Lời giải:
$A=x^2+2y^2+3x-y+6$
$\Leftrightarrow x^2+3x+(2y^2-y+6-A)=0(*)$
Coi đây là PT bậc 2 ẩn $x$
Vì $A$ xác định nên $(*)$ luôn có nghiệm.
$\Rightarrow \Delta'=9-4(2y^2-y+6-A)\geq 0$
$\Leftrightarrow A\geq 8y^2-4y+15$
Mà $8y^2-4y+15=8(y-\frac{1}{4})^2+\frac{29}{2}\geq \frac{29}{2}$
$\Rightarrow A\geq \frac{29}{2}$ hay $A_{\min}=\frac{29}{2}$
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\(B=\frac{x^2-1}{x^2+1}=1-\frac{2}{x^2+1}\)
$x^2\geq 0\Rightarrow x^2+1\geq 1\Rightarrow \frac{2}{x^2+1}\leq 2$
$\Rightarrow B=1-\frac{2}{x^2+1}\geq 1-2=-1$
Vậy $B_{\min}=-1$
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ĐK: $x\neq 1$
\(C=\frac{x^2-3x+3}{x^2-2x+1}=\frac{x^2-2x+1-(x-1)+1}{x^2-2x+1}=1-\frac{1}{x-1}+\frac{1}{(x-1)^2}\)
\(=\left(\frac{1}{x-1}-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}\)
Vậy $C_{\min}=\frac{3}{4}$
\(C=\frac{x^2-3x+3}{x^2-2x+1}=\frac{x^2-2x+1-x+1+1}{\left(x-1\right)^2}\)
\(=\frac{\left(x-1\right)^2-\left(x-1\right)+1}{\left(x-1\right)^2}=1-\frac{1}{x-1}+\frac{1}{\left(x-1\right)^2}\)
Đặt \(\frac{1}{x-1}=c\)
\(\Rightarrow\) \(C=c^2-c+1\)
\(=c^2-2.c.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(c-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) \(\forall c\)
Vậy GTNN của C là \(\frac{3}{4}\)
Dấu '' = '' xảy ra khi \(c=\frac{1}{2}\Leftrightarrow\frac{1}{x-1}=\frac{1}{2}\Leftrightarrow3\)