\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)

Vì \(\left|x+3\right|\)và \(\left(y-1\right)^{2018}\)\(\ge0\forall x;y\)

\(\Rightarrow A\ge4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy.....

\(C=4-\left|3x-5\right|-\left|5y+8\right|\)

\(C=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)

Lí luận như câu a) ta có :

\(C\le4\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}\)

Vậy,...........

\(A=\left|x+3\right|+\left(y-1\right)^{2018}-4\)

Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x;y\)

\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy \(A_{min}=-4\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

\(C=4-\left|3x-5\right|-\left|5y+8\right|\)

Ta có: \(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-\left|3x-5\right|\le0\forall x\\-\left|5y+8\right|\le0\forall y\end{cases}}\)

\(\Rightarrow4-\left|3x-5\right|-\left|5y+8\right|\le4\forall x;y\)

\(C=4\Leftrightarrow\hept{\begin{cases}-\left|3x-5\right|=0\\-\left|5y+8\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}}\)

Vậy \(C_{max}=4\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{8}{5}\end{cases}}\)

Tham khảo nhé~

Ta có: \(\hept{\begin{cases}\left|x+3\right|\ge0\forall x\\\left(y-1\right)^{2018}\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}\ge0\forall x,y\)

\(\Rightarrow\left|x+3\right|+\left(y-1\right)^{2018}-4\ge-4\forall x,y\)

\(\Rightarrow A\ge-4\)

\(A=-4\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-1\right)^{2018}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+3=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

Vậy MinA=-4\(\Leftrightarrow\)x=-3: y=1

Ta có: \(C=4-\left|3x-5\right|-\left|5y+8\right|\)

\(=4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\)

Vì\(\hept{\begin{cases}\left|3x-5\right|\ge0\forall x\\\left|5y+8\right|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left|3x-5\right|+\left|5y+8\right|\ge0\forall x,y\)

\(\Rightarrow-\left(\left|3x-5\right|+\left|5y+8\right|\right)\le0\forall x,y\)

\(\Rightarrow4-\left(\left|3x-5\right|+\left|5y+8\right|\right)\ge4\forall x,y\)

\(\Rightarrow C\ge4\)

\(C=4\Leftrightarrow\hept{\begin{cases}\left|3x-5\right|=0\\\left|5y+8\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}3x-5=0\\5y+8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-8}{5}\end{cases}}}\)

Vậy MaxC=4\(\Leftrightarrow\)x=\(\frac{5}{3}\): y=\(\frac{-8}{5}\)

a: A=-(x-7)^2-888<=-888

Dấu = xảy ra khi x=7

b: \(B=\left|2x-1\right|+\left|y-5\right|+\dfrac{8}{3}>=\dfrac{8}{3}\)

Dấu = xảy ra khi x=1/2 và y=5

c: \(C=\left(x+3\right)^2+\left|2y-5\right|-232>=-232\)

Dấu = xảy ra khi x=-3 và y=5/2

a) Vì \(\sqrt{x-5}\) ≥0

⇒ \(\sqrt{x-5}+7\) ≥ 7

Min A=7⇔x-5=0

             ⇔x=5

b) Vì \(\sqrt{3x-5}\) ≥0

⇒ 8-\(\sqrt{3x-5}\) ≤8

Max=8⇔3x-5\(=\)0

           ⇔\(x=\dfrac{5}{3}\)

\(C=-x^2-3x+4\)

\(\Rightarrow C=-\left(x^2+3x\right)+4\)

\(\Rightarrow C=-\left(x^2+3x+\dfrac{9}{4}-\dfrac{9}{4}\right)+4\)

\(\Rightarrow C=-\left(x^2+3x+\dfrac{9}{4}\right)+4+\dfrac{9}{4}\)

\(\Rightarrow C=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\left(-\left(x+\dfrac{3}{2}\right)^2\le0,\forall x\right)\)

\(\Rightarrow Max\left(C\right)=\dfrac{25}{4}\left(tạix=-\dfrac{3}{2}\right)\)

MAX_C = 25/4 khi x =-3/2

a) Để A lớn nhất thì \(\frac{15}{4.\left|3x+7\right|+3}\) lớn nhất hay 4.|3x + 7| + 3 nhỏ nhất

Có: \(4.\left|3x+7\right|+3\ge3\forall x\)

Dấu "=" xảy ra khi |3x + 7| = 0

=> 3x + 7 = 0

=> 3x = -7

\(\Rightarrow x=\frac{-7}{3}\)

Với x = \(\frac{-7}{3}\) thay vào đề bài ta được A = 10

Vậy \(A_{Max}=10\) khi x = \(\frac{-7}{3}\)

b) Để B lớn nhất thì \(\frac{21}{8.\left|15x-21\right|+7}\) lớn nhất hay 8.|15x - 21| + 7 nhỏ nhất

Có: \(8.\left|15x-21\right|+7\ge7\forall x\)

Dấu "=" xảy ra khi |15x - 21| = 0

=> 15x - 21 = 0

=> 15x = 21

\(\Rightarrow x=\frac{21}{15}=\frac{7}{5}\)

Với \(x=\frac{7}{5}\) thay vảo đề bài ta tìm được B = \(\frac{8}{3}\)

Vậy \(B_{Max}=\frac{8}{3}\) khi x = \(\frac{7}{5}\)

c) Có: \(\begin{cases}\left|x+1\right|\ge x+1\\\left|3x-4\right|\ge4-3x\\\left|2x-1\right|\ge2x-1\end{cases}\)\(\forall x\)

\(\Rightarrow C\ge\left(x+1\right)+\left(4-3x\right)+\left(2x-1\right)+5\)

hay \(C\ge9\)

Dấu "=" xảy ra khi \(\begin{cases}x+1\ge0\\3x-4\le0\\2x-1\ge0\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\3x\le4\\2x\ge1\end{cases}\)\(\Rightarrow\begin{cases}x\ge-1\\x\le\frac{3}{4}\\x\ge\frac{1}{2}\end{cases}\)\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{4}\)

Vậy \(C_{Max}=9\) khi \(\frac{1}{2}\le x\le\frac{3}{4}\)