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# номер 120 гдз 8 класс алгебра Муравин Муравин

1)$$\left ( \frac{2ab}{a^{2}-b^{2}}+\frac{a-b}{2a+2b} \right )\cdot \frac{2a}{a+b}+\frac{b}{b-a}$$ 1)$$\frac{2ab}{a^{2}-b^{2}}+\frac{a-b}{2\left ( a+b \right )}=$$ $$=\frac{4ab+\left ( a-b \right )^{2}}{2\cdot \left ( a^{2}-b^{2} \right )}=$$ $$=\frac{4ab+a^{2}-2ab+b^{2}}{2\cdot \left ( a^{2}-b^{2} \right )}=$$ $$=\frac{a^{2}+2ab+b^{2}}{2\cdot \left ( a^{2}-b^{2} \right )}=$$ $$=\frac{\left ( a+b \right )^{2}}{2\cdot \left ( a-b \right )\left ( a+b \right )}=$$ $$=\frac{a+b}{2\cdot \left ( a-b \right )}$$ 2)$$\frac{a+b}{2\cdot \left ( a-b \right )}\cdot \frac{2a}{a+b}=\frac{a}{a-b}$$ 3)$$\frac{a}{a-b}+\frac{b}{b-a}=\frac{a}{a-b}-\frac{b}{a-b}=$$ $$=\frac{a-b}{a-b}=1$$ 2)$$\left ( \frac{b}{b^{2}-36}-\frac{b-6}{b^{2}+6b} \right ):\frac{2b-6}{b^{2}+6b}-\frac{b}{b-6}$$ 1)$$\frac{b}{b^{2}-36}-\frac{b-6}{b\left ( b+6 \right )}=$$ $$=\frac{b^{2}-\left ( b-6 \right )^{2}}{b\left ( b^{2}-36 \right )}=$$ $$=\frac{b^{2}-b^{2}+12b-36}{b\left ( b^{2}-36 \right )}=$$ $$=\frac{6\left ( 2b-6 \right )}{b\left ( b^{2}-36 \right )}=$$ $$=\frac{12\left ( b-3 \right )}{b\left ( b^{2}-36 \right )}$$ 2)$$\frac{6\left ( 2b-6 \right )}{b\left ( b^{2}-36 \right )}:\frac{2b-6}{b^{2}+6b}=$$ $$=\frac{6\left ( 2b-6 \right )}{b\left ( b^{2}-36 \right )}\cdot \frac{b\left ( b+6 \right )}{2b-6}=$$ $$=\frac{6b\left ( b+6 \right )}{b\left ( b-6 \right )\left ( b+6 \right )}=\frac{6}{b-6}$$ 3)$$\frac{6}{b-6}-\frac{b}{b-6}=$$ $$=\frac{6-b}{b-6}=-1$$ 3)$$\left ( \frac{b+8}{b^{2}-8b}+\frac{b+24}{64-b^{2}} \right ):\frac{1}{b}-\frac{b}{b+8}$$ 1)$$\frac{b+8}{b\left ( b-8 \right )}-\frac{b+24}{b^{2}-64}=$$ $$=\frac{\left ( b+8 \right )^{2}-b^{2}-24b}{b\left ( b^{2}-64 \right )}=$$ $$=\frac{b^{2}+16b+64-b^{2}-24b}{b\left ( b^{2}-64 \right )}=$$ $$=\frac{64-8b}{b\left ( b^{2}-64 \right )}=\frac{8\left ( 8-b \right )}{b\left ( b^{2}-64 \right )}=$$ $$=\frac{-8\left ( b-8 \right )}{b\left ( b-8 \right )\left ( b+8 \right )}=-\frac{8}{b\left ( b+8 \right )}$$ 2)$$-\frac{8}{b\left ( b+8 \right )}:\frac{1}{b}=$$ $$=-\frac{8}{b\left ( b+8 \right )}\cdot \frac{b}{1}=-\frac{8}{b+8}$$ 3)$$-\frac{8}{b+8}-\frac{b}{b+8}=$$ $$=\frac{-8-b}{b+8}=-\frac{\left ( 8+b \right )}{b+8}=-1$$ 4)$$\left ( \frac{a+6}{2a-12}-\frac{18}{a^{2}-36} \right )\cdot \frac{a+6}{a+12}+\frac{1,5a-12}{a-6}$$ 1)$$\frac{a+6}{2\left ( a-6 \right )}-\frac{18}{a^{2}-36}=$$ $$=\frac{\left ( a+6 \right )^{2}-36}{2\left ( a^{2}-36 \right )}=$$ $$=\frac{a^{2}+12a+36-36}{2\left ( a^{2}-36 \right )}=$$ $$=\frac{a\left ( a+12 \right )}{2\left ( a^{2}-36 \right )}$$ 2)$$\frac{a\left ( a+12 \right )}{2\cdot \left ( a+6 \right )\left ( a-6 \right )}\cdot \frac{a+6}{a+12}=$$ $$=\frac{a}{2\left ( a-6 \right )}$$ 3)$$\frac{a}{2\left ( a-6 \right )}+\frac{1,5a-12}{a-6}=$$ $$=\frac{a+3a-24}{2\left ( a-6 \right )}=\frac{4a-24}{2\left ( a-6 \right )}=$$ $$=\frac{4\left ( a-6 \right )}{2\left ( a-6 \right )}=2$$ 5)$$\frac{y-13x}{5y-15x}+\frac{x^{3}-y^{3}}{9x^{3}-xy^{2}}:\frac{x^{2}+xy+y^{2}}{3x^{2}+xy}$$ 1)$$\frac{x^{3}-y^{3}}{9x^{3}-xy^{2}}:\frac{x^{2}+xy+y^{2}}{3x^{2}+xy}=$$ $$=\frac{x^{3}-y^{3}}{x\left ( 9x^{2}-y^{2} \right )}\cdot \frac{x\left ( 3x+y \right )}{x^{2}+xy+y^{2}}=$$ $$=\frac{\left ( x-y \right )\left ( x^{2}+xy+y^{2} \right )}{x\left ( 9x^{2}-y^{2} \right )}\cdot \frac{x\left ( 3x+y \right )}{x^{2}+xy+y^{2}}=$$ $$=\frac{\left ( x-y \right )\cdot x\left ( 3x+y \right )}{x\left ( 3x-y \right )\left ( 3x+y \right )}=$$ $$=\frac{x-y}{3x-y}$$ 2)$$\frac{y-13x}{5y-15x}+\frac{x-y}{3x-y}=$$ $$=\frac{y-13x}{5\left ( y-3x \right )}-\frac{x-y}{y-3x}=$$ $$=\frac{y-13x-5x+5y}{5\left ( y-3x \right )}=$$ $$=\frac{6y-18x}{5\left ( y-3x \right )}=$$ $$=\frac{6\left ( y-3x \right )}{5\left ( y-3x \right )}=\frac{6}{5}$$
ошибка в условии
9x³ - xy должно быть 9x³ - xy²
6)$$\frac{a^{3}-4ax^{2}}{a^{3}+x^{3}}\cdot \frac{a^{2}-ax+x^{2}}{a^{2}-2ax}+\frac{a-x}{2a+2x}$$ 1)$$\frac{a\left ( a^{2}-4x^{2} \right )\cdot \left ( a^{2}-ax+x^{2} \right )}{\left ( a+x \right )\left ( a^{2}-ax+x^{2} \right )\cdot a\left ( a-2x \right )}=$$ $$=\frac{\left ( a-2x \right )\left ( a+2x \right )}{\left ( a+x \right )\left ( a-2x \right )}=\frac{a+2x}{a+x}$$ 2)$$\frac{a+2x}{a+x}+\frac{a-x}{2\left ( a+x \right )}=$$ $$=\frac{2a+4x+a-x}{2\left ( a+x \right )}=\frac{3a+3x}{2\left ( a+x \right )}=$$ $$=\frac{3\left ( a+x \right )}{2\left ( a+x \right )}=\frac{3}{2}$$