Math Question and Answers


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 208819 by Ismoiljon_008 last updated on 23/Jun/24

	Answered by Ismoiljon_008 last updated on 23/Jun/24

	$h e l p p l e a s e$
	Answered by A5T last updated on 23/Jun/24

	$4 a^{4} + 1 = {(2 a^{2})}^{2} + 1^{2} = {(2 a^{2} + 1)}^{2} - {(2 a)}^{2}$ $= \underset{a}{\underset{⏟}{(2 a^{2} - 2 a + 1)}} \underset{b}{\underset{⏟}{(2 a^{2} + 1 + 2 a)}}$ $4 {(a + 1)}^{4} + 1 = {[2 {(a + 1)}^{2}]}^{2} + 1 = {[2 {(a + 1)}^{2} + 1]}^{2} - 4 {(a + 1)}^{2}$ $= [2 {(a + 1)}^{2} + 1 - 2 (a + 1)] [2 {(a + 1)}^{2} + 1 + 2 (a + 1)]$ $= \underset{b}{\underset{⏟}{[2 a^{2} + 2 a + 1]}} \underset{c}{\underset{⏟}{[2 a^{2} + 6 a + 5]}}$ $S o, 4 a^{4} + 1 = a b \Rightarrow 4 {(a + 1)}^{4} + 1 = b c$ $Q u e s t i o n \Rightarrow \frac{a b \times c d \times e f \times g h \times i j}{b c \times d e \times f g \times h i \times j k}$ $= \frac{a}{k} = \frac{2 {(1)}^{2} - 2 (1) + 1}{2 {(9)}^{2} + 6 (9) + 5} = \frac{1}{221}$
	Commented by Ismoiljon_008 last updated on 24/Jun/24

	$t h a n k y o u v e r y m u c h$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum