№ 7.22 Алгебра = № 12.22 Математика
Спростіть вираз:
1. $\frac{1+\frac{4}{m}}{1-\frac{4}{m}};$
2. $\frac{\frac{3p+m}{m}-1}{\frac{3p-m}{m}+1};$
3. $\frac{\frac{1}{4t}+\frac{1}{t}}{\frac{1}{4t^2}+\frac{1}{t^2}};$
4. $\frac{1-\frac{1}{x}}{x-\frac{2x-1}{x}};$
5. $\frac{\frac{m}{2-m}+\frac{2+m}{m}}{\frac{m}{2+m}+\frac{2-m}{m}};$
6. $\frac{\frac{1}{x-2}+\frac{1}{x+2}}{\frac{1}{x-2}-\frac{1}{x+2}}.$
Розв'язок:
1. $\frac{1+\frac{4}{m}}{1-\frac{4}{m}}=\frac{\frac{m+4}{m}}{\frac{m-4}{m}}=\frac{m+4}{m}\cdot\frac{m}{m-4}=$
$= \frac{m\left(m+4\right)}{m\left(m-4\right)}=\frac{m+4}{m-4};$
2. $\frac{\frac{3p+m}{m}-1}{\frac{3p-m}{m}+1}=\frac{\frac{3p+m-m}{m}}{\frac{3p-m+m}{m}}=$
$= \frac{\frac{3p}{m}}{\frac{3p}{m}}=\frac{3p}{m}\cdot\frac{m}{3p}=1;$
3. $\frac{\frac{1}{4t}+\frac{1}{t}}{\frac{1}{4t^2}+\frac{1}{t^2}}=\frac{\frac{1+4}{4t}}{\frac{1+4}{4t^2}}=$
$= \frac{5}{4t}:\frac{5}{4t^2}=\frac{5}{4t}\cdot\frac{4t^2}{5}=$
$= \frac{4t^2}{4t}=t;$
4. $\frac{1-\frac{1}{x}}{x-\frac{2x-1}{x}}=\frac{\frac{x-1}{x}}{\frac{x^2-\left(2x-1\right)}{x}}=$
$= \frac{x-1}{x}\cdot\frac{x}{x^2-2x+1}=\frac{x\left(x-1\right)}{x\left(x-1\right)^2}=$
$= \frac{1}{x-1};$
5. $\frac{\frac{m}{2-m}+\frac{2+m}{m}}{\frac{m}{2+m}+\frac{2-m}{m}}=\frac{\frac{m^2+\left(2-m\right)\left(2+m\right)}{m\left(2-m\right)}}{\frac{m^2+\left(2-m\right)\left(2+m\right)}{m\left(2+m\right)}}=$
$= \frac{m^2+4-m^2}{m\left(2-m\right)}\cdot\frac{m\left(2+m\right)}{m^2+4-m^2}=$
$= \frac{4m\left(2+m\right)}{4m\left(2-m\right)}=\frac{2+m}{2-m};$
6. $\frac{\frac{1}{x-2}+\frac{1}{x+2}}{\frac{1}{x-2}-\frac{1}{x+2}}=\frac{\frac{\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}}{\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}}=$
$= \frac{2x}{x^2-4}\cdot\frac{x^2-4}{4}=\frac{2x\left(x^2-4\right)}{4\left(x^2-4\right)}=\frac{x}{2}.$