Question

Tìm min:

a) H = x² - 4x + 16

b) K = 2x² + 9y² - 6xy - 8x - 12y + 2018

Answer 1

a) H=x² - 4x +16

<=> H=x² -4x + 4 + 12

<=> H=(x-2)² +12 \(\ge12\)

Vậy Min H = 12

Dấu "=" xảy ra khi x=2

Answer 2

\(K=x^2-6xy+9y^2+4\left(x-3y\right)+4+x^2-12x+36+1978\)

\(K=\left(x-3y\right)^2+4\left(x-3y\right)+2^2+\left(x-6\right)^2+1978\)

\(K=\left(x-3y+2\right)^2+\left(x-6\right)^2+1978\ge1978\)

Vậy Min K =1978

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-3y+2=0\\x-6=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=\dfrac{8}{3}\\x=6\end{matrix}\right.\)

Answer 3

Cái phần cuối câu K nó bị gì đó , nếu không thấy thì bạn tự giải tiếp nhen

Answer 4

a) \(H=x^2-4x+16\)

\(\Leftrightarrow H=\left(x^2-2.x.2+2^2\right)+12\)

\(\Leftrightarrow H=\left(x-2\right)^2+12\)

Vậy GTNN của \(H=12\) khi \(x-2=0\Leftrightarrow x=2\)

b) \(K=2x^2+9y^2-6xy-8x-12y+2018\)

\(\Leftrightarrow K=x^2+x^2+9y^2-6xy+4x-12x-12y+36+1982\)

\(\Leftrightarrow K=\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+\left(x^2-12x+36\right)+1982\)

\(\Leftrightarrow K=\left[x^2-2.x.3y+\left(3y\right)^2\right]+\left(4x-12y\right)+\left(x^2-2.x.6+6^2\right)+1982\)

\(\Leftrightarrow K=\left(x-3y\right)^2+4\left(x-3y\right)+\left(x-6\right)^2+1982\)

\(\Leftrightarrow K=\left[\left(x-3y\right)^2+2.\left(x-3y\right).2+2^2\right]+\left(x-6\right)^2+1978\)

\(\Leftrightarrow K=\left(x-3y+2\right)^2+\left(x-6\right)^2+1978\)

Vậy GTNN của \(K=1978\) khi \(\left\{{}\begin{matrix}x-3y+2=0\\x-6=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}6-3y+2=0\\x=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{3}\\x=6\end{matrix}\right.\)