Опростяване на дробни изрази

[Гост]

Ако може да ми дадете решенията на задачи : 29, 32, 35, 36, 39, 41 и 42 те са в файловете 1 и 2. Трябват ми до утре следобяд. Благодаря много !

[ammornil]

29

[tex]M=\frac{2a-1}{2a}-\frac{1}{2a-4a^2}-\frac{2a}{2a-1}, \hspace{12} a\ne 0, a \ne \pm \frac{1}{2} \\
\vspace{10} \\
M=\frac{2a-1}{2a}-\frac{1}{-2a.(2a-1)}-\frac{2a}{2a-1}=\frac{(2a-1)^2+1-(2a)^2}{2a.(2a-1)} \\
M=\frac{\cancel{4a^2}-4a+1+1\cancel{-4a^2}}{2a.(2a-1)}=\frac{-4a+2}{2a.(2a-1)}=\frac{-\cancel{2}.\cancel{(2a-1)}}{\cancel{2}a.\cancel{(2a-1)}}=-\frac{1}{a}[/tex]

[ammornil]

32

[tex]M=\frac{a^2+x^2+2ax}{2a^2}.\left(\frac{a}{x-a}-\frac{a}{x+a}\right), \hspace{12} a \ne 0, a \ne \pm x \\
\vspace{10}\\
M=\frac{a^2+2ax+x^2}{2a^2}\hspace{4}.\hspace{4}\frac{a.(x+a)-a.(x-a)}{(x-a).(x+a)}=\frac{(a+x)^2}{2.a^2}\hspace{4}.\hspace{4}\frac{a.[x+a-(x-a)]}{(x^2-a^2)}=\frac{(a+x)^2}{2.a^2}\hspace{4}.\hspace{4}\frac{a.(\cancel{x}+a\cancel{-x}+a)}{(x^2-a^2)} \\
M=\frac{(a+x)^2}{\cancel{2.a^2}}\hspace{4}.\hspace{4}\frac{\cancel{2.a^2}}{x^2-a^2}=\frac{(a+x)^2}{x^2-a^2}[/tex]

[ammornil]

35

[tex]M=\left(\frac{6a+c}{a^2-6ac}+\frac{6a-c}{a^2+6ac}\right).\frac{a^2-36c^2}{a^2+c^2}, \hspace{12} a \ne 0, a\ne \pm 6c \\
\vspace{10} \\
M=\left(\frac{6a+c}{a.(a-6c)}+\frac{6a-c}{a.(a+6c)}\right).\frac{a^2-(6c)^2}{a^2+c^2}=\frac{(6a+c).(a+6c)+(6a-c).(a-6c)}{a.(a-6c).(a+6c)}.\frac{a^2-(6c)^2}{a^2+c^2} \\
\vspace{10} \\
M=\frac{6a^2\cancel{+36ac}\cancel{+ac}+6c^2+6a^2\cancel{-36ac}\cancel{-ac}+6c^2}{a.\cancel{(a^2-36c^2)}}.\frac{\cancel{a^2-36c^2}}{a^2+c^2}=\frac{12a^2+12c^2}{a.(a^2+c^2)}=\frac{12.\cancel{(a^2+c^2)}}{a.\cancel{(a^2+c^2)}}=\frac{12}{a}[/tex]

[Гост]

Благодаря за 3те задачи !

[ammornil]

36
[tex]M=\frac{9n^2-16m^2}{7n}.\left(\frac{3m-4n}{4m^2-3mn}-\frac{3m+4n}{4m^2+3mn}\right), \hspace{12} n\ne0, m\ne0, m \ne \pm \frac{3}{4}n \\
\vspace{10} \\
M=\frac{(3n)^2-(4m)^2}{7n}.\left(\frac{3m-4n}{m.(4m-3n)}-\frac{3m+4n}{m(4m+3n)}\right)=\frac{(3n)^2-(4m)^2}{7n}.\hspace{4}\frac{(3m-4n).(4m+3n)-(3m+4n).(4m-3n)}{m.(4m-3n).(4m+3n)}\\
\vspace{10} \\
M=\frac{(3n)^2-(4m)^2}{7n}.\hspace{4}\frac{\cancel{12m^2}+9mn-16mn\cancel{-12n^2}\cancel{-12m^2}+9mn-16mn\cancel{+12n^2}}{m.[(4m)^2-(3n)^2]} \\
\vspace{10} \\
M=\frac{\cancel{(3n)^2-(4m)^2}}{7n}.\hspace{4}\frac{-14mn}{-m.\cancel{[(3n)^2-(4m)^2-]}}=\frac{\left,2 \\ \cancel{-14} \cancel{mn}\right,}{\cancel{-7}{\cancel{mn}}}=2 \\
\vspace{10} \\[/tex]

