: Розділ 1. Раціональні вирази...стор. № 15-55

Сторінка 38

240. Спростіть:

1) $5-\frac{5x+2y}{2x}\cdot\left(\frac{1}{5x-2y}+\frac{1}{5x+2y}\right)$;
2) $\left(5-\frac{5x+2y}{2x}\right)\cdot\left(\frac{1}{5x-2y}+\frac{1}{5x+2y}\right)$;
3) $\left(5-\frac{5x+2y}{2x}\right)\cdot\frac{1}{5x-2y}+\frac{1}{5x+2y}$.

Розв'язок:

1) 1)
$\frac{1}{5x-2y}+\frac{1}{5x+2y} =$
$= \frac{5x+2y+5x-2y}{(5x-2y)(5x+2y)} =$
$= \frac{10x}{(5x-2y)(5x+2y)}$
2)
$\frac{5x+2y}{2x}\cdot\frac{10x}{(5x-2y)(5x+2y)} =$
$= \frac{5}{5x-2y}$
3)
$5-\frac{5}{5x-2y} =$
$= \frac{5(5x-2y)-5}{5x-2y} =$
$= \frac{25x-10y-5}{5x-2y}$

2) 1)
$5-\frac{5x+2y}{2x} =$
$= \frac{10x-5x-2y}{2x} =$
$= \frac{5x-2y}{2x}$
2)
$\frac{1}{5x-2y}+\frac{1}{5x+2y} =$
$= \frac{10x}{(5x-2y)(5x+2y)}$
3)
$\frac{5x-2y}{2x}\cdot\frac{10x}{(5x-2y)(5x+2y)} =$
$= \frac{5}{5x+2y}$

3) 1)
$5-\frac{5x+2y}{2x} =$
$= \frac{5x-2y}{2x}$
2)
$\frac{5x-2y}{2x}\cdot\frac{1}{5x-2y} =$
$= \frac{1}{2x}$
3)
$\frac{1}{2x}+\frac{1}{5x+2y} =$
$= \frac{5x+2y+2x}{2x(5x+2y)} =$
$= \frac{7x+2y}{2x(5x+2y)}$

Відповідь: 1) $\frac{25x-10y-5}{5x-2y}$;
2) $\frac{5}{5x+2y}$;
3) $\frac{7x+2y}{2x(5x+2y)}$.

241. Виконайте дії:

1) $\left(\frac{2x}{2x+y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)$;
2) $\frac{4x^2+24x+36}{x-3}\cdot\left(\frac{x}{3x-9}-\frac{3}{x^2+3x}+\frac{x^2+9}{27-3x^2}\right)$;
3) $\left(\frac{a-b}{a^2+ab}-\frac{1}{a^2-b^2}\cdot\frac{(b-a)^2}{a+b}\right):\frac{a-b}{a^2+ab}$;
4) $\left(\frac{a+2}{2-a}-\frac{2-a}{2+a}-\frac{4a^2}{a^2-4}\right):\left(\frac{1}{a^3+a^2}-\frac{1-a}{a^2}-1\right)$.

Розв'язок:

1) 1)
$\frac{2x}{2x+y}-\frac{4x^2}{(2x+y)^2} =$
$= \frac{2x(2x+y)-4x^2}{(2x+y)^2} =$
$= \frac{2xy}{(2x+y)^2}$
2)
$\frac{2x}{(2x-y)(2x+y)}-\frac{1}{2x-y} =$
$= \frac{2x-(2x+y)}{(2x-y)(2x+y)} =$
$= \frac{-y}{(2x-y)(2x+y)}$
3)
$\frac{2xy}{(2x+y)^2}:\frac{-y}{(2x-y)(2x+y)} =$
$= -\frac{2xy(2x-y)(2x+y)}{(2x+y)^2\cdot y} =$
$= -\frac{2x(2x-y)}{2x+y} =$
$= \frac{2x(y-2x)}{2x+y}$

