Question

[Dương Bảo Yến]

Giúp mk luôn vs ạ!! Mk đang cần gấp!!

Answer 1

[Trần Đức Huy]

a)

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

\(=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-2-\left(\frac{1}{2}\right)^2\)

\(=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)

Mà\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

Vậy \(C_{Min}=-\frac{9}{4}\)khi và chỉ khi\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

b)

\(D=x^2+y^2+x-6y+5\)

\(=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+y^2-2.y.3+3^2+5-\left(\frac{1}{2}\right)^2-3^2\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\)

Mà\(\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\ge-\frac{17}{4}\)

Vậy \(D_{Min}=-\frac{17}{4}\)khi và chỉ khi \(\hept{\begin{cases}\left(x+\frac{1}{2}\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

c)

\(E=x^2+10y^2-6xy-10y+26\)

\(=x^2-2.x.3y+\left(3y\right)^2+y^2-2.y.5+5^2+26-5^2\)

\(=\left(x-3y\right)^2+\left(y-5\right)^2+1\)

Mà\(\left(x-3y\right)^2+\left(y-5\right)^2\ge0\Rightarrow\left(x-3y\right)^2+\left(y-5\right)^2+1\ge1\)

Vậy \(E_{Min}=1\)khi và chỉ khi\(\hept{\begin{cases}\left(x-3y\right)^2=0\\\left(y-5\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=15\\y=5\end{cases}}}\)