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26 tháng 10 2019

Ta có: 

\(1^4+\frac{1}{4}=\left(1^2-1+\frac{1}{2}\right)\left(1^2+1+\frac{1}{2}\right)=\frac{1}{2}.\left(2+\frac{1}{2}\right)\)

\(2^4+\frac{1}{4}=\left(2^2-2+\frac{1}{2}\right)\left(2^2+2+\frac{1}{2}\right)=\left(2+\frac{1}{2}\right).\left(6+\frac{1}{2}\right)\)

\(3^4+\frac{1}{4}=\left(3^2-3+\frac{1}{2}\right)\left(3^2+3+\frac{1}{2}\right)=\left(6+\frac{1}{2}\right).\left(12+\frac{1}{2}\right)\)

\(4^4+\frac{1}{4}=\left(4^2-4+\frac{1}{2}\right)\left(4^2+4+\frac{1}{2}\right)=\left(12+\frac{1}{2}\right).\left(20+\frac{1}{2}\right)\)

...

\(19^4+\frac{1}{4}=\left(19^2-19+\frac{1}{2}\right)\left(19^2+19+\frac{1}{2}\right)=\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)\)

\(20^4+\frac{1}{4}=\left(20^2-20+\frac{1}{2}\right)\left(20^2+20+\frac{1}{2}\right)=\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)\)

=> \(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(19^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(20^4+\frac{1}{4}\right)}\)

\(=\frac{\frac{1}{2}\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)...\left(342+\frac{1}{2}\right).\left(380+\frac{1}{2}\right)}{\left(2+\frac{1}{2}\right)\left(6+\frac{1}{2}\right)\left(12+\frac{1}{2}\right)\left(20+\frac{1}{2}\right)...\left(380+\frac{1}{2}\right).\left(420+\frac{1}{2}\right)}\)

\(=\frac{\frac{1}{2}}{420+\frac{1}{2}}=\frac{1}{841}\)

15 tháng 10 2016

a) \(x^4-2x^3+2x-1\)

\(=x^4-x^3-x^3+2x-2+1\)

\(=\left(x^4-x^3\right)+\left(2x-2\right)-\left(x^3-1\right)\)

\(=x^3\left(x-1\right)+2\left(x-1\right)-\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x-1\right)\left(x^3+2-x^2-x-1\right)\)

\(=\left(x-1\right)\left(x^3-x^2-x+1\right)\)

\(=\left(x-1\right)\left[\left(x^3-x^2\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left[x^2\left(x-1\right)-\left(x-1\right)\right]\)

\(=\left(x-1\right)\left(x^2-1\right)\left(x-1\right)\)

\(=\left(x-1\right)^2\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^3\left(x+1\right)\)

b) \(x^4+2x^3+2x^2+2x+1\)

\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)\)

\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)\)

\(=\left(x^2+1\right)\left(x^2+1+2x\right)\)

\(=\left(x^2+1\right)\left(x+1\right)^2\)

4 tháng 8 2017

a, \(x^4-x^3-x^3+x^2-x^2+x+x-1\)\(1\)

=\(x^3\left(x-1\right)+x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)

=\(\left(x-1\right)\left(x^3+x^2-x+1\right)\)

b, \(\left(ab-1\right)^2+\left(a+b\right)^2\)

=\(a^2b^2-2ab+1+a^2+2ab+b^2\)

=\(a^2b^2+a^2+b^2+1\)

=\(a^2\left(b^2+1\right)+\left(b^2+1\right)\)

=\(\left(b^2+1\right)\left(a^2+1\right)\)

c,\(x^4+2x^3+2x^2+2x+1\)

=\(x^4+x^3+x^3+x^2+x^2+x+x+1\)

=\(x^3\left(x+1\right)+x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\)

=\(\left(x+1\right)\left(x^3+x^2+x+1\right)\)

=\(\left(x+1\right)^2\left(x^2+1\right)\)