}{2}\Leftrightarrow y=\frac{-1}{4}\)

Vậy GTNN là -1/16 khi y=1/4

\(xy+x+y=2\)

\(\Leftrightarrow x\left(y+1\right)+y+1=3\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=3\)

Ta có bảng giá trị: 

Chọn D

• puvi9176
• 16/01/2021

mx−2+m=3xmx−2+m=3x

a) Phương trình nhận x=12x=12 làm nghiệm

→m⋅12−2+m=3⋅12→m⋅12−2+m=3⋅12

→32m=72→32m=72

→m=73→m=73

b) mx−2+m=3xmx−2+m=3x

→(m−3)x=2−m→(m−3)x=2−m

Phương trình có nghiệm duy nhất

→m−3≠0→m−3≠0

→m≠3→m≠3

Khi đó:

THAM KHẢO

Answer:

Tìm được giá trị nhỏ nhất thôi chứ nhỉ?

\(x^2y^2+x^2-2xy-4x+11\)

\(=\left(x^2y^2-2xy+1\right)+\left(x^2-4x+4\right)+6\)

\(=[\left(xy\right)^2-2xy.1+1^2]+\left(x^2-2x.2+2^2\right)+6\)

\(=\left(xy-1\right)^2+\left(x-2\right)^2+6\)

Mà \(\left(xy-1\right)^2\ge0;\left(x-2\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(xy-1\right)^2+\left(x-2\right)^2+6\ge6\forall x;y\)

Dấu "=" xảy ra khi: \(xy=1;x=2\Rightarrow y=\frac{1}{2}\)

\(x-2\sqrt{xy}+2y-2\sqrt{y}+3\)

\(=\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{y}+1\right)+2\)

\(=\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-1\right)^2+2\)\(\ge2\)

dấu "=" xảy ra khi x=y=1

cảm ơn

\(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{12}+\frac{x+4}{13}\)

\(\Leftrightarrow\frac{x+1}{15}+\frac{x+2}{14}-\frac{x+3}{12}-\frac{x+4}{13}=0\)

\(\Leftrightarrow\frac{x+1}{15}+1+\frac{x+2}{14}+1-\frac{x+3}{12}+1-\frac{x+4}{13}+1=0\)

\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+13}{12}-\frac{x+16}{13}=0\)

\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{13}\right)=0\)

\(\Leftrightarrow x=-16\) (vì \(\frac{1}{15}+\frac{1}{14}-\frac{1}{12}-\frac{1}{13}>0\))

Vậy: \(S=\left\{-16\right\}\)