Решение №16074: \(\left ( 6a^{2}+5a-1+\frac{a+4}{a+1} \right ):\left ( 3a-2+\frac{3}{a+1} \right )=\frac{\left ( a+1 \right )\left ( 6a^{2}+5a-1 \right )+a+4}{a+1}:\frac{\left ( a+1 \right )\left ( 3a-2 \right )+3}{a+1}=\frac{6a^{3}+11a^{2}+5a+3}{a+1}\cdot \frac{a+1}{3a^{2}+a+1}=\frac{6a^{3}+2a^{2}+2a+9a^{2}+3a+3}{3a^{2}+a+1}=\frac{2a\left ( 3a^{2}+a+1 \right )+3\left ( 3a^{2}+a+1 \right )}{3a^{2}+a+1}=2a+3\)

Ответ: \(2a+3\)

Решение №16075: \(\frac{x^{-6}-64}{4+2x^{-1}+x^{-2}}\cdot \frac{x^{2}}{4-\frac{4}{x}+\frac{1}{x^{2}}}-\frac{4x^{2}\left ( 2x+1 \right )}{1-2x}=\frac{\frac{1}{x^{6}-64}}{4+\frac{2}{x}+\frac{1}{x^{2}}}\cdot \frac{x^{2}}{\frac{4x^{2}-4x+1}{x^{2}}}-\frac{4x^{2}\left ( 2x+1 \right )}{1-2x}=\frac{\frac{1-64x^{6}}{x^{6}}}{\frac{4x^{2}+2x+1}{x^{2}}}\cdot \frac{x^{4}}{\left ( 2x-1 \right )^{2}}-\frac{4x^{2}\left ( 2x+1 \right )}{1-2x}=\frac{\left ( 1-4x^{2} \right )\left ( 1+4x^{2}+16x^{4} \right )}{\left ( 4x^{2}+2x+1 \right )\left ( 1-2x \right )^{2}}-\frac{4x^{2}\left ( 2x+1 \right )}{1-2x}=\frac{\left ( 1+2x \right )\left ( 1+4x^{2}+16x^{4} \right )-4x^{2}\left ( 2x+1 \right )\left ( 4x^{2}+2x+1 \right )}{\left ( 4x^{2}+2x+1 \right )\left ( 1-2x \right )}=\frac{\left ( 1+2x \right )\left ( 1-2x \right )\left ( 1+2x+4x^{2} \right )}{\left ( 4x^{2}+2x+1 \right )\left ( 1-2x \right )}=1+2x\)

Ответ: \(1+2x\)

Решение №16076: \(\frac{2b+a-\frac{4a^{2}-b^{2}}a{}}{b^{3}+2ab^{2}-3a^{2}b}\cdot \frac{a^{3}-2a^{2}b^{2}+ab^{3}}{a^{2}-b^{2}}=\frac{\frac{2ab-a^{2}-4a^{2}+b^{2}}{a}}{b\left ( b^{2}+2ab-3a^{2} \right )}\cdot \frac{ab\left ( a^{2}-2ab+b^{2} \right )}{\left ( a-b \right )\left ( a+b \right )}=\frac{\left ( a^{2}+2ab+b^{2} \right )-4a^{2}}{ab\left ( b+3a \right )\left ( b-a \right )}\cdot \frac{ab\left ( a-b \right )^{2}}{\left ( a-b \right )\left ( a+b \right )}=\frac{\left ( a+b \right )^{2}-4a^{2}}{-\left ( b+3a \right )\left ( a+b \right )}=-\frac{b-a}{a+b}=\frac{a-b}{a+b}\)

Ответ: \(\frac{a-b}{a+b}\)

Решение №16077: \(\frac{\left ( \frac{1}{a}+\frac{1}{b}-\frac{2c}{ab} \right )\left ( a+b+2c \right )}{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{2}{ab}-\frac{4c^{2}}{a^{2}b^{2}}}=\frac{\frac{a+b-2c}{ab}\left ( a+b+2c \right )}{\frac{a^{2}+2ab+b^{2}-4c^{2}}{a^{2}b^{2}}}=\frac{\left ( a+b-2c \right )\left ( a+b+2c \right )a^{2}b^{2}}{\left ( \left ( a+b \right )^{2}-\left ( 2c \right )^{2} \right )ab}=\frac{\left ( a+b-2c \right )\left ( a+b+2c \right )ab}{\left ( a+b-2c \right )\left ( a+b+2c \right )}=ab=7.4\cdot \frac{5}{37}=\frac{37}{5}\cdot \frac{5}{37}=1\)

Ответ: 1

Решение №16078: \(\left ( \frac{1}{t^{2}+3t+2}+\frac{2t}{t^{2}+4t+3}+\frac{1}{t^{2}+5t+6} \right )^{2}\cdot \frac{\left ( t-3 \right )^{2}+12t}{2}=\left ( \frac{t+3+2t\left ( t+2 \right )+t+1}{\left ( t+1 \right )\left ( t+2 \right )\left ( t+3 \right )} \right )^{2}\cdot \frac{t^{2}+6t+9}{2}=\frac{\left ( 2\left ( t+2 \right )\left ( t+1 \right ) \right )^{2}\left ( t+3 \right )^{2}}{2\left ( \left ( t+1 \right )\left ( t+2 \right )\left ( t+3 \right ) \right )^{2}}=2\)

