Question Number 158724 by HongKing last updated on 08/Nov/21 $$\mathrm{let}\:\:\mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{0}\:\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{ax}\:+\:\mathrm{by}\:+\:\mathrm{cz}\:=\:\mathrm{a}}\\{\mathrm{bx}\:+\:\mathrm{cy}\:+\:\mathrm{az}\:=\:\mathrm{b}}\\{\mathrm{cx}\:+\:\mathrm{ay}\:+\:\mathrm{bz}\:=\:\mathrm{c}}\end{cases} \\ $$$$ \\ $$ Answered by ajfour last updated on 08/Nov/21 $${x}+{y}+{z}=\mathrm{1} \\…
Question Number 93177 by ar247 last updated on 11/May/20 $${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{47} \\ $$$$\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=… \\ $$ Commented by ar247 last updated on 11/May/20 $${help} \\…
Question Number 158711 by HongKing last updated on 07/Nov/21 Commented by Rasheed.Sindhi last updated on 08/Nov/21 $$\mathrm{1000000} \\ $$ Commented by HongKing last updated on…
Question Number 93170 by ar247 last updated on 11/May/20 $${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{27} \\ $$$$\sqrt{{x}}+\frac{\mathrm{1}}{\:\sqrt{{x}}}=…. \\ $$ Commented by mr W last updated on 11/May/20 $$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}}…
Question Number 93173 by I want to learn more last updated on 11/May/20 $$\mathrm{If}\:\:\:\:\:\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{3}} \:\:=\:\:\frac{\mathrm{a}_{\mathrm{n}\:\:−\:\:\mathrm{1}} }{\mathrm{a}_{\mathrm{n}\:\:+\:\:\mathrm{1}} }\:,\:\:\:\:\mathrm{and}\:\:\:\mathrm{a}_{\mathrm{0}} \:\:=\:\:\mathrm{1},\:\:\:\mathrm{a}_{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\mathrm{find}\:\:\:\mathrm{a}_{\mathrm{n}} \\ $$ Answered by…
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Question Number 158698 by ajfour last updated on 07/Nov/21 $$\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \right)=\mathrm{4}{c}^{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 158687 by mahdipoor last updated on 07/Nov/21 $$\forall{x}\in{R}\:\:{f}\left({x}\right)={f}\:'\:\left({x}\right)\:\:\:\:\:\:{and}\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${prove}\:\:{f}\left({a}+{b}\right)={f}\left({a}\right)×{f}\left({b}\right) \\ $$ Answered by mr W last updated on 07/Nov/21 $${y}'=\frac{{dy}}{{dx}}={y} \\ $$$$\frac{{dy}}{{y}}={dx}…
Question Number 93138 by i jagooll last updated on 11/May/20 $$\begin{cases}{\mathrm{18x}^{\mathrm{2}} =\mathrm{3y}\left(\mathrm{1}+\mathrm{9x}^{\mathrm{2}} \right)}\\{\mathrm{18y}^{\mathrm{2}} =\mathrm{3z}\left(\mathrm{1}+\mathrm{9y}^{\mathrm{2}} \right)}\\{\mathrm{18z}^{\mathrm{2}} =\mathrm{3x}\left(\mathrm{1}+\mathrm{9z}^{\mathrm{2}} \right)}\end{cases} \\ $$ Answered by john santu last updated…
Question Number 158671 by mahdipoor last updated on 07/Nov/21 $$\forall{x}\in{R}\:;\:{f}\left({x}\right)={f}\:'\left({x}\right) \\ $$$${prove}\:{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right) \\ $$ Commented by mr W last updated on 07/Nov/21 $${generally} \\ $$$${for}\:{f}\left({x}+{y}\right)={f}\left({x}\right){f}\left({y}\right)\:{we}\:{have}\:…