Упр.488 ГДЗ Муравин 7 класс (Алгебра)

1) (2ab-bc)/(a+b+c) при a = 1/2, b = 1/3, c = -7/12;
2) (p-k(p+k))/(2p-k) при p = 1,5, k = 0,5;
3) (a^2 -4ab+3b^2)/(3a-2b) при a = -2 1/3, b = -3;
4) (x^2 -3xy+5y^2)/(4y-x) при x = -2, y = -0,2;
5) (a^3 -b^3)/(a^2 +b(a+b)) при a = -1/2, b = -3/5;
6) (x^2 — y(x-y))/(x^3 +y^3) при x = 3/4, y = -2/3.

1. $$\frac{2ab-bc}{a+b+c}$$ при $$a=\frac12,\ b=\frac13,\ c=-\frac{7}{12}$$

   $$2ab-bc=2\cdot \frac12 \cdot \frac13-\frac13\cdot \left(-\frac{7}{12}\right)=\frac13+\frac{7}{36}=\frac{19}{36}$$

   $$a+b+c=\frac12+\frac13-\frac{7}{12}=\frac{6}{12}+\frac{4}{12}-\frac{7}{12}=\frac{3}{12}=\frac14$$

   $$\frac{2ab-bc}{a+b+c}=\frac{19}{36}:\frac14=\frac{19}{36}\cdot 4=\frac{19}{9}=2\frac19$$

2. $$\frac{p-k(p+k)}{2p-k}$$ при $$p=1{,}5,\ k=0{,}5$$

   $$p-k(p+k)=1{,}5-0{,}5\cdot(1{,}5+0{,}5)=1{,}5-0{,}5\cdot 2=1{,}5-1=0{,}5$$

   $$2p-k=2\cdot 1{,}5-0{,}5=3-0{,}5=2{,}5$$

   $$\frac{p-k(p+k)}{2p-k}=\frac{0{,}5}{2{,}5}=\frac15=0{,}2$$

3. $$\frac{a^2-4ab+3b^2}{3a-2b}$$ при $$a=-2\frac13,\ b=-3$$

   $$a^2-4ab+3b^2=(a-2b)^2-b^2=(a-3b)(a-b)$$

   Подставим значения:
   $$a=-\frac73,\quad b=-3$$

   $$a^2-4ab+3b^2=\left(-\frac73\right)^2-4\cdot\left(-\frac73\right)\cdot(-3)+3\cdot(-3)^2$$

   $$=\frac{49}{9}-28+27=\frac{49}{9}-1=\frac{40}{9}$$

   $$3a-2b=3\cdot\left(-\frac73\right)-2\cdot(-3)=-7+6=-1$$

   $$\frac{a^2-4ab+3b^2}{3a-2b}=\frac{40}{9}:(-1)=-\frac{40}{9}=-4\frac49$$

4. $$\frac{x^2-3xy+5y^2}{4y-x}$$ при $$x=-2,\ y=-0{,}2$$

   $$x^2-3xy+5y^2=(-2)^2-3\cdot(-2)\cdot(-0{,}2)+5\cdot(-0{,}2)^2$$

   $$=4-1{,}2+5\cdot 0{,}04=4-1{,}2+0{,}2=3$$

   $$4y-x=4\cdot(-0{,}2)-(-2)=-0{,}8+2=1{,}2$$

   $$\frac{x^2-3xy+5y^2}{4y-x}=\frac{3}{1{,}2}=\frac{30}{12}=\frac52=2{,}5$$

5. $$\frac{a^3-b^3}{a^2+b(a+b)}$$ при $$a=-\frac12,\ b=-\frac35$$

   $$a^2+b(a+b)=a^2+ab+b^2$$

   Тогда
   $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

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

   $$a-b=-\frac12-\left(-\frac35\right)=-\frac12+\frac35=-\frac{5}{10}+\frac{6}{10}=\frac{1}{10}=0{,}1$$

6. $$\frac{x^2-y(x-y)}{x^3+y^3}$$ при $$x=\frac34,\ y=-\frac23$$

   $$x^2-y(x-y)=x^2-xy+y^2$$

   $$x^3+y^3=(x+y)(x^2-xy+y^2)$$

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

   $$x+y=\frac34-\frac23=\frac{9}{12}-\frac{8}{12}=\frac{1}{12}$$

   $$\frac{1}{x+y}=\frac{1}{1/12}=12$$

Ответ

1) $$2\frac19$$; 2) $$0{,}2$$; 3) $$-4\frac49$$; 4) $$2{,}5$$; 5) $$0{,}1$$; 6) $$12$$.