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\(\left(\frac{x+2}{2x-4}-\frac{x-2}{2x+4}+\frac{8}{x^2-4}\right):\frac{4}{x-2}\)
\(=\left(\frac{x+2}{2\left(x-2\right)}-\frac{x-2}{2\left(x+2\right)}+\frac{8}{\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)
\(=\left(\frac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}-\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\frac{8x}{x\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)
\(=\left(\frac{x^2+4x+4-x^2+4x-4+8x}{2\left(x-2\right)\left(x+2\right)}\right):\frac{4}{x-2}\)
\(=\frac{16x}{2\left(x-2\right)\left(x+2\right)}.\frac{x-2}{4}=\frac{16x}{8\left(x+2\right)}=\frac{2x}{x+2}\)
\(\left(1+\frac{1}{x}\right)\left(1+\frac{1}{x+1}\right)\left(1+\frac{1}{x+2}\right)...\left(1+\frac{1}{x+99}\right)\)
\(=\frac{x+1}{x}.\frac{x+2}{x+1}.\frac{x+3}{x+2}...\frac{x+100}{x+99}=\frac{\left(x+1\right)\left(x+2\right)\left(x+3\right)...\left(x+100\right)}{x\left(x+1\right)\left(x+2\right)...\left(x+99\right)}=\frac{x+100}{x}\)
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 )
\(\frac{\left(x^3+1\right)\left(x^6+1\right)}{x^{24}+1}.\frac{\left(x^{12}+1\right)\left(x^{24}+1\right)}{x^{24}-1}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{24}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^{12}-1\right)}\)
\(=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^6-1\right)}=\frac{\left(x^3+1\right)\left(x^6+1\right)\left(x^{12}+1\right)\left(x^{24}+1\right)}{\left(x^{24}+1\right)\left(x^{12}+1\right)\left(x^6+1\right)\left(x^3+1\right)\left(x^3-1\right)}\)
\(=\frac{1}{x^3-1}\)