Задание 520

Решение:

• \({\largeа)}\ \frac{1}{2a-2}+\frac{2}{4a-4}=\frac{1}{ 2(a-1)}+\frac{2}{ 4(a-1)}=\frac{1}{ 2(a-1)}+\frac{1}{ 2(a-1)}=\frac{1+1}{ 2(a-1)}=\frac{2}{ 2(a-1)}=\frac{1}{a-1}\)

• \({\largeб)}\ \frac{7a}{3x+3}-\frac{a}{6x+6}=\frac{7a^{\backslash2}}{ 3(x+1)\phantom{^{\backslash2}}}-\frac{a^{\backslash1}}{ 6(x+1)\phantom{^{\backslash1}}}=\frac{7a\cdot2}{ 3(x+1)\cdot2}-\frac{a\cdot1}{ 6(x+1)\cdot1}=\frac{14a}{ 6(x+1)}-\frac{a}{ 6(x+1)}=\frac{14a-a}{ 6(x+1)}=\frac{13a}{ 6(x+1)}\)

• \({\largeв)}\ \frac{2m}{4m+4n}+\frac{4n}{8m+8n}=\frac{2m}{ 4(m+n)}+\frac{4n}{ 8(m+n)}=\frac{m}{ 2(m+n)}+\frac{n}{ 2(m+n)}=\frac{m+n}{ 2(m+n)}=\frac{1}{2}\)

• \({\largeг)}\ \frac{2p}{10p-10q}-\frac{3q}{15p-15q}=\frac{2p}{ 10(p-q)}-\frac{3q}{ 15(p-q)}=\frac{p}{ 5(p-q)}-\frac{q}{ 5(p-q)}=\frac{p-q}{ 5(p-q)}=\frac{1}{5}\)

• \({\largeд)}\ \frac{2x}{ax+bx}+\frac{3y}{ay+by}=\frac{2x}{ x(a+b)}+\frac{3y}{ y(a+b)}=\frac{2}{a+b}+\frac{3}{a+b}=\frac{2+3}{a+b}=\frac{5}{a+b}\)

• \({\largeе)}\ \frac{y}{ax-bx}-\frac{x}{ay-by}=\frac{y^{\backslash{y}}}{ x(a-b)\phantom{^{\backslash{y}}}}-\frac{x^{\backslash{x}}}{ y(a-b)\phantom{^{\backslash{x}}}}=\frac{y\cdot{y}}{ x(a-b)\cdot{y}}-\frac{x\cdot{x}}{ y(a-b)\cdot{x}}=\frac{y^2}{ xy(a-b)}-\frac{x^2}{ xy(a-b)}=\frac{y^2-x^2}{ xy(a-b)}\)

• \({\largeж)}\ \frac{1}{2x^2y-xy}+\frac{2}{y-2xy}=\frac{1}{ xy(2x-1)}+\frac{2}{ y(1-2x)}=\frac{1^{\backslash1}}{ xy(2x-1)\phantom{^{\backslash1}}}-\frac{2^{\backslash{x}}}{ y(2x-1)\phantom{^{\backslash{x}}}}=\frac{1\cdot1}{ xy(2x-1)\cdot1}-\frac{2\cdot{x}}{ y(2x-1)\cdot{x}}=\frac{1}{ xy(2x-1)}-\frac{2x}{ xy(2x-1)}=\frac{1-2x}{ xy(2x-1)}={-}\frac{2x-1}{ xy(2x-1)}={-}\frac{1}{xy}\)

• \({\largeз)}\ \frac{3}{3m^2n-6mn^2}-\frac{2}{4mn-2m^2}=\frac{3}{ 3mn(m-2n)}-\frac{2}{ 2m(2n-m)}=\frac{1^{\backslash1}}{ mn(m-2n)\phantom{^{\backslash1}}}+\frac{1^{\backslash{n}}}{ m(m-2n)\phantom{^{\backslash{n}}}}=\frac{1\cdot1}{ mn(m-2n)\cdot1}+\frac{1\cdot{n}}{ m(m-2n)\cdot{n}}=\frac{1}{ mn(m-2n)}+\frac{n}{ mn(m-2n)}=\frac{1+n}{ mn(m-2n)}\)

• \({\largeи)}\ \frac{15}{10p^3q-15p^2q^2}-\frac{6q}{9pq^3-6p^2q^2}=\frac{15}{ 5p^2q(2p-3q)}-\frac{6q}{ 3pq^2(3q-2p)}=\frac{3^{\backslash1}}{ p^2q(2p-3q)\phantom{^{\backslash1}}}+\frac{2^{\backslash{p}}}{ pq(2p-3q)\phantom{^{\backslash{p}}}}=\frac{3\cdot1}{ p^2q(2p-3q)\cdot1}+\frac{2\cdot{p}}{ pq(2p-3q)\cdot{p}}=\frac{3}{ p^2q(2p-3q)}+\frac{2p}{ p^2q(2p-3q)}=\frac{3+2p}{ p^2q(2p-3q)}\)

• \({\largeк)}\ \frac{3b}{2a^3b-8a^2b^2}-\frac{5a}{12a^3b-3a^4}=\frac{3b}{ 2a^2b(a-4b)}-\frac{5a}{ 3a^3(4b-a)}=\frac{3^{\backslash3}}{ 2a^2(a-4b)\phantom{^{\backslash3}}}+\frac{5^{\backslash2}}{ 3a^2(a-4b)\phantom{^{\backslash2}}}=\frac{3\cdot3}{ 2a^2(a-4b)\cdot3}+\frac{5\cdot2}{ 3a^2(a-4b)\cdot2}=\frac{9}{ 6a^2(a-4b)}+\frac{10}{ 6a^2(a-4b)}=\frac{9+10}{ 6a^2(a-4b)}=\frac{19}{ 6a^2(a-4b)}\)