\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\Leftrightarrow\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\Leftrightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\Leftrightarrow a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+b^2xy+2abxy\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\Leftrightarrow\left(ay-bx\right)^2\ge0\)*đúng*

Đẳng thức xảy ra khi ^a/_b = ^x/_y

12x² - 4x( 3x - 5 ) = 10x - 18 < đã sửa >

<=> 12x² - 12x² + 20x = 10x - 18

<=> 20x = 10x - 18

<=> 20x - 10x = -18

<=> 10x = -18

<=> x = -18/10 = -9/5

\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)

ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)

\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)

\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)

\(=\frac{77x+175}{15x^2+150x+375}\)

                                                                 Bài làm

Với x = 3 thì : 

Đặt \(A=x^2-4x+6=x^2+2\cdot2x+2\cdot2+2=\left(x+2\right)^2+2\ge2\forall x\)

\(\Rightarrow\text{ Khi }x=3\text{ thì }Min_A=\left(x+2\right)^2+2=5^2+2=27\)

x² - 4x + 6 = ( x² - 4x + 4 ) + 2 = ( x - 2 )² + 2 ≥ 2 ∀ x

Dấu "=" xảy ra khi x = 2

=> GTNN của biểu thức = 2 <=> x = 2

Ta có: \(\left(x-1\right)^2+\left(x-3\right)^2\)

\(=x^2-2x+1+x^2-6x+9\)

\(=2x^2-8x+10\)

\(=2\left(x^2-4x+4\right)+2\)

\(=2\left(x-2\right)^2+2\ge2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(2\left(x-2\right)^2=0\Rightarrow x=2\)

Vậy Min = 2 khi x = 2

                                                               Bài giải

Đặt \(A=\left(x-1\right)^2+\left(x-3\right)^2=x^2-2x+1+x^2-6x+9=2x^2-8x+10\)

                                              \(=2\left(x^2-4x+4\right)+2=2\left(x-2\right)^2+2\ge2\forall x\)

Dấu " = " xảy ra khi \(\left(x-2\right)^2=0\text{ }\Leftrightarrow\text{ }x=2\)

Vậy \(Min_A=2\text{ khi }x=2\)

                                                          Bài giải

Đặt \(A=x^2-4x+6=x^2-2\cdot2x+2^2+2=\left(x-2\right)^2+2\ge2\)

\(\Rightarrow\text{ Với }x\ge3\text{ }\text{thì }A_{min}\text{ khi }\left(x-2\right)^2_{min}\Rightarrow\text{ }x\text{ nhỏ nhất }\Rightarrow\text{ }x=3\)

Vậy với \(x=3\text{ thì }Min_A=3\)

Bài này tìm được min thôi

Ta có: \(2x^2+x=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)-\frac{1}{8}=2\left(x+\frac{1}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(2\left(x+\frac{1}{4}\right)^2=0\Rightarrow x=-\frac{1}{4}\)

Vậy Min = -1/8 khi x = -1/4