2) 1)
$\frac{x}{3x-9}-\frac{3}{x^2+3x}+\frac{x^2+9}{27-3x^2} =$
$= \frac{x}{3(x-3)}-\frac{3}{x(x+3)}-\frac{x^2+9}{3(x-3)(x+3)} =$
$= \frac{x^2(x+3)-9(x-3)-x(x^2+9)}{3x(x-3)(x+3)} =$
$= \frac{x^3+3x^2-9x+27-x^3-9x}{3x(x-3)(x+3)} =$
$= \frac{3x^2-18x+27}{3x(x-3)(x+3)} =$
$= \frac{3(x-3)^2}{3x(x-3)(x+3)} =$
$= \frac{x-3}{x(x+3)}$
2)
$\frac{4(x+3)^2}{x-3}\cdot\frac{x-3}{x(x+3)} =$
$= \frac{4(x+3)}{x} =$
$= \frac{4x+12}{x}$

3) 1)
$\frac{1}{a^2-b^2}\cdot\frac{(b-a)^2}{a+b} =$
$= \frac{(a-b)^2}{(a-b)(a+b)(a+b)} =$
$= \frac{a-b}{(a+b)^2}$
2)
$\frac{a-b}{a(a+b)}-\frac{a-b}{(a+b)^2} =$
$= \frac{(a-b)(a+b)-a(a-b)}{a(a+b)^2} =$
$= \frac{(a-b)(a+b-a)}{a(a+b)^2} =$
$= \frac{b(a-b)}{a(a+b)^2}$
3)
$\frac{b(a-b)}{a(a+b)^2}:\frac{a-b}{a(a+b)} =$
$= \frac{b(a-b)\cdot a(a+b)}{a(a+b)^2(a-b)} =$
$= \frac{b}{a+b}$

4) 1)
$\frac{a+2}{2-a}-\frac{2-a}{2+a}+\frac{4a^2}{(2-a)(2+a)} =$
$= \frac{(a+2)^2-(2-a)^2+4a^2}{(2-a)(2+a)} =$
$= \frac{a^2+4a+4-4+4a-a^2+4a^2}{(2-a)(2+a)} =$
$= \frac{4a(a+2)}{(2-a)(2+a)} =$
$= \frac{4a}{2-a}$
2)
$\frac{1}{a^2(a+1)}-\frac{1-a}{a^2}-1 =$
$= \frac{1-(1-a)(a+1)-a^2(a+1)}{a^2(a+1)} =$
$= \frac{1-1+a^2-a^3-a^2}{a^2(a+1)} =$
$= \frac{-a^3}{a^2(a+1)} =$
$= -\frac{a}{a+1}$
3)
$\frac{4a}{2-a}:\left(-\frac{a}{a+1}\right) =$
$= -\frac{4a(a+1)}{(2-a)\cdot a} =$
$= \frac{4(a+1)}{a-2}$

Відповідь: 1) $\frac{2x(y-2x)}{2x+y}$;
2) $\frac{4x+12}{x}$;
3) $\frac{b}{a+b}$;
4) $\frac{4(a+1)}{a-2}$.

242. Спростіть вираз:

1) $\left(\frac{a^2}{a+b}-\frac{a^3}{a^2+b^2+2ab}\right):\left(\frac{a}{a+b}+\frac{a^2}{b^2-a^2}\right)$;
2) $\left(\frac{y}{4y+16}-\frac{y^2+16}{4y^2-64}-\frac{4}{y^2-4y}\right)\cdot\frac{3y^2-24y+48}{y+4}$;
3) $\left(\frac{1}{4x^2-y^2}:\frac{2x+y}{(y-2x)^2}-\frac{2x-y}{4x^2+2xy}\right)\cdot\frac{(2x+y)^2}{y^2}$;
4) $\left(\frac{a^2}{b^3-ab^2}+\frac{a-b}{b^2}-\frac{2}{b}\right):\left(\frac{a+b}{b-a}-\frac{b-a}{b+a}-\frac{4a^2}{a^2-b^2}\right)$.