Ответ: 2

Решение №16079: \(\left ( 2-x+4x^{2}+\frac{5x^{2}-6x+3}{x-1} \right ):\left ( 2x+1+\frac{2x}{x-1} \right )=\frac{\left ( 4x^{2}-x+2 \right )\left ( x-1 \right )+5x^{2}-6x+3}{x-1}:\frac{\left ( 2x+1 \right )\left ( x-1 \right )+2x}{x-1}=\frac{\left ( x^{3}+1 \right )+\left ( 3x^{3} -3x\right )}{\left ( x^{2}-1 \right )+\left ( x^{2}+x \right )}=\frac{4x^{2}-4x+1}{2x-1}=\frac{\left ( 2x-1 \right )^{2}}{2x-1}=2x-1\)

Ответ: \(2x-1\)

Решение №16080: \(\left ( \frac{2-b}{b-1} +2\frac{a-1}{a-2}\right ):\left ( b\frac{a-1}{b-1}+a\frac{2-b}{a-2} \right )=\frac{ab-2}{\left ( b-1 \right )\left ( a-2 \right )}\cdot \frac{\left ( b-1 \right )\left ( a-2 \right )}{a^{2}b-ab^{2}-2a+2b}=\frac{ab-2}{ab\left ( a-b \right )-2\left ( a-b \right )}=\frac{ab-2}{ab\left ( a-b \right )-2\left ( a-b \right )}=\frac{ab-2}{\left ( a-b \right )\left ( ab-2 \right )}=\frac{1}{a-b}=\frac{1}{\sqrt{2}+0.8-\sqrt{2}+0.2}=1\)

Ответ: 1

Решение №16081: \(\frac{1+\left ( a+x \right )^{-1}}{1-\left ( a+x \right )^{-1}}\cdot \left ( 1-\frac{1-\left ( a^{2}+x^{2} \right )}{2ax} \right )=\frac{1+\frac{1}{a+x}}{1-\frac{1}{a+x}}\cdot \frac{2ax-1+a^{2}+x^{2}}{2ax}=\frac{\frac{a+x+1}{a+x}}{\frac{a+x-1}{a+x}}\cdot \frac{a^{2}+2ax+x^{2}-1}{2ax}=\frac{a+x+1}{a+x-1}\cdot \frac{\left ( a+x \right )^{2}-1}{2ax}=\frac{\left ( a+x+1 \right )\left ( a+x+1 \right )\left ( a+x-1 \right )}{\left ( a+x-1 \right )2ax}=\frac{\left ( a+x+1 \right )^{2}}{2ax}=\frac{\left ( a+\frac{1}{a-1}+1 \right )^{2}}{\frac{2a}{a-1}}=\frac{a^{4}}{\left ( a-1 \right )^{2}}\cdot \frac{a-1}{2a}=\frac{a^{3}}{2\left ( a-1 \right )}\)

Ответ: \(\frac{a^{3}}{2\left ( a-1 \right )}\)

Решение №16082: \(\left ( \frac{a}{b} +\frac{b}{a}+2\right )\left ( \frac{a+b}{2a}-\frac{b}{a+b} \right ):\left ( \left ( a+2b+\frac{b^{2}}{a} \right )\left ( \frac{a}{a+b}+ \frac{b}{a-b}\right ) \right )=\frac{a^{2}+2ab+b^{2}}{ab}\cdot \frac{a^{2}+2ab+b^{2}-2ab}{2a\left ( a+b \right )}:\left ( \frac{a^{2}+2ab+b^{2}}{a}\cdot \frac{a^{2}-ab+ab+b^{2}}{\left ( a+b \right )\left ( a-b \right )} \right )=\frac{\left ( a+b \right )^{2}\left ( a^{2}+b^{2} \right )}{2a^{2}b\left ( a+b \right )}\cdot \frac{a\left ( a+b \right )\left ( a-b \right )}{\left ( a+b \right )^{2}\left ( a^{2}+b^{2} \right )}=\frac{a-b}{2ab}=\frac{0.75-\frac{4}{3}}{2\cdot 0.75\cdot \frac{4}{3}}=-\frac{7}{24}\)

Ответ: \(-\frac{7}{24}\)

Решение №16083: \(\frac{\left ( ab^{-1}+a^{-1}b+1 \right )\left ( a^{-1}-b^{-1} \right )^{2}}{a^{2}b^{-2}+a^{-2}b^{2}-\left ( ab^{-1}+a^{-1}b \right )}=\frac{\left ( \frac{a}{b}+\frac{b}{a}+1 \right )\left ( \frac{1}{a}-\frac{1}{b} \right )^{2}}{\frac{a^{2}}{b^{2}}+\frac{a^{2}}{b^{2}}-\left ( \frac{a}{b}+\frac{b}{a} \right )}=\frac{\left ( a^{2}+ab+b^{2} \right )\left ( a-b \right )^{2}}{a^{3}b^{3}}\cdot \frac{a^{2}b^{2}}{\left ( a^{4}-a^{3}b \right )-\left ( ab^{3}-b^{4} \right )}=\frac{\left ( a^{2}+ab+b^{2} \right )\left ( a-b \right )^{2}}{ab\left ( a-b \right )\left ( a^{3}-b^{3} \right )}=\frac{1}{ab}\)

Ответ: \(\frac{1}{ab}\)