Упр.963 ГДЗ Никольский Потапов 7 класс (Алгебра)
б) ((x-1)/(x2-x+1))+ (4x+5)/(x3+1)):(2-x)/(4×2-4x+4) при x=0,6;
в) (8a3-27b3)/((3b+2a)2-6ab) при a=2,5;b=-1*2/3;
г) (64a3+8b3)/((2a-b)2+2ab) при a=0,25;b=1*7/8.
а)
$$\left(\frac{x-2}{x^2-2x+4}-\frac{6x-13}{x^3+8}\right):\frac{15-5x}{2x^3+16}$$
$$=\left(\frac{x-2}{x^2-2x+4}-\frac{6x-13}{(x+2)(x^2-2x+4)}\right):\frac{5(3-x)}{2(x^3+8)}$$
$$=\frac{(x-2)(x+2)-(6x-13)}{(x+2)(x^2-2x+4)}\cdot \frac{2(x+2)(x^2-2x+4)}{5(3-x)}$$
$$=\frac{x^2-4-6x+13}{(x+2)(x^2-2x+4)}\cdot \frac{2(x+2)(x^2-2x+4)}{5(3-x)}$$
$$=\frac{x^2-6x+9}{1}\cdot \frac{2}{5(3-x)}=\frac{2(3-x)^2}{5(3-x)}=\frac{2(3-x)}{5}.$$При $$x=3{,}5$$:
$$\frac{2(3-3{,}5)}{5}=\frac{2\cdot(-0{,}5)}{5}=-\frac15=-0{,}2.$$
б)
$$\left(\frac{x-1}{x^2-x+1}+\frac{4x+5}{x^3+1}\right):\frac{2-x}{4x^2-4x+4}$$
$$=\left(\frac{x-1}{x^2-x+1}+\frac{4x+5}{(x+1)(x^2-x+1)}\right):\frac{2-x}{4(x^2-x+1)}$$
$$=\frac{(x-1)(x+1)+4x+5}{(x+1)(x^2-x+1)}\cdot \frac{4(x^2-x+1)}{2-x}$$
$$=\frac{x^2-1+4x+5}{x+1}\cdot \frac{4}{2-x}=\frac{x^2+4x+4}{x+1}\cdot \frac{4}{2-x}$$
$$=\frac{(x+2)^2}{x+1}\cdot \frac{4}{2-x}=\frac{4(x+2)^2}{(x+1)(2-x)}.$$При $$x=0{,}6$$:
$$\frac{4(0{,}6+2)^2}{(0{,}6+1)(2-0{,}6)}=\frac{4\cdot 2{,}6^2}{1{,}6\cdot 1{,}4}=\frac{4\cdot 6{,}76}{2{,}24}=\frac{169}{14}=12\frac{1}{14}.$$
в)
$$\frac{8a^3-27b^3}{(3b+2a)^2-6ab}$$
$$=\frac{(2a)^3-(3b)^3}{(2a+3b)^2-6ab}$$
$$=\frac{(2a-3b)(4a^2+6ab+9b^2)}{4a^2+12ab+9b^2-6ab}$$
$$=\frac{(2a-3b)(4a^2+6ab+9b^2)}{4a^2+6ab+9b^2}=2a-3b.$$При $$a=2{,}5$$, $$b=-1\frac{2}{3}=-\frac53$$:
$$2a-3b=2\cdot 2{,}5-3\cdot\left(-\frac53\right)=5+5=10.$$
г)
$$\frac{64a^3+8b^3}{(2a-b)^2+2ab}$$
$$=\frac{(4a)^3+(2b)^3}{(2a-b)^2+2ab}$$
$$=\frac{(4a+2b)(16a^2-8ab+4b^2)}{4a^2-4ab+b^2+2ab}$$
$$=\frac{2(2a+b)\cdot 4(4a^2-2ab+b^2)}{4a^2-2ab+b^2}=8(2a+b).$$При $$a=-0{,}25$$, $$b=1\frac78=\frac{15}{8}$$:
$$8(2a+b)=8\left(2\cdot(-0{,}25)+\frac{15}{8}\right)=8\left(-\frac12+\frac{15}{8}\right)=8\cdot\frac{11}{8}=11.$$
Ответ
а) $$-0{,}2$$; б) $$12\frac{1}{14}$$; в) $$10$$; г) $$11$$.
