x/5×3/7 , a×a³, c/4×c/5, t/9 × t³

Вопрос школьника по предмету Математика

Ответ:

Пошаговое объяснение:

0,2а+5)³=0,008а³+3*0,04а²+3*25а+125=0,008а³+0,12а²+125а+125

(4-0,5b)³=64-24b+3b²-0.125b³

(0.6c-5)³=0.216c³-5.4c²+15c-125

(1/2 d -2)³=1/8 d³ — 1.5d²+6d-8

(1/3 t+3)³=1/27 t³ +t²+9t+27

(2-1/4 k)³=8 — 3k+3/8 k-1/64 k³

(4x+.0.1y)³=64x³+4.8x²y+0.12xy²+0.001y³

(0.2a+30b)³=0.008a³+3.6a²b+540ab²+27000b³

(1/7 a — 7c)³=1/343 a³ — 3/7 a²c+21ac²-343c³

(0.3b-10c)³=0.027b³-2.7b²c+90bc²-1000c³

(0.5x-2y)³=0.125x³-1.5x²y+6xy²-8y³

(2/9 n+9/2 m)³=8/729 n³ +2/3 n²m+27/2 nm²+729/8 m³