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Category: Algebra

log-2x-1-x-2-1-log-x-1-3-2x-5-18-2x-

Question Number 143329 by Huy last updated on 13/Jun/21 $$\mathrm{log}_{\mathrm{2x}+\mathrm{1}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{log}_{\mathrm{x}+\mathrm{1}} \left(\mathrm{3}\sqrt{\mathrm{2x}+\mathrm{5}}+\mathrm{18}\right)=\mathrm{2x} \\ $$ Answered by Olaf_Thorendsen last updated on 13/Jun/21 $$\mathrm{log}_{\mathrm{2}{x}+\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{log}_{{x}+\mathrm{1}}…

Question-12257

Question Number 12257 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17 Answered by sma3l2996 last updated on 17/Apr/17 $$\left({a}+{b}−\mathrm{2}{ab}\right)^{\mathrm{2}} +\left({a}+{b}−\mathrm{1}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{4}{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\mathrm{4}{ab}^{\mathrm{2}} +\mathrm{2}{a}\left({b}−\mathrm{2}{ab}\right)\right)+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}}…

Question-12255

Question Number 12255 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17 Answered by mrW1 last updated on 16/Apr/17 $$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)\left({x}+\mathrm{4}\right)+\mathrm{1} \\ $$$$=\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({x}+\frac{\mathrm{5}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}\right)+\mathrm{1} \\ $$$$=\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \right]\left[\left({x}+\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \right]+\mathrm{1}…

f-x-ln-1-ln-x-f-x-ln-f-x-Find-x-

Question Number 143293 by Huy last updated on 12/Jun/21 $${f}\left({x}\right)={ln}\left(\mathrm{1}+{ln}\left({x}\right)\right). \\ $$$${f}\left({x}\right)={ln}\left({f}'\left({x}\right)\right).{Find}\:{x} \\ $$ Answered by Olaf_Thorendsen last updated on 12/Jun/21 $${f}\left({x}\right)\:=\:\mathrm{ln}\left(\mathrm{1}+\mathrm{ln}{x}\right) \\ $$$${f}'\left({x}\right)\:=\:\frac{\frac{\mathrm{1}}{{x}}}{\mathrm{1}+\mathrm{ln}{x}} \\…

Question-143287

Question Number 143287 by enter last updated on 12/Jun/21 Answered by Huy last updated on 12/Jun/21 $$\mathrm{Dieu}\:\mathrm{kien}:\:\mathrm{x}\geqslant−\mathrm{1} \\ $$$$\mathrm{Phuong}\:\mathrm{trinh}\:\mathrm{tuong}\:\mathrm{duong}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{5}\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)} \\ $$$$<=>\mathrm{2}\left(\mathrm{x}^{\mathrm{2}}…