Задание 538

Решение:

• \({\largeа)}\ \frac{a-1}{2a}\cdot\left(\frac{a+3}{a+1}-\frac{a^2-5}{a^2-1}\right)=\frac{a-1}{2a}\cdot\left(\frac{a+3^{\backslash{a\ -\ 1}}}{a+1\phantom{^{\backslash{a\ -\ 1}}}}-\frac{a^2-5^{\backslash1}}{ (a+1)(a-1)\phantom{^{\backslash1}}}\right)=\frac{a-1}{2a}\cdot\frac{ (a+3)(a-1)-(a^2-5)\cdot1}{ (a+1)(a-1)}=\frac{a-1}{2a}\cdot\frac{a^2-a+3a-3-a^2+5}{ (a+1)(a-1)}=\frac{a-1}{2a}\cdot\frac{2a+2}{ (a+1)(a-1)}=\frac{a-1}{2a}\cdot\frac{ 2(a+1)}{ (a+1)(a-1)}=\frac{1}{a}\)

• \({\largeб)}\ \left(\frac{c+3^{\backslash{c\ +\ 3}}}{c-3\phantom{^{\backslash{c\ +\ 3}}}}-\frac{c^{\backslash{c\ -\ 3}}}{c+3\phantom{^{\backslash{c\ -\ 3}}}}\right)\cdot\frac{c-3}{c+1}=\frac{ (c+3)(c+3)-c(c-3)}{ (c+3)(c-3)}\cdot\frac{c-3}{c+1}=\frac{c^2+6c+9-c^2+3c}{ (c+3)(c-3)}\cdot\frac{c-3}{c+1}=\frac{9c+9}{ (c+3)(c-3)}\cdot\frac{c-3}{c+1}=\frac{ 9(c+1)}{ (c+3)(c-3)}\cdot\frac{c-3}{c+1}=\frac{9}{c+3}\)

• \({\largeв)}\ \left(\frac{14+a^2}{a^2-4}-\frac{a-4}{a+2}\right)\cdot\frac{a-2}{6}=\left(\frac{14+a^{2\backslash1}}{ (a+2)(a-2)\phantom{^{\backslash1}}}-\frac{a-4^{\backslash{a\ -\ 2}}}{a+2\phantom{^{\backslash{a\ -\ 2}}}}\right)\cdot\frac{a-2}{6}=\frac{ (14+a^2)\cdot1-(a-4)(a-2)}{ (a+2)(a-2)}\cdot\frac{a-2}{6}=\frac{ 14+a^2-(a^2-2a-4a+8)}{ (a+2)(a-2)}\cdot\frac{a-2}{6}=\frac{14+a^2-a^2+6a-8}{ (a+2)(a-2)}\cdot\frac{a-2}{6}=\frac{6a+6}{ (a+2)(a-2)}\cdot\frac{a-2}{6}=\frac{ 6(a+1)}{ (a+2)(a-2)}\cdot\frac{a-2}{6}=\frac{a+1}{a+2}\)

• \({\largeг)}\ \left(\frac{a^{\backslash{a\ +\ 4}}}{a-4\phantom{^{\backslash{a\ +\ 4}}}}-\frac{a-4^{\backslash{a\ -\ 4}}}{a+4\phantom{^{\backslash{a\ -\ 4}}}}\right)\cdot\frac{a+4}{4}=\frac{ a(a+4)-(a-4)(a-4)}{ (a+4)(a-4)}\cdot\frac{a+4}{4}=\frac{ a^2+4a-(a^2-8a+16)}{ (a+4)(a-4)}\cdot\frac{a+4}{4}=\frac{a^2+4a-a^2+8a-16}{ (a+4)(a-4)}\cdot\frac{a+4}{4}=\frac{12a-16}{ (a+4)(a-4)}\cdot\frac{a+4}{4}=\frac{ 4(3a-4)}{ (a+4)(a-4)}\cdot\frac{a+4}{4}=\frac{3a-4}{a-4}\)

• \({\largeд)}\ \left(\frac{y+1^{\backslash{y\ +\ 1}}}{y-1\phantom{^{\backslash{y\ +\ 1}}}}-\frac{y-1^{\backslash{y\ -\ 1}}}{y+1\phantom{^{\backslash{y\ -\ 1}}}}\right)\cdot\frac{y+1}{4y}=\frac{ (y+1)(y+1)-(y-1)(y-1)}{ (y+1)(y-1)}\cdot\frac{y+1}{4y}=\frac{ y^2+2y+1-(y^2-2y+1)}{ (y+1)(y-1)}\cdot\frac{y+1}{4y}=\frac{y^2+2y+1-y^2+2y-1}{ (y+1)(y-1)}\cdot\frac{y+1}{4y}=\frac{4y}{ (y+1)(y-1)}\cdot\frac{y+1}{4y}=\frac{1}{y-1}\)

• \({\largeе)}\ \left(\frac{1+a^{\backslash{1\ +\ a}}}{1-a\phantom{^{\backslash{1\ +\ a}}}}-\frac{1-a^{\backslash{1\ -\ a}}}{1+a\phantom{^{\backslash{1\ -\ a}}}}\right):\frac{2a}{1-a}=\frac{ (1+a)(1+a)-(1-a)(1-a)}{ (1+a)(1-a)}\cdot\frac{1-a}{2a}=\frac{ 1+2a+a^2-(1-2a+a^2)}{ (1+a)(1-a)}\cdot\frac{1-a}{2a}=\frac{1+2a+a^2-1+2a-a^2}{ (1+a)(1-a)}\cdot\frac{1-a}{2a}=\frac{4a}{ (1+a)(1-a)}\cdot\frac{1-a}{2a}=\frac{2}{1+a}\)

• \({\largeж)}\ \frac{4y}{y-1}\cdot\left(\frac{y^{\backslash{y}}}{8\phantom{^{\backslash{y}}}}-\frac{1^{\backslash2y}}{4\phantom{^{\backslash2y}}}+\frac{1^{\backslash1}}{8y\phantom{^{\backslash1}}}\right)=\frac{4y}{y-1}\cdot\frac{y\cdot{y}-1\cdot2y+1\cdot1}{8y}=\frac{4y}{y-1}\cdot\frac{y^2-2y+1}{8y}=\frac{4y}{y-1}\cdot\frac{ (y-1)^2}{8y}=\frac{y-1}{2}\)

• \({\largeз)}\ \left(\frac{a^{\backslash3a}}{8\phantom{^{\backslash3a}}}+\frac{1^{\backslash8a}}{3\phantom{^{\backslash8a}}}+\frac{1^{\backslash4}}{6a\phantom{^{\backslash4}}}\right):\frac{a+1}{12a}=\frac{a\cdot3a+1\cdot8a+1\cdot4}{24a}\cdot\frac{12a}{a+1}=\frac{3a^2+8a+4}{24a}\cdot\frac{12a}{a+1}=\frac{3a^2+8a+4}{ 2(a+1)}=\frac{3a^2+8a+4}{2a+2}\)