[ammornil]

39


[tex]M=\left(1+\frac{x^2-xy+y^2}{x^2-y^2}\right)\hspace{4}.\hspace{4}\left(\frac{3x}{2x-y}-\frac{2x+y}{x}\right), \hspace{12} x\ne0, y\ne \pm x, y \ne 2x \\
\vspace{12} \\
M=\left(\frac{x^2\cancel{-y^2}+x^2-xy\cancel{+y^2}}{x^2-y^2}\right)\hspace{4}.\hspace{4}\left(\frac{3x^2-(2x+y).(2x-y)}{x.(2x-y)}\right)=\frac{2x^2-xy}{x^2-y^2}\hspace{4}.\hspace{4}\frac{3x^2-[(2x)^2-y^2]}{x.(2x-y)} \\
\vspace{12} \\
M=\frac{\cancel{x.(2x-y)}}{x^2-y^2}\hspace{4}.\hspace{4}\frac{3x^2-4x^2+y^2}{\cancel{x.(2x-y)}}=\frac{-x^2+y^2}{x^2-y^2}=\frac{-1.\cancel{(x^2-y^2)}}{\cancel{x^2-y^2}}=-1[/tex]

[ammornil]

41


[tex]M=\left(a+\frac{a-b}{a+b}-b\right)\hspace{4}:\hspace{4}\left(\frac{2a+1}{a^2-b^2}+1\right), \hspace{12} a \ne \pm b, a \ne b-1, a \ne -b-1 \\
\vspace{12} \\
M=\left[(a-b)+\frac{a-b}{a+b}\right]\hspace{4}:\hspace{4}\left(\frac{2a+1+a^2-b^2}{a^2-b^2}\right)=\frac{(a-b).(a+b)+(a-b)}{a+b}\hspace{4}:\hspace{4}\frac{(a^2+2.a.1+1^2)-b^2}{a^2-b^2} \\
\vspace{12} \\
M=\frac{(a-b)(a+b+1)}{a+b}\hspace{4}:\hspace{4}\frac{(a+1)^2-b^2}{a^2-b^2}=\frac{(a-b)(a+b+1)}{a+b}\hspace{4}.\hspace{4}\frac{(a-b).(a+b)}{(a+1-b)(a+1+b)} \\
\vspace{12} \\
M=\frac{(a-b)^2.\cancel{(a+b+1)}.\cancel{(a+b)}}{\cancel{(a+b)}.(a-b+1).\cancel{(a+b+1)}}=\frac{(a-b)^2}{a-b+1}[/tex]

[ammornil]

42


[tex]M=\left(3+\frac{12ab+49b^2}{12a^2-32ab}\right)\hspace{4}.\hspace{4}\left[\frac{1}{6a-7b}-\frac{9b}{(7b-6a)^2}\right], \hspace{12} a\ne0, a\ne\frac{7}{6}b, a\ne\frac{8}{3}b \\
\vspace{12} \\
M=\frac{a^2-96ab+12ab+49b^2}{4a(3a-8b)}\hspace{4}.\hspace{4}\left[\frac{-1}{7b-6a}-\frac{9b}{(7b-6a)^2}\right]=\frac{49b^2-84ab+36a^2}{4a(3a-8b)}\hspace{4}.\hspace{4}\frac{-1.(7b-6a)-9b}{(7b-6a)^2} \\
\vspace{12} \\
M=\frac{(7b)^2-2.7b.6a+(6a)^2}{4a(3a-8b)}\hspace{4}.\hspace{4}\frac{-7b+6a-9b}{(7b-6a)^2}=\frac{\cancel{(7b-6a)^2}}{4a(3a-8b)}\hspace{4}.\hspace{4}\frac{6a-16b}{\cancel{(7b-6a)^2}} \\
\vspace{12} \\
M=\frac{\cancel{2}.\cancel{(3a-8b)}}{\cancel{2}.2.a.\cancel{(3a-8b)}}=\frac{1}{2a}[/tex]