![1) \\ \sqrt[3]{27} = \sqrt[3]{ {3}^{3} } = 3 \\ \sqrt[4]{16} = \sqrt[4]{ {2}^{4} } = 2 \\ \sqrt[3]{ - \frac{1}{8} } = \sqrt[3]{ - { \frac{1}{2} }^{3} } = - \frac{1}{2}](/tpl/images/1578/6979/23fcc.png)
![2) \\ \sqrt[3]{8} - \sqrt[4]{81} = \sqrt[3]{ {2}^{3} } - \sqrt[4]{ {3}^{4} } = 2 - 3 = - 1](/tpl/images/1578/6979/cd16d.png)
![\sqrt[3]{32 {a}^{6} {b}^{7} } = \sqrt[3]{ {2}^{3} \times {2}^{2} \times {( {a}^{2} )}^{3} \times {( {b}^{2} )}^{3} \times b} = 2 {a}^{2} {b}^{2} \sqrt[3]{4b}](/tpl/images/1578/6979/0cc07.png)
![\sqrt[3]{81} - {49}^{0.5} \times \sqrt[3]{24} = \sqrt[3]{ {3}^{3} \times 3} - \sqrt{ {7}^{2} } \times \sqrt[3]{ {2}^{3} \times 3} = 3 \sqrt[3]{3} - 7 \times 2 \sqrt[3]{3} = 3 \sqrt[3]{3} - 14 \sqrt[3]{3} = 11 \sqrt[3]{3}](/tpl/images/1578/6979/bd08e.png)
![\sqrt[3]{4 \sqrt{4 {m}^{6} } } = \sqrt[3]{4 \sqrt{ {2}^{2} {( {m}^{3} )}^{2} } } = \sqrt[3]{4 \times 2 \times {m}^{3} } = \sqrt[3]{ {2}^{3} {m}^{3} } = 2m](/tpl/images/1578/6979/a9bf2.png)
x - 3h kolesnie velosipedi;
y - 2h kolesnie;
togda po usloviu mozhno sostavit' sistemu:
x+y=20;togda x=20-y;
3x+2y=55;ili 3(20-y)+2y=55;
60-3y+2y=55,togda y=5.
Otvet:5 2h kolesnih velosipedov.
Esli reshat' bez uravnenii:
1)dopustim vse velosipedi trehkolesnie,togda bi bilo 20*3=60 koles;
no ih 55,togda,2)esli uvelichivat' chislo dvuhkolesnih velosipedov na 1, budet umen'shatsia chislo trehkolesnih na 1,a chislo koles umenshitsia na 3-2=1.
3) no nam nado umenshit chislo koles s 60 do 55,znachit dvuhkolesnih velosipedov 60-55=5(no etot sposob reshenia bolshoi i ne ochen' udobnii dlia ponimania).