Упр.201 ГДЗ Колягин Ткачёва 7 класс (Алгебра)
1) 2,3a-0,7a+3,6a-1;
2) 0,48b+3+0,52b-3,7b;
3) 1/3 x+1/2 x-1/6 a-5/6 a+2;
4) 5/6 y-1/3 b-1/6 y+2/3 b-3;
5) 2,1m+n-3,2m+2n+1,1m-n;
6) 5,7p-2,7q+0,3p+0,8q+1,9q-p.
Вычислить:
1) (35/48)^3•(6/7)^3•(1 3/5)^2;
2) (14/15)^4•(3/7)^4•(2,5)^3;
3) (5^3/6^2 )^4•(2/5)^5•(3/5)^7;
4) (7^4/?15?^2 )^3•(5/7)^6•(3/7)^5.
$$2{,}3a-0{,}7a+3{,}6a-1=(2{,}3-0{,}7+3{,}6)a-1=5{,}2a-1.$$
$$0{,}48b+3+0{,}52b-3{,}7b=(0{,}48+0{,}52-3{,}7)b+3=-2{,}7b+3.$$
$$\frac13x+\frac12x-\frac16a-\frac56a+2=\left(\frac13+\frac12\right)x-\left(\frac16+\frac56\right)a+2$$
$$=\frac56x-a+2.$$$$\frac56y-\frac13b-\frac16y+\frac23b-3=\left(\frac56-\frac16\right)y+\left(\frac23-\frac13\right)b-3$$
$$=\frac23y+\frac13b-3.$$$$2{,}1m+n-3{,}2m+2n+1{,}1m-n=(2{,}1-3{,}2+1{,}1)m+(n+2n-n)$$
$$=0m+2n=2n.$$$$5{,}7p-2{,}7q+0{,}3p+0{,}8q+1{,}9q-p=(5{,}7+0{,}3-1)p+(-2{,}7+0{,}8+1{,}9)q$$
$$=5p+0q=5p.$$
$$\left(\frac{35}{48}\right)^3\cdot\left(\frac67\right)^3\cdot\left(1\frac35\right)^2 =\left(\frac{35}{48}\cdot\frac67\right)^3\cdot\left(\frac85\right)^2$$
$$=\left(\frac58\right)^3\cdot\left(\frac85\right)^2 =\frac{5^3\cdot 8^2}{8^3\cdot 5^2} =\frac58.$$$$\left(\frac{14}{15}\right)^4\cdot\left(\frac37\right)^4\cdot(2{,}5)^3 =\left(\frac{14}{15}\cdot\frac37\right)^4\cdot\left(\frac52\right)^3$$
$$=\left(\frac25\right)^4\cdot\left(\frac52\right)^3 =\frac{2^4\cdot 5^3}{5^4\cdot 2^3} =\frac25=0{,}4.$$$$\left(\frac{5^3}{6^2}\right)^4\cdot\left(\frac25\right)^5\cdot\left(\frac35\right)^7 =\frac{5^{12}}{6^8}\cdot\frac{2^5}{5^5}\cdot\frac{3^7}{5^7}$$
$$=\frac{5^{12}\cdot 2^5\cdot 3^7}{(2\cdot 3)^8\cdot 5^{12}} =\frac{2^5\cdot 3^7}{2^8\cdot 3^8} =\frac1{2^3\cdot 3} =\frac1{24}.$$$$\left(\frac{7^4}{15^2}\right)^3\cdot\left(\frac57\right)^6\cdot\left(\frac37\right)^5 =\frac{7^{12}}{15^6}\cdot\frac{5^6}{7^6}\cdot\frac{3^5}{7^5}$$
$$=\frac{7^{12}\cdot 5^6\cdot 3^5}{(3\cdot 5)^6\cdot 7^{11}} =\frac{7}{3} =2\frac13.$$
Ответ
1) $$5{,}2a-1$$; 2) $$-2{,}7b+3$$; 3) $$\frac56x-a+2$$; 4) $$\frac23y+\frac13b-3$$; 5) $$2n$$; 6) $$5p$$; 7) $$\frac58$$; 8) $$0{,}4$$; 9) $$\frac1{24}$$; 10) $$2\frac13$$.