A B C H K I F E

a) Tứ giác AHKI là hình vuông \(\Rightarrow S_{AHKI}=AH^2=2^2=4\left(cm^2\right)\)

b) Xét \(\Delta ABH\)và \(\Delta AFI\)có:

 +) \(\widehat{AIF}=\widehat{AHB}=90^o\)

+) \(AH=AI\)( vì \(AHKI\)là hình vuông )

+) \(\widehat{BAH}=\widehat{IAF}\)( cùng phụ với \(\widehat{HAC}\))

\(\Rightarrow\Delta ABH=\Delta AFI\left(g.c.g\right)\)\(\Rightarrow AB=AF\)

Xét tứ giác \(ABEF\)có: \(BE//AF\), \(AB//EF\), \(\widehat{BAC}=90^o\), \(AB=AF\)

\(\Rightarrow ABEF\)là hình vuông ( đpcm )

\(\frac{\left(x^3+1\right)\left(x^6+1\right)}{x^{24}+1}.\frac{\left(x^{12}+1\right)\left(x^{24}+1\right)}{x^{24}-1}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{24}-1\right)}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^{12}-1\right)}\)

\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^6-1\right)}=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^3+1\right)\left(x^3-1\right)}\)

\(=\frac{1}{x^3-1}\)

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hi bạn 2k7 à

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