400 + 400 + 200 + 10

= 800 + 200 + 10

= 1000 + 10

= 1010

________________________

900 x 1 + 3 x 1 + 8 x 1

= ( 900 + 3 + 8 ) x 1

= ( 903 + 8 ) x 1

= 911 x 1

= 911

400 + 400 + 200 + 10

= 800 + 200 + 10

= 1000 + 10

= 1010

900 x 1 + 3 x 1 + 8 x 1

= 900 + 3 + 8

= 903 + 8

= 911

k nha ^_^

a) \(A=9x^2-2x+15\)

\(A=9x^2-2x+\frac{1}{9}+\frac{134}{9}\)

\(A=\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\)

Có: \(\left(3x+\frac{1}{3}\right)^2\ge0\Rightarrow\left(3x+\frac{1}{3}\right)^2+\frac{134}{9}\ge\frac{134}{9}\)

Dấu '=' xảy ra khi: \(\left(3x+\frac{1}{3}\right)^2=0\Rightarrow3x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{9}\)

Vậy: \(Min_A=\frac{134}{9}\) tại \(x=-\frac{1}{9}\)

b) \(B=3x^2+x+1\)

\(B=3x^2+x+\frac{1}{12}+\frac{11}{12}\)

\(B=\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\)

Có: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2\ge0\Rightarrow\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2+\frac{11}{12}\ge\frac{11}{12}\)

Dấu '=' xảy ra khi: \(\left(\sqrt{3}x+\sqrt{\frac{1}{12}}\right)^2=0\Rightarrow\sqrt{3}x+\sqrt{\frac{1}{12}}=0\Rightarrow x=-\frac{1}{6}\)

Vậy: \(Min_B=\frac{11}{12}\) tại \(x=-\frac{1}{6}\)

c) \(C=x^2-6y+4x+y^2+38\)

\(C=\left(x^2+4x+4\right)+\left(y^2-6y+9\right)+25\)

\(C=\left(x+2\right)^2+\left(y-3\right)^2+25\)

Có: \(\left(x+2\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+25\ge25\)

Dấu = xảy ra khi: \(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2=0\\y-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

Vậy: \(Min_C=25\) tại \(\hept{\begin{cases}x=-2\\y=3\end{cases}}\)

b)5x^2+9y^2-12xy-6x+9=0

=>4x^2-12xy+9y^2+x^2-6x+9=0

=>(2x-3y)^2+(x-3)^2=0

=>2x-3y=0 và x-3=0

=>y=2 và x=3

112 nha k nha

=12
tk cho mk nha