[tex]\displaystyle 1)\frac{5}{a}+\frac{a-5}{a+2}=\frac{5(a+2)+a(a-5)}{a(a+2)}=\frac{5a+10+a^2-5a}{a^2+2a}=\frac{10+a^2}{a^2+2a}\\ \\2)\frac{2x^2}{x^2-4}-\frac{2x}{x+2}=\frac{2x^2}{(x-2)(x+2)}-\frac{2x}{x+2}=\frac{2x^2-2x(x-2)}{(x-2)(x+2)}=\frac{2x^2-2x^2+4x}{x^2-4}=\\\frac{4x}{x^2-4}[/tex]
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[tex]\displaystyle 1)\frac{5}{a}+\frac{a-5}{a+2}=\frac{5(a+2)+a(a-5)}{a(a+2)}=\frac{5a+10+a^2-5a}{a^2+2a}=\frac{10+a^2}{a^2+2a}\\ \\2)\frac{2x^2}{x^2-4}-\frac{2x}{x+2}=\frac{2x^2}{(x-2)(x+2)}-\frac{2x}{x+2}=\frac{2x^2-2x(x-2)}{(x-2)(x+2)}=\frac{2x^2-2x^2+4x}{x^2-4}=\\\frac{4x}{x^2-4}[/tex]