Tìm GTLN của biểu thức A=3-x^2

Các câu kia thì dễ rồi làm 2 câu cuối thôi!

Bài 2:

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

\(C=\left(x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{11}{4}\right)^2+\left(y^2+\dfrac{1}{2}y+\dfrac{1}{2}y+\dfrac{1}{4}-\dfrac{9}{4}\right)^2\)

\(C=\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]^2+\left[\left(y+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]^2\)

Với mọi giá trị của \(x;y\in R\) thì:

\(\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\\\left(y+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]^2\ge\dfrac{121}{16}\\\left[\left(y+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]\ge\dfrac{81}{16}\end{matrix}\right.\)

\(\Rightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\right]^2+\left[\left(y+\dfrac{1}{2}\right)^2-\dfrac{9}{4}\right]^2\ge\dfrac{101}{8}\)

Hay \(C\ge\dfrac{101}{8}\) với mọi giá trị của \(x;y\in R\)

Để \(C=\dfrac{101}{8}\) thì \(\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=\dfrac{11}{4}\\\left(y+\dfrac{1}{2}\right)^2-\dfrac{9}{4}=-\dfrac{9}{4}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy......................

\(D=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+2017\)

\(D=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+2017\)

\(D=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)+2017\)

\(D=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2017\)

\(D=\left(x^2+5x\right)^2-6^2+2017=\left(x^2+5x\right)^2+1981\)

Với mọi giá trị của \(x\in R\) thì:

\(\left(x^2+5x\right)^2+1981\ge1981\)

Hay \(D\ge1981\)với mọi giá trị của \(x\in R\)

Để \(D=1981\) thì \(\left(x^2+5x\right)^2=0\)

\(\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy...............

Chúc bạn học tốt!!!