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Category: Algebra

x-4x-16x-64x-4-2019-x-3-x-1-x-

Question Number 128850 by ruwedkabeh last updated on 10/Jan/21 $$\sqrt{{x}+\sqrt{\mathrm{4}{x}+\sqrt{\mathrm{16}{x}+\sqrt{\mathrm{64}{x}+…+\sqrt{\mathrm{4}^{\mathrm{2019}} {x}+\mathrm{3}}}}}}−\sqrt{{x}}=\mathrm{1} \\ $$$${x}=? \\ $$ Answered by mindispower last updated on 10/Jan/21 $$\Rightarrow\sqrt{\mathrm{4}{x}+……+\sqrt{\mathrm{4}^{\mathrm{2009}} {x}+\mathrm{3}}}=\mathrm{1}+\mathrm{2}\sqrt{{x}} \\…

find-some-of-all-real-x-such-that-4x-2-15x-17-x-2-4x-12-5x-2-16x-18-2x-2-5x-13-

Question Number 63291 by aliesam last updated on 02/Jul/19 $${find}\:{some}\:{of}\:{all}\:{real}\:{x}\:{such}\:{that} \\ $$$$ \\ $$$$\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{15}{x}+\mathrm{17}}{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{12}}\:=\:\frac{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{16}{x}+\mathrm{18}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}+\mathrm{13}} \\ $$ Commented by Prithwish sen last…

Prove-that-abc-1-a-1-b-1-c-1-for-0-a-b-c-1-

Question Number 63247 by Joel122 last updated on 01/Jul/19 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\sqrt{{abc}}\:+\:\sqrt{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)}\:\leqslant\:\mathrm{1} \\ $$$$\mathrm{for}\:\mathrm{0}\:\leqslant\:{a},{b},{c}\:\leqslant\:\mathrm{1} \\ $$ Commented by Tony Lin last updated on 01/Jul/19 $$\because\:{abc}+\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)\left(\mathrm{1}−{c}\right)\geqslant…

fine-the-number-thats-divided-by-3-4-5-and-9-and-produce-the-remanider-with-order-1-3-5-and-7-

Question Number 128715 by Study last updated on 09/Jan/21 $${fine}\:{the}\:{number}\:{thats}\:{divided}\:{by}\: \\ $$$$\mathrm{3},\mathrm{4},\mathrm{5}\:{and}\:\mathrm{9}\:{and}\:{produce}\:{the}\:{remanider} \\ $$$${with}\:{order}\:\mathrm{1},\mathrm{3},\mathrm{5}\:{and}\:\mathrm{7}?? \\ $$ Commented by mr W last updated on 10/Jan/21 $${do}\:{you}\:{mean}:…

if-P-x-1-2x-3-Q-x-x-2-find-P-1-4-Q-x-

Question Number 128712 by Study last updated on 09/Jan/21 $${if}\:{P}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{3}} \centerdot{Q}\left({x}\right)+{x}^{\mathrm{2}} \\ $$$${find}\:\frac{{P}\left(\mathrm{1}\right)−\mathrm{4}}{{Q}\left({x}\right)}=?? \\ $$ Commented by Adel last updated on 13/Jan/21 $$\mathrm{p}\left(\mathrm{1}\right)=\boldsymbol{\mathrm{p}}\left(\mathrm{2}−\mathrm{1}\right)=\mathrm{2}×\mathrm{2}^{\mathrm{3}} ×\boldsymbol{\mathrm{Q}}\left(\mathrm{x}\right)+\mathrm{2}^{\mathrm{2}} \\…

solve-for-x-x-x-x-16-x-2-but-how-to-use-Lambert-W-function-

Question Number 63175 by Tawa1 last updated on 30/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:\mathrm{x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \:=\:\:\mathrm{16} \\ $$$$\mathrm{x}\:=\:\mathrm{2},\:\:\:\:\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{Lambert}\:\mathrm{W}\:\mathrm{function} \\ $$ Commented by mr W last updated on 30/Jun/19 $${i}\:{remember}\:{you}\:{have}\:{asked}\:{the}\:{same}…

Find-the-set-of-values-of-x-which-satisfy-the-inequalities-2-x-1-1-x-and-x-2-3x-2-lt-0-

Question Number 63162 by Rio Michael last updated on 29/Jun/19 $${Find}\:{the}\:{set}\:{of}\:{values}\:{of}\:{x}\:{which}\:{satisfy}\:{the}\:{inequalities}\: \\ $$$$\frac{\mathrm{2}}{{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{{x}}\:\:{and}\:\:{x}^{\mathrm{2}} −\mid\mathrm{3}{x}\mid+\mathrm{2}<\mathrm{0} \\ $$ Commented by Prithwish sen last updated on 30/Jun/19 $$\frac{\mathrm{2}}{\mathrm{x}−\mathrm{1}}\leqslant\frac{\mathrm{1}}{\mathrm{x}}\:\Rightarrow\mathrm{x}\leqslant−\mathrm{1}…

how-we-can-convert-0-9-to-q-p-

Question Number 128684 by Study last updated on 09/Jan/21 $${how}\:{we}\:{can}\:{convert}\:\mathrm{0}.\overset{−} {\mathrm{9}}\:{to}\:\frac{{q}}{{p}}? \\ $$ Answered by liberty last updated on 09/Jan/21 $$\:\mathrm{let}\:\mathrm{r}\:=\:\mathrm{0},\mathrm{99999999999}… \\ $$$$\:\:\:\:\:\:\:\mathrm{10r}\:=\:\mathrm{9}.\mathrm{9999999999}… \\ $$$$\mathrm{substract}\:\Rightarrow\:\mathrm{9r}\:=\:\mathrm{9}\:\Rightarrow\:\mathrm{r}\:=\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\:\frac{\mathrm{2}}{\mathrm{2}}=\frac{\mathrm{n}}{\mathrm{n}}\:;\:\mathrm{n}\neq\:\mathrm{0}…