Question

помогите пж нужно срочно​

Answer 1

Решение

$1) (0,6+2x)^3=(\frac{3}{5} +2x)^3=\frac{27}{125} +\frac{54}{25} x+\frac{36}{5} x^2+8x^3.\\ 2) (-x+5)^3=(5-x)^3=125-75x+15x^2-x^3=-x^3+15x^2-75x+125.\\ 3) (-9a+4b)^3=(4b-9a)^3=64b^4-432ab^2+972a^2b-729a^3.\\ 4) (-11x-7y)^3=(-11x)^3-3*(-11x)^2*7y+3*(-11)x*(7y)^2-(7y)^3=-1331x^3-3*121x^2*7y+3*(-11)x*49y^2-343y^3=-1331x^3-2541x^2y-1617xy^2-343y^3.\\ 5) (2a+b^4)^3=8a^3+3*4a^2b^4+3*2ab^8+b^{12}=8a^3+12a^2b^4+6ab^8+b^{12}.\\ 6)(-3p+q^3)^3=(q^3-3p)^3=q^9-9pq^6+27p^2q^3-27p^3.$

$7) (c^2-0,7c^3)^3=c^6-2,1c^7+1,47c^8-0,343c^9.$

Формулы

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.\\ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.$

Answer 2

Ответ:

1)(0,6+2x)3=(53+2x)3=12527+2554x+536x2+8x3.2)(−x+5)3=(5−x)3=125−75x+15x2−x3=−x3+15x2−75x+125.3)(−9a+4b)3=(4b−9a)3=64b4−432ab2+972a2b−729a3.4)(−11x−7y)3=(−11x)3−3∗(−11x)2∗7y+3∗(−11)x∗(7y)2−(7y)3=−1331x3−3∗121x2∗7y+3∗(−11)x∗49y2−343y3=−1331x3−2541x2y−1617xy2−343y3.5)(2a+b4)3=8a3+3∗4a2b4+3∗2ab8+b12=8a3+12a2b4+6ab8+b12.6)(−3p+q3)3=(q3−3p)3=q9−9pq6+27p2q3−27p3.

7) (c^2-0,7c^3)^3=c^6-2,1c^7+1,47c^8-0,343c^9.7)(c2−0,7c3)3=c6−2,1c7+1,47c8−0,343c9.

Формулы

\begin{gathered}(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3.\\ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3.\end{gathered}(a+b)3=a3+3a2b+3ab2+b3.(a−b)3=a3−3a2b+3ab2−b3.