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VT
27 tháng 6 2018
\(\left(a+b\right)^0=1\)
\(\left(a+b\right)^1=a+b\)
\(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\)
\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(\left(a+b\right)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)
Tổng quát:
\(\left(a+b\right)^n=C_0a^n+C_1a^{n-1}b+...+C_nb^n\)
Trong đó : C0, C1, ..., Cn là các hệ số trong tam giác cân Paxcan:
(a + b)^0 1 (a + b)^1 1 1 (a + b)^2 1 2 1 (a + b)^3 1 3 3 1 (a + b)^4 1 4 6 4 1 (a + b)^5 1 5 10 10 5 1 (a + b)^6 1 6 15 20 15 6 1 ........... ...........
Chúc bn học tốt <3
26 tháng 10 2019
Ta có:
\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)
\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)
\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)
\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)
...
\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)
\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)
=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)
\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)
5 tháng 10 2017
Bài 2 :
a ) \(A=\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(A=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
\(A=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right)\)
\(A=\left(a+b\right)^2+\left(a+c\right)^2+\left(b+c\right)^2\)
\(\left(a+b\right)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5\)\(b^5\)
Sorry, mk nhầm
\(\left(a+b\right)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+\)\(5ab^4+b^5\)