\(\text{a)Để C đạt GTNN}\)

\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)

\(\Rightarrow C\ge-10\)

\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)

b)\(\text{Để D đạt GTLN}\)

=>(2x-3)²+5 đạt GTNN

Mà (2x-3)²\(\ge\)5

\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)

Ta có:

|x+7|\(\ge\)0

Dấu "=" xảy ra \(\Leftrightarrow\)|x+7|=0

\(\Leftrightarrow\)x+7=0

\(\Leftrightarrow\)x=-7

Thay x=-7 vào M ta được:

Min_D=\(\frac{-15.\left(-7\right)-68}{3.\left(-7\right)+12}\)

=\(\frac{105-68}{-21+12}\)

=\(\frac{37}{-9}\)

Vậy Min_D=\(\frac{37}{-9}\)\(\Leftrightarrow\)x=-7.

Bạn cho mình hỏi là chỉ a,b > 0 hay cả a,b,c > 0 vậy

Ta có :

\(C=-\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)

Ta có : | x + 4 | \(\ge\)0 ; ( y - 1.3 )¹⁰⁴ \(\ge\)0 

\(\Rightarrow\) | x + 4 |  + ( y - 1.3 )¹⁰⁴ \(\ge\)0​​​​

\(\Rightarrow\)| x + 4 |  + ( y - 1.3 )¹⁰⁴ ​+ 18 \(\ge\)18

Dấu " = " xảy ra khi \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)

\(\Rightarrow\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\le\frac{2}{18}=\frac{1}{9}\)

\(\Rightarrow\)GTLN của ​\(\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)là \(\frac{1}{9}\)

\(\Rightarrow\)\(-\frac{2}{\left|x+4\right|+\left(y-1.3\right)^{104}+18}\)có GTNN của ​\(\frac{1}{9}\)

Vậy C_min = \(\frac{1}{9}\)khi \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)

​​​

a.\(A=\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\)

Ta có: \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)

          \(\left|y-\frac{14}{3}\right|\ge0\forall x\)

    \(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|\ge0\forall x\)

   \(\Rightarrow\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\)

Dấu = xảy ra khi :

        \(\frac{x}{5}+\frac{23}{2}=0\Leftrightarrow\frac{x}{5}=-\frac{23}{2}\Leftrightarrow x=-\frac{115}{2}\)

         \(y-\frac{14}{3}=0\Leftrightarrow y=\frac{14}{3}\)

Vậy ..............

Ta có:

a) \(\left|\frac{x}{5}+\frac{23}{2}\right|\ge0\forall x\)

   \(\left|y-\frac{14}{3}\right|\ge0\forall y\)

=> \(\left|\frac{x}{5}+\frac{23}{2}\right|+\left|y-\frac{14}{3}\right|+2019\ge2019\forall x;y\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\frac{x}{5}+\frac{23}{2}=0\\y-\frac{14}{3}=0\end{cases}}\) <=> \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)

Vậy Min của A = 2019 tại \(\hept{\begin{cases}x=-\frac{115}{2}\\y=\frac{14}{3}\end{cases}}\)

câu b tượng tự 

\(b,B=\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{30}-11\)

Ta có : \(\left[x-\frac{5}{4}\right]^{20}\ge0\forall x\)

\(\left[y-\frac{4}{3}\right]^{30}\ge0\forall y\)

\(\Leftrightarrow\left[x-\frac{5}{4}\right]^{20}+\left[y-\frac{4}{3}\right]^{20}-11\ge-11\forall x,y\)

Dấu " = " xảy ra : \(\hept{\begin{cases}\left[x-\frac{5}{4}\right]^{20}=0\\\left[y-\frac{4}{3}\right]^{20}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-\frac{5}{4}=0\\y-\frac{4}{3}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=\frac{4}{3}\end{cases}}\)

Vậy : ...