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1.\(A=\frac{2x^2-16x+41}{x^2-8x+22}\) \(=\frac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\frac{3}{\left(x-4\right)^2+6}\ge\frac{1}{2}\)
Dấu '' = '' xảy ra khi x = 4.
Vậy MinA= \(\frac{1}{2}\) tại x = 4.
Bài 2: \(a,\frac{7x-1}{2x^2+6x}=\frac{7x-1}{2x\left(x+3\right)}=\frac{\left(7x-1\right)\left(x-3\right)}{2x\left(x+3\right)\left(x-3\right)}\)
\(\frac{5-3x}{x^2-9}=\frac{5-3x}{\left(x-3\right)\left(x+3\right)}=\frac{\left(5-3x\right)2x}{2x\left(x-3\right)\left(x+3\right)}\)
\(b,\frac{x+1}{x-x^2}=\frac{x+1}{x\left(1-x\right)}=-\frac{x+1}{x\left(x+1\right)}=-\frac{2\left(x-1\right)\left(x+1\right)}{2x\left(x-1\right)^2}\)
\(\frac{x+2}{2-4x+2x^2}=\frac{x+2}{2\left(x-1\right)^2}=\frac{2x\left(x+2\right)}{2x\left(x-1\right)^2}\)
\(c,\frac{4x^2-3x+5}{x^3-1}=\frac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{2x}{x^2+x+1}=\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\frac{6}{x-1}=\frac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(d,\frac{7}{5x}=\frac{7.2\left(2y-x\right)\left(2y+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{4}{x-2y}=-\frac{4}{2y-x}=-\frac{4.2.5x\left(2x+x\right)}{2.5x\left(2y-x\right)\left(2y+x\right)}\)
\(\frac{x-y}{8y^2-2x^2}=\frac{x-y}{2\left(4y^2-x^2\right)}=\frac{x-y}{2\left(2y-x\right)\left(2y+x\right)}=\frac{5x\left(x-y\right)}{2.5x.\left(2y-x\right)\left(2y+x\right)}\)
\(C=\left[\frac{x^2.\left(x^2-4\right)+4x^2}{x^2-4}\right].\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x.\left(x^2-4\right)}.\frac{x^2-4}{x-2}\right]\)
\(C=\frac{x^4-4x^2+4x^2}{x^2-4}.\left[\frac{x}{2.\left(x-2\right)}+\frac{2-2x}{x\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2}{2x.\left(x-2\right)}+\frac{\left(2-2x\right).2}{2x.\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\left[\frac{x^2+4-4x}{2x.\left(x-2\right)}\right]\)
\(C=\frac{x^4}{x^2-4}.\frac{\left(x-2\right)^2}{2x.\left(x-2\right)}\)
\(C=\frac{x^4}{\left(x-2\right).\left(x+2\right)}.\frac{\left(x-2\right).\left(x-2\right)}{2x.\left(x-2\right)}\)
\(C=\frac{x^3}{\left(x+2\right).2}\)
Bạn chưa rút gọn hết, Despacito.