Реши номера 121,122,123,124,125,126,127(реши их пожалуйста как я писал,там каждый номер по 1 примеру решено,отталкиваясь от такого решения напиши также)

Краткие ответы

• 121: а) \(\dfrac{1}{axy}\); б) \(\dfrac{4}{x^2}\).
• 122: а) \(\dfrac{y-4}{6x}\); б) \(-\dfrac{3b}{a+b}\).
• 123: а) \(\dfrac{7(a-1)}{a}\); б) \(\dfrac{5b}{b-2c}\); в) \(\dfrac{(x+2)(x+3)}6\); г) \(\dfrac{(y-5)(y-6)}4\).
• 124: а) \(1\); б) при \(x=0{,}5\): \(-\dfrac{10}{9}\), при \(x=-1{,}5\): \(-\dfrac{2}{21}\).
• 125: а) \(\dfrac{2(a-b)}{a(a+b)}\); б) \(\dfrac{b(x-5)}{a(x+5)}\).
• 126: а) \(\dfrac{3(x-y)}2\); б) \(\dfrac{ax}{7(x-y)}\); в) \((x-y)^2\); г) \(\dfrac{a-1}{(a+1)^2}\).
• 127: а) \(\dfrac{(x-5)(x-4)}6\); б) \(\dfrac{(a+1)(a+2b)}{12}\); в) \(\dfrac{3(y-5)}{2(y+6)}\); г) \(\dfrac{b+2}{9b}\).

Главное правило: сначала разложить многочлены на множители, затем сократить одинаковые множители, а не слагаемые.

121

а)

\[ \frac{xy}{a^2+a^3}\cdot\frac{a+a^2}{x^2y^2} = \frac{xy}{a^2(a+1)}\cdot \frac{a(a+1)}{x^2y^2} = \frac{1}{axy}. \]

б)

\[ \frac{6a}{x^2-x}\cdot\frac{2x-2}{3ax} = \frac{6a}{x(x-1)}\cdot \frac{2(x-1)}{3ax} = \frac{4}{x^2}. \]

122

а)

\[ \frac{y^2-16}{10xy}\cdot\frac{5y}{3y+12} = \frac{(y-4)(y+4)}{10xy}\cdot \frac{5y}{3(y+4)} = \frac{y-4}{6x}. \]

б)

\[ \frac{b-a}{a}\cdot\frac{3ab}{a^2-b^2} = \frac{-(a-b)}{a}\cdot \frac{3ab}{(a-b)(a+b)} = -\frac{3b}{a+b}. \]

123

а)

\[ \frac{a^2-1}{a-b}\cdot\frac{7a-7b}{a^2+a} = \frac{(a-1)(a+1)}{a-b}\cdot \frac{7(a-b)}{a(a+1)} = \frac{7(a-1)}a. \]

б)

\[ \frac{b^2+2bc}{b+3}\cdot\frac{5b+15}{b^2-4c^2} = \frac{b(b+2c)}{b+3}\cdot \frac{5(b+3)}{(b-2c)(b+2c)} = \frac{5b}{b-2c}. \]

в)

\[ \frac{(x+3)^2}{2x-4}\cdot\frac{x^2-4}{3x+9} = \frac{(x+3)^2}{2(x-2)}\cdot \frac{(x-2)(x+2)}{3(x+3)} = \frac{(x+3)(x+2)}6. \]

г)

Так как \((5-y)^2=(y-5)^2\), получаем:

\[ \frac{(5-y)^2}{2y+12}\cdot\frac{y^2-36}{2y-10} = \frac{(y-5)^2}{2(y+6)}\cdot \frac{(y-6)(y+6)}{2(y-5)} = \frac{(y-5)(y-6)}4. \]

124

а)

\[ \frac{5mn-m}{4m+n}\cdot\frac{16m^2-n^2}{5n-1} = \frac{m(5n-1)}{4m+n}\cdot \frac{(4m-n)(4m+n)}{5n-1} = m(4m-n). \]