Розв'язок:

1) 1)
$\frac{a^2}{a+b}-\frac{a^3}{(a+b)^2} =$
$= \frac{a^2(a+b)-a^3}{(a+b)^2} =$
$= \frac{a^2b}{(a+b)^2}$
2)
$\frac{a}{a+b}-\frac{a^2}{(a-b)(a+b)} =$
$= \frac{a(a-b)-a^2}{(a-b)(a+b)} =$
$= \frac{-ab}{(a-b)(a+b)}$
3)
$\frac{a^2b}{(a+b)^2}:\frac{-ab}{(a-b)(a+b)} =$
$= -\frac{a^2b(a-b)(a+b)}{(a+b)^2\cdot ab} =$
$= -\frac{a(a-b)}{a+b} =$
$= \frac{a(b-a)}{a+b}$

2) 1)
$\frac{y}{4y+16}-\frac{y^2+16}{4y^2-64}-\frac{4}{y^2-4y} =$
$= \frac{y}{4(y+4)}-\frac{y^2+16}{4(y-4)(y+4)}-\frac{4}{y(y-4)} =$
$= \frac{y^2(y-4)-y(y^2+16)-16(y+4)}{4y(y-4)(y+4)} =$
$= \frac{y^3-4y^2-y^3-16y-16y-64}{4y(y-4)(y+4)} =$
$= \frac{-4(y^2+8y+16)}{4y(y-4)(y+4)} =$
$= -\frac{(y+4)^2}{y(y-4)(y+4)} =$
$= -\frac{y+4}{y(y-4)}$
2)
$-\frac{y+4}{y(y-4)}\cdot\frac{3(y-4)^2}{y+4} =$
$= -\frac{3(y-4)}{y} =$
$= \frac{12-3y}{y}$

3) 1)
$\frac{1}{(2x-y)(2x+y)}\cdot\frac{(2x-y)^2}{2x+y} =$
$= \frac{2x-y}{(2x+y)^2}$
2)
$\frac{2x-y}{(2x+y)^2}-\frac{2x-y}{2x(2x+y)} =$
$= \frac{2x(2x-y)-(2x-y)(2x+y)}{2x(2x+y)^2} =$
$= \frac{(2x-y)(2x-2x-y)}{2x(2x+y)^2} =$
$= \frac{-y(2x-y)}{2x(2x+y)^2}$
3)
$\frac{-y(2x-y)}{2x(2x+y)^2}\cdot\frac{(2x+y)^2}{y^2} =$
$= -\frac{2x-y}{2xy} =$
$= \frac{y-2x}{2xy}$

4) 1)
$\frac{a^2}{b^2(b-a)}+\frac{a-b}{b^2}-\frac{2}{b} =$
$= \frac{a^2+(a-b)(b-a)-2b(b-a)}{b^2(b-a)} =$
$= \frac{a^2-a^2+2ab-b^2-2b^2+2ab}{b^2(b-a)} =$
$= \frac{4ab-3b^2}{b^2(b-a)} =$
$= \frac{4a-3b}{b(b-a)}$
2)
$\frac{a+b}{b-a}-\frac{b-a}{b+a}+\frac{4a^2}{(b-a)(b+a)} =$
$= \frac{(a+b)^2-(b-a)^2+4a^2}{(b-a)(b+a)} =$
$= \frac{4ab+4a^2}{(b-a)(b+a)} =$
$= \frac{4a(a+b)}{(b-a)(b+a)} =$
$= \frac{4a}{b-a}$
3)
$\frac{4a-3b}{b(b-a)}:\frac{4a}{b-a} =$
$= \frac{(4a-3b)(b-a)}{b(b-a)\cdot 4a} =$
$= \frac{4a-3b}{4ab}$

Відповідь: 1) $\frac{a(b-a)}{a+b}$;
2) $\frac{12-3y}{y}$;
3) $\frac{y-2x}{2xy}$;
4) $\frac{4a-3b}{4ab}$.

