Упр.206 ГДЗ Муравин 7 класс (Алгебра)

1) 2с + 4,7с — 3,4с — 1,3;
2) 7р — 5,5р — 8,7р + 9,1;
3) 2,6x — 3,8y — 1,2y + 1,4x + 5у;
4) 3,2k + 4,3р — 1,8k — 2,4k + 1,7р;
5) 2/3 x — 2/5 y — 5/12 x — 7/20 y — 1/6 x;
6) 7/15 a — 11/18 b — 1/20 a — 28/45 b + 1/15 b — 5/12 a.

1. $$2c+4{,}7c-3{,}4c-1{,}3=(2+4{,}7-3{,}4)c-1{,}3=3{,}3c-1{,}3.$$
2. $$7p-5{,}5p-8{,}7p+9{,}1=(7-5{,}5-8{,}7)p+9{,}1=-7{,}2p+9{,}1.$$
3. $$2{,}6x-3{,}8y-1{,}2y+1{,}4x+5y=(2{,}6+1{,}4)x-(3{,}8+1{,}2-5)y=4x.$$
4. $$3{,}2k+4{,}3p-1{,}8k-2{,}4k+1{,}7p=(3{,}2-1{,}8-2{,}4)k+(4{,}3+1{,}7)p=-k+6p.$$
5. $$\frac{2}{3}x-\frac{2}{5}y-\frac{5}{12}x-\frac{7}{20}y-\frac{1}{6}x= \left(\frac{2}{3}-\frac{5}{12}-\frac{1}{6}\right)x-\left(\frac{2}{5}+\frac{7}{20}\right)y.$$
   $$\left(\frac{2}{3}-\frac{5}{12}-\frac{1}{6}\right)=\frac{8}{12}-\frac{5}{12}-\frac{2}{12}=\frac{1}{12},$$
   $$\left(\frac{2}{5}+\frac{7}{20}\right)=\frac{8}{20}+\frac{7}{20}=\frac{15}{20}=\frac{3}{4}.$$
   Значит,
   $$\frac{2}{3}x-\frac{2}{5}y-\frac{5}{12}x-\frac{7}{20}y-\frac{1}{6}x=\frac{1}{12}x-\frac{3}{4}y.$$
6. $$\frac{7}{15}a-\frac{11}{18}b-\frac{1}{20}a-\frac{28}{45}b+\frac{1}{15}b-\frac{5}{12}a= \left(\frac{7}{15}-\frac{1}{20}-\frac{5}{12}\right)a-\left(\frac{11}{18}+\frac{28}{45}-\frac{1}{15}\right)b.$$
   $$\left(\frac{7}{15}-\frac{1}{20}-\frac{5}{12}\right)=\frac{28}{60}-\frac{3}{60}-\frac{25}{60}=0,$$
   $$\left(\frac{11}{18}+\frac{28}{45}-\frac{1}{15}\right)=\frac{55}{90}+\frac{56}{90}-\frac{6}{90}=\frac{105}{90}=\frac{7}{6}.$$
   Тогда
   $$\frac{7}{15}a-\frac{11}{18}b-\frac{1}{20}a-\frac{28}{45}b+\frac{1}{15}b-\frac{5}{12}a=-\frac{7}{6}b.$$

Ответ

1) $$3{,}3c-1{,}3$$;
2) $$-7{,}2p+9{,}1$$;
3) $$4x$$;
4) $$-k+6p$$;
5) $$\frac{1}{12}x-\frac{3}{4}y$$;
6) $$-\frac{7}{6}b$$.