Подставим \(m=\dfrac14,\ n=-3\):

\[ m(4m-n) = \frac14\left(4\cdot\frac14-(-3)\right) = \frac14(1+3)=1. \]

б)

\[ \frac{(x+2)^2}{3x+9}\cdot\frac{2x+6}{x^2-4} = \frac{(x+2)^2}{3(x+3)}\cdot \frac{2(x+3)}{(x-2)(x+2)} = \frac{2(x+2)}{3(x-2)}. \]

При \(x=0{,}5\):

\[ \frac{2(0{,}5+2)}{3(0{,}5-2)} = \frac{5}{-4{,}5} = -\frac{10}{9}. \]

При \(x=-1{,}5\):

\[ \frac{2(-1{,}5+2)}{3(-1{,}5-2)} = \frac{1}{-10{,}5} = -\frac{2}{21}. \]

125

а)

\[ \frac{a^2-b^2}{a^2-3a}\cdot \frac{2a-6}{b^2+2ab+a^2} = \frac{(a-b)(a+b)}{a(a-3)}\cdot \frac{2(a-3)}{(a+b)^2} = \frac{2(a-b)}{a(a+b)}. \]

б)

\[ \frac{bx+3b}{x^2-25}\cdot \frac{25-10x+x^2}{ax+3a} = \frac{b(x+3)}{(x-5)(x+5)}\cdot \frac{(x-5)^2}{a(x+3)} = \frac{b(x-5)}{a(x+5)}. \]

126

а)

\[ \frac{mx^2-my^2}{2m+8}\cdot \frac{3m+12}{my+mx} = \frac{m(x-y)(x+y)}{2(m+4)}\cdot \frac{3(m+4)}{m(x+y)} = \frac{3(x-y)}2. \]

б)

\[ \frac{ax+ay}{x^2-2xy+y^2}\cdot \frac{x^2-xy}{7x+7y} = \frac{a(x+y)}{(x-y)^2}\cdot \frac{x(x-y)}{7(x+y)} = \frac{ax}{7(x-y)}. \]

в)

\[ \frac{x^3-y^3}{x+y}\cdot \frac{x^2-y^2}{x^2+xy+y^2} = \frac{(x-y)(x^2+xy+y^2)}{x+y}\cdot \frac{(x-y)(x+y)}{x^2+xy+y^2} = (x-y)^2. \]

г)

\[ \frac{a^2-1}{a^3+1}\cdot \frac{a^2-a+1}{a^2+2a+1} = \frac{(a-1)(a+1)} {(a+1)(a^2-a+1)} \cdot \frac{a^2-a+1}{(a+1)^2} = \frac{a-1}{(a+1)^2}. \]

127

а)

\[ \frac{x^2-10x+25}{3x+12}\cdot \frac{x^2-16}{2x-10} = \frac{(x-5)^2}{3(x+4)}\cdot \frac{(x-4)(x+4)}{2(x-5)} = \frac{(x-5)(x-4)}6. \]

б)

\[ \frac{1-a^2}{4a+8b}\cdot \frac{a^2+4ab+4b^2}{3-3a} = \frac{(1-a)(1+a)}{4(a+2b)}\cdot \frac{(a+2b)^2}{3(1-a)} = \frac{(a+1)(a+2b)}{12}. \]

в)

\[ \frac{y^2-25}{y^2+12y+36}\cdot \frac{3y+18}{2y+10} = \frac{(y-5)(y+5)}{(y+6)^2}\cdot \frac{3(y+6)}{2(y+5)} = \frac{3(y-5)}{2(y+6)}. \]

г)

\[ \frac{b^3+8}{18b^2+27b}\cdot \frac{2b+3}{b^2-2b+4} = \frac{(b+2)(b^2-2b+4)} {9b(2b+3)} \cdot \frac{2b+3}{b^2-2b+4} = \frac{b+2}{9b}. \]