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b: Ta có: \(x^2-x+5\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{19}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\forall x\)
\(\Leftrightarrow\dfrac{2022}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{19}{4}}\le\dfrac{8088}{19}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)
\(\dfrac{x+2}{x^2+3x-10}=\dfrac{x+2}{\left(x+5\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(2x-3\right)\left(x+5\right)\left(x-2\right)}\)
\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)
b: \(\dfrac{1}{-x^2+3x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}=\dfrac{-\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)}\)
\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\dfrac{1}{-x^2+4x-3}=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)
c: \(\dfrac{3}{x^3-1}=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x}{x-1}=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)