Question

(a-2b)^5 разложить по формуле бинома Ньютона и упростить

Answer 1

1⋅((a)^2(−2b))+2⋅((a)(−2b))+1⋅((a)(−2b)^2)

a^2−4ab+4b^2

Answer 2

To expand the expression $(a-2b)^5$ using the binomial theorem, we can use the formula:

$\binom{n}{k} \cdot a^{n-k} \cdot b^k$

where $\binom{n}{k}$ is the binomial coefficient and is given by $\frac{n!}{k!(n-k)!}$.

For our case, $n = 5$ and $k$ will vary from 0 to 5. Let's calculate the coefficients and expand the expression:

(a-2b)^5 & = \binom{5}{0} \cdot a^5 \cdot (-2b)^0 + \binom{5}{1} \cdot a^4 \cdot (-2b)^1 + \binom{5}{2} \cdot a^3 \cdot (-2b)^2 \\ & + \binom{5}{3} \cdot a^2 \cdot (-2b)^3 + \binom{5}{4} \cdot a^1 \cdot (-2b)^4 + \binom{5}{5} \cdot a^0 \cdot (-2b)^5 \end{align*}\] Now, let's simplify each term: \[\begin{align*} \binom{5}{0} \cdot a^5 \cdot (-2b)^0 & = 1 \cdot a^5 \cdot 1 = a^5 \\ \binom{5}{1} \cdot a^4 \cdot (-2b)^1 & = 5 \cdot a^4 \cdot (-2b) = -10a^4b \\ \binom{5}{2} \cdot a^3 \cdot (-2b)^2 & = 10 \cdot a^3 \cdot (4b^2) = 40a^3b^2 \\ \binom{5}{3} \cdot a^2 \cdot (-2b)^3 & = 10 \cdot a^2 \cdot (-8b^3) = -80a^2b^3 \\ \binom{5}{4} \cdot a^1 \cdot (-2b)^4 & = 5 \cdot a^1 \cdot (16b^4) = 80ab^4 \\ \binom{5}{5} \cdot a^0 \cdot (-2b)^5 & = 1 \cdot a^0 \cdot (-32b^5) = -32b^5 \end{align*}\] Now, we can combine all the simplified terms: \[(a-2b)^5 = a^5 - 10a^4b + 40a^3b^2 - 80a^2b^3 + 80ab^4 - 32b^5\] So, the expanded and simplified form of \((a-2b)^5\) is \(a^5 - 10a^4b + 40a^3b^2 - 80a^2b^3 + 80ab^4 - 32b^5\).