\(A=\frac{2x^2+4x+6}{x^2+2x-3}+\frac{x-2}{x-1}+\frac{3}{x+3}\)

<=>\(A=\frac{2x^2+4x+6}{x\left(x-1\right)+3\left(x-1\right)}+\frac{\left(x-2\right)\left(x+3\right)+3\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)

<=>\(A=\frac{2x^2+4x+6}{\left(x+3\right)\left(x-1\right)}+\frac{x^2+x-6+3x-3}{\left(x-1\right)\left(x+3\right)}\)

<=>\(A=\frac{2x^2+4x+6+x^2+x-6+3x-3}{\left(x-1\right)\left(x+3\right)}\)

<=>\(A=\frac{3x^2+8x-3}{\left(x-1\right)\left(x+3\right)}=\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\frac{3x-1}{x-1}\)

A B C D E F H 10 8 17

a, Xét tam giác HFB và tam giác HEC 

^BHF = ^CHE ( đối đỉnh )

^HFB = ^HEC = 90⁰

Vậy tam giác HFB ~ tam giác HEC ( g.g ) (1) 

\(\Rightarrow\frac{HB}{HC}=\frac{HF}{HE}\)( tỉ số đồng dạng ) \(\Rightarrow HB.HE=HF.HC\)

b, Xét tam giác HFB và tam giác AEB ta có : 

^B _ chung 

^HFB = ^AEB = 90⁰

Vậy tam giác HFB ~ tam giác AEB (2) 

Từ (1) ; (2) suy ra : tam giác AEB ~ tam giác HEC 

\(\Rightarrow\frac{AE}{HE}=\frac{EB}{EC}\)( tỉ số đồng dạng ) \(\Rightarrow AE.EC=EB.HE\)

Ta có :\(\frac{x}{x^2+x+1}=\frac{1}{4}\)<=> x² + x + 1 = 4x <=> x² = 3x - 1

Suy ra : x³ = x ( 3x - 1 ) = 3x² - x = 3 ( 3x - 1 ) - x = 8x - 3

x⁴ = ( 8x - 3 ) x = 8x² - 3x = 8 ( 3x - 1 ) - 3x = 21x - 8

x⁵ = ( 21x - 8 ) x = 21x² - 8x = 21 ( 3x - 1 ) - 8x = 55x - 21

Áp dụng kq trên ta có : 

\(M=\frac{55x-21-\left(21x-8\right)-2\left(8x-3\right)-3\left(3x-1\right)+11x+4}{21x-8+8x-3-6\left(3x-1\right)-7x+5}=\frac{20x}{4x}=5\)

\(A=\frac{\left(1^4+4\right).\left(5^4+4\right).\left(9^4+4\right).....\left(21^4+4\right)}{\left(3^4+4\right).\left(7^4+4\right).\left(11^4+4\right).....\left(23^4+4\right)}\)

\(\Leftrightarrow A=\frac{\left(1+4\right).\left(4^2+1\right).\left(6^2+1\right).\left(8^2+1\right).\left(10^2+1\right)....\left(20^2+1\right).\left(22^2+1\right)}{\left(2^2+1\right).\left(4^2+1\right).\left(6^2+1\right).\left(8^2+1\right).\left(10^2+1\right).\left(12^2+1\right)....\left(24^2+1\right)}\)

\(\Leftrightarrow A=\frac{1+4}{\left(2^2+1\right).\left(24^2+1\right)}=\frac{5}{5.\left(24^2+1\right)}=\frac{1}{24^2+1}=\frac{1}{577}\)

Sai bạn nhé bạn phải cm là n^4+4=\(\left(\left(n-1\right)^2+1\right).\left(\left(n+1\right)^2+1\right)\))