Viết 13 hằng đẳng thức đánh nhớ theo cách cụ thể 03/07/2021 Bởi aihong Viết 13 hằng đẳng thức đánh nhớ theo cách cụ thể
`(a+b)^2=a^2+2ab+b^2` `(a-b)^2=a^2-2ab+b^2` `a^2-b^2=(a-b)(a+b)` `(a+b)^3=a^3+3a^2b+3ab^2+b^3` `(a-b)^3=a^3-3a^2b+3ab^2-b^3` `a^3+b^3=(a-b)(a^2-ab+b^2)` `a^3-b^3=(a+b)(a^2+ab+b^2)` `(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc` `(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc` `(a+b+c)^3=a^3+b^3+c^3+3(a-b)(-b-c)(a-c)` `a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc` `(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)` `(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc` Bình luận
1,$(a+b)²=a²+2ab+b²$ 2,$(a-b)²=a²-2ab+b²$ 3,$a^2-b^2=(a+b)(a-b)$ 4,$(a+b)^3=a^3+3a²b+3ab²+b^3$ 5,$(a-b)^3=a^3-3a²b+3ab²-b^3$ 6,$a^3+b^3=(a+b)(a^2-ab+b^2)$ 7,$a^3-b^3=(a-b)(a^2+ab+b^2)$ 8,$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+….+ab^{n-2}+b^{n-1})$ 9,$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+….-ab^{n-2}+b^{n-1})$ (n lẻ) 10,$(a_{1}+a_{2}+…+a_{n})^2=a_{1}^2+a_{2}^2+….+a_{n}^2+2(a_{1}a_{2}+a_{1}a_{3}+…+a_{1}a_{n}+a_{2}a_{3}+…+a_{n-1}a_{n}$ 11,$(a+b)^n=C_{n}^0a^n+C_{n}a^{n-1}b+…+C_{n}^nb^n$ $C_{n}^k=\frac{n!}{k!(n-k)!}$ 12,$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ac-bc-ab)$ 13,$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)$ Bình luận
`(a+b)^2=a^2+2ab+b^2`
`(a-b)^2=a^2-2ab+b^2`
`a^2-b^2=(a-b)(a+b)`
`(a+b)^3=a^3+3a^2b+3ab^2+b^3`
`(a-b)^3=a^3-3a^2b+3ab^2-b^3`
`a^3+b^3=(a-b)(a^2-ab+b^2)`
`a^3-b^3=(a+b)(a^2+ab+b^2)`
`(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc`
`(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc`
`(a+b+c)^3=a^3+b^3+c^3+3(a-b)(-b-c)(a-c)`
`a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)+3abc`
`(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)`
`(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc`
1,$(a+b)²=a²+2ab+b²$
2,$(a-b)²=a²-2ab+b²$
3,$a^2-b^2=(a+b)(a-b)$
4,$(a+b)^3=a^3+3a²b+3ab²+b^3$
5,$(a-b)^3=a^3-3a²b+3ab²-b^3$
6,$a^3+b^3=(a+b)(a^2-ab+b^2)$
7,$a^3-b^3=(a-b)(a^2+ab+b^2)$
8,$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+….+ab^{n-2}+b^{n-1})$
9,$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+….-ab^{n-2}+b^{n-1})$ (n lẻ)
10,$(a_{1}+a_{2}+…+a_{n})^2=a_{1}^2+a_{2}^2+….+a_{n}^2+2(a_{1}a_{2}+a_{1}a_{3}+…+a_{1}a_{n}+a_{2}a_{3}+…+a_{n-1}a_{n}$
11,$(a+b)^n=C_{n}^0a^n+C_{n}a^{n-1}b+…+C_{n}^nb^n$
$C_{n}^k=\frac{n!}{k!(n-k)!}$
12,$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ac-bc-ab)$
13,$(a+b+c)^3-a^3-b^3-c^3=3(a+b)(b+c)(c+a)$