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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
TQ
3
13 tháng 12 2020
8922 + 892.216 + 1082
= 8922 + 2.892.108 + 1082
= ( 892 + 108 )2
= 10002 = 1 000 000
362 + 262 - 52.36
= 362 - 2.36.26 + 262
= ( 36 - 26 )2
= 102 = 100
TB
0
\(\frac{\left(x^3+1\right)\left(x^6+1\right)}{x^{24}+1}.\frac{\left(x^{12}+1\right)\left(x^{24}+1\right)}{x^{24}-1}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{24}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^{12}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^6-1\right)}=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^3+1\right)\left(x^3-1\right)}\)
\(=\frac{1}{x^3-1}\)
hi nha bạn 2k7 à