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\(\dfrac{-0{,}9}{-3{,}6}=0{,}25\)

а) \(x=7\); б) \(a=0\)

\(\dfrac{a^2+ab}{ab}=\dfrac{a+b}{b}\)

а) \(\dfrac{2b^2}{b^2-4}\); б) \(\dfrac{2-c^2}{c}\)

а) \(\dfrac{y}{4}\); б) \(\dfrac{4x^3}{y}\)

\(b-\dfrac{a+b}{2}=\dfrac{b-a}{2}\)

\(R=\dfrac{R_1R_2}{R_1+R_2}\)

\(\dfrac{a^3}{3c}:\dfrac{a^2}{3bc}=ab\)

\(\dfrac{1-4(a+b)}{(a-b)(4(a+b)-1)}=\dfrac{1}{b-a}\)

\(4x\cdot\dfrac{x+1}{x}=4(x+1)\)

\[ \frac{1}{(a-b)(a-c)}-\frac{1}{(a-b)(b-c)}+\frac{1}{(a-c)(b-c)} =\frac{(b-c)-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}=0. \]