b) Ta có :

\(\left(x-8\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x-8\right)^2+2019\ge2019;\forall x\)

Hay\(B\ge2019;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-8\right)^2=0\)

                         \(\Leftrightarrow x=8\)

Vậy MIN B=2019 \(\Leftrightarrow x=8\)

c) Vì \(\hept{\begin{cases}-|20-x|\le0;\forall x\\-|70+y|\le0;\forall y\end{cases}}\)

\(\Rightarrow-|20-x|-|70+y|\le0;\forall x,y\)

\(\Rightarrow90-|20-x|-|70+y|\le90-0;\forall x,y\)

Hay \(C\le90;\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}|20-x|=0\\|70+y|=0\end{cases}}\)

                       \(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)

Vậy MAX C=90 \(\Leftrightarrow\hept{\begin{cases}x=20\\y=-70\end{cases}}\)

A=|2-2|+|y+5|-10

 \(=\left|y+5\right|-10\ge-10\forall x\in R\)

Dấu "=" xảy ra <=> \(\left|y+5\right|=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\) 

Vậy A_min =-10 tại y=-5

2. \(P=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\) (BĐT Cauchy-Schwarz) 

\(=\dfrac{1}{2}\)

\(\Rightarrow P_{min}=\dfrac{1}{2}\) khi \(\dfrac{x}{y+z}=\dfrac{y}{z+x}=\dfrac{z}{x+y}\Rightarrow x=y=z=\dfrac{1}{3}\)

1, đặt \(x^2+x=t\)

=>\(A=t\left(t-4\right)=t^2-4t=t^2-4t+4-4\)

\(=>A=\left(t-2\right)^2-4\ge-4\) dấu"=' xảy ra\(t=2\)

\(=>x^2+x=2< =>x^2+x-2=0\)

\(< =>x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{9}{4}=0\)

\(< =>\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{3}{2}\right)^2=0< =>\left(x-1\right)\left(x+2\right)=0\)

\(=>\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\) vậy Amin=-4<=>\(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

B2

\(=>P=\dfrac{x^2}{y+z}+\dfrac{y+z}{4}+\dfrac{y^2}{x+z}+\dfrac{x+z}{4}+\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\)

\(-\left(\dfrac{y+z+x+z+x+y}{4}\right)\)

áp dụng BDT AM-GM

\(=>\dfrac{x^2}{y+z}+\dfrac{y+z}{4}\ge2\sqrt{\dfrac{x^2}{4}}=x^{ }\left(1\right)\)

\(\)tương tự \(=>\dfrac{y^2}{x+z}+\dfrac{x+z}{4}\ge y\left(2\right)\)

\(=>\dfrac{z^2}{x+y}+\dfrac{x+y}{4}\ge z\left(3\right)\)

(1)(2)(3) \(=>P\ge x+y+z-\dfrac{1}{2}.x+y+z=1-\dfrac{1}{2}=\dfrac{1}{2}\)

dấu"=" xảy ra<=>x=y=z=1/3

1)a Ta có: \(A=\left|x+19\right|+\left|y-5\right|+1890\)

\(\hept{\begin{cases}\left|x+19\right|\ge0\\\left|y-5\right|\ge0\end{cases}\Rightarrow\left|x+19\right|+\left|y-5\right|+1890\ge1890}\)

Vậy giá trị A nhỏ nhất = 1890 <=> x=-19; y= 5

2) a.   \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+...+\left(x+99\right)=2019\)

           \(\left(1+3+5+...+99\right)+\left(x+x+x+...+x\right)=2019\)

Rồi bn tính tổng của dãy số cách đều nha. Công thức: (Số cuối+ Số đầu). Số số hạng: 2 

3) Ta có: \(A^2=b\left(a-c\right)-c\left(a-b\right)\)

              \(A^2=ab-bc-ac+bc\)

             \(A^2=\left(-bc+bc\right)+\left(ab-ac\right)\)

            \(A^2=0+a\left(b-c\right)\)

           \(A^2=-20.\left(-5\right)=100\)

      \(\Rightarrow A=10\)

Chúc bạn năm mới vui vẻ nha! Happy new year !

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 : ...

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

\(A_{min}=-3\) khi \(x=\dfrac{1}{2}\)

Vì \(\left(\frac{1}{2}x-5\right)^{10}\ge0\)và \(\left(y^2-\frac{1}{4}\right)^{20}\ge0\)

nên \(\left(\frac{1}{2}x-5\right)^{10}+\left(y^2-\frac{1}{4}\right)^{20}=0\)

<=>\(\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)<=>\(\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

Ta có:\(\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}\ge0\forall x\\\left\{y^2-\frac{1}{4}\right\}^{20}\ge0\forall y\end{cases}}\)

Mà \(\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}\le0\)

\(\Rightarrow\left\{\frac{1}{2}x-5\right\}^{10}+\left\{y^2-\frac{1}{4}\right\}^{20}=0\)

\(\Leftrightarrow\hept{\begin{cases}\left\{\frac{1}{2}x-5\right\}^{10}=0\\\left\{y^2-\frac{1}{4}\right\}^{20}=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}}\)

Vậy \(x=10;y=\pm\frac{1}{2}\)

Xét  \(\left(\frac{1}{2}x-5\right)^{20}\ge0\)

         \(\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

\(\Rightarrow\)  \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}\ge0\)

mà  \(\left(\frac{1}{2}x-5\right)^{20}+\left(y^2-\frac{1}{4}\right)^{10}=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2}x-5=0\\y^2-\frac{1}{4}=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x=5\\y^2=\frac{1}{4}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=10\\y=\pm\frac{1}{2}\end{cases}}\)

x=10;y=1/2

x = 10

y = \(\frac{1}{2}\)

nha

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