243. Спростіть вираз:

1) $\left(\frac{a^2-5a}{a^2-10a+25}+\frac{25}{a^2-25}\right)\cdot\frac{5+a}{125-a^3}$;
2) $\left(\frac{x^2+y^2}{x}+y\right):\left(\frac{x^3-y^3}{x^2+y^2}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\right)$;
3) $\left(\frac{8a^2+2a}{8a^3-1}-\frac{2a+1}{4a^2+2a+1}\right)\cdot\left(1+\frac{2a+1}{2a}-\frac{4a^2+10a}{4a^2+2a}\right)$;
4) $\frac{x^3+27}{x-1}\cdot\left(\frac{x-3}{x^2-3x+9}+\frac{x+9}{x^3+27}\right):\frac{x^2+x}{x^2-1}$.

Розв'язок:

1) 1)
$\frac{a(a-5)}{(a-5)^2}+\frac{25}{(a-5)(a+5)} =$
$= \frac{a}{a-5}+\frac{25}{(a-5)(a+5)} =$
$= \frac{a(a+5)+25}{(a-5)(a+5)} =$
$= \frac{a^2+5a+25}{(a-5)(a+5)}$
2)
$\frac{a^2+5a+25}{(a-5)(a+5)}\cdot\frac{a+5}{(5-a)(25+5a+a^2)} =$
$= \frac{1}{(a-5)(5-a)} =$
$= -\frac{1}{(a-5)^2}$

2) 1)
$\frac{x^2+y^2}{x}+y =$
$= \frac{x^2+xy+y^2}{x}$
2)
$\frac{x^3-y^3}{x^2+y^2}\cdot\left(\frac{1}{x^2}+\frac{1}{y^2}\right) =$
$= \frac{x^3-y^3}{x^2+y^2}\cdot\frac{y^2+x^2}{x^2y^2} =$
$= \frac{(x-y)(x^2+xy+y^2)}{x^2y^2}$
3)
$\frac{x^2+xy+y^2}{x}:\frac{(x-y)(x^2+xy+y^2)}{x^2y^2} =$
$= \frac{(x^2+xy+y^2)\cdot x^2y^2}{x(x-y)(x^2+xy+y^2)} =$
$= \frac{xy^2}{x-y}$

3) 1)
$\frac{8a^2+2a}{(2a-1)(4a^2+2a+1)}-\frac{2a+1}{4a^2+2a+1} =$
$= \frac{8a^2+2a-(2a+1)(2a-1)}{(2a-1)(4a^2+2a+1)} =$
$= \frac{8a^2+2a-4a^2+1}{(2a-1)(4a^2+2a+1)} =$
$= \frac{4a^2+2a+1}{(2a-1)(4a^2+2a+1)} =$
$= \frac{1}{2a-1}$
2)
$1+\frac{2a+1}{2a}-\frac{2a(2a+5)}{2a(2a+1)} =$
$= 1+\frac{2a+1}{2a}-\frac{2a+5}{2a+1} =$
$= \frac{2a(2a+1)+(2a+1)^2-2a(2a+5)}{2a(2a+1)} =$
$= \frac{4a^2+2a+4a^2+4a+1-4a^2-10a}{2a(2a+1)} =$
$= \frac{4a^2-4a+1}{2a(2a+1)} =$
$= \frac{(2a-1)^2}{2a(2a+1)}$
3)
$\frac{1}{2a-1}\cdot\frac{(2a-1)^2}{2a(2a+1)} =$
$= \frac{2a-1}{2a(2a+1)}$

4) 1)
$\frac{x-3}{x^2-3x+9}+\frac{x+9}{(x+3)(x^2-3x+9)} =$
$= \frac{(x-3)(x+3)+x+9}{x^3+27} =$
$= \frac{x^2-9+x+9}{x^3+27} =$
$= \frac{x^2+x}{x^3+27}$
2)
$\frac{x^3+27}{x-1}\cdot\frac{x^2+x}{x^3+27} =$
$= \frac{x^2+x}{x-1}$
3)
$\frac{x^2+x}{x-1}:\frac{x^2+x}{x^2-1} =$
$= \frac{x^2+x}{x-1}\cdot\frac{(x-1)(x+1)}{x^2+x} =$
$= x+1$

Відповідь: 1) $-\frac{1}{(a-5)^2}$;
2) $\frac{xy^2}{x-y}$;
3) $\frac{2a-1}{2a(2a+1)}$;
4) $x+1$.