cho 1/a+1/b+1/c=1/a+b+c. cmr 1/a3+1/b3+1/c3=1/a3+b3+c3
K
Khách
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NM
2
XO
13 tháng 12 2020
\(\frac{\left(x^3+1\right)\left(x^6+1\right)}{x^{24}+1}.\frac{\left(x^{12}+1\right)\left(x^{24}+1\right)}{x^{24}-1}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{24}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^{12}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^6-1\right)}=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^3+1\right)\left(x^3-1\right)}\)
\(=\frac{1}{x^3-1}\)
H
1
H
1
ND
0
PT
1
13 tháng 12 2020
\(=x^2+4y^2+4xy+x^2-6x+9+1=\left(x+2y\right)^2+\left(x-3\right)^2+1\)
Ta có: \(\left(x+2y\right)^2\ge0;\left(x-3\right)^2\ge0\left(\forall x;y\right)\)
\(\Rightarrow\left(x+2y\right)^2+\left(x-3\right)^2+1\ge1>0\forall x;y\)
=> đpcm