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Category: Algebra

a-1-b-b-2-b-1-c-c-2-c-ab-a-2-b-2-solve-for-a-b-c-

Question Number 55888 by behi83417@gmail.com last updated on 05/Mar/19 $$\:\:\:\boldsymbol{\mathrm{a}}=\mathrm{1}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\mathrm{b}}=\mathrm{1}+\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{c}}^{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \\ $$$$\:\:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\::\:\:\boldsymbol{\mathrm{a}},\:\:\boldsymbol{\mathrm{b}},\:\:\boldsymbol{\mathrm{c}}. \\ $$ Answered by MJS last updated on…

x-4-2-2-x-2-x-2-2-0-x-

Question Number 121423 by Jamshidbek2311 last updated on 08/Nov/20 $${x}^{\mathrm{4}} −\mathrm{2}\sqrt{\mathrm{2}}{x}^{\mathrm{2}} −{x}+\mathrm{2}−\sqrt{\mathrm{2}}=\mathrm{0}\:\:\:{x}=? \\ $$ Answered by 2004 last updated on 08/Nov/20 $$\sqrt{\mathrm{2}}=\mathrm{t} \\ $$$$\mathrm{t}^{\mathrm{2}} −\mathrm{t}\left(\mathrm{2x}^{\mathrm{2}}…

a-2-1-b-2-b-2-c-2-b-4-ab-c-solve-for-a-b-c-

Question Number 55887 by behi83417@gmail.com last updated on 05/Mar/19 $$\:\:\:{a}^{\mathrm{2}} +\mathrm{1}={b}^{\mathrm{2}} \\ $$$$\:\:\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={b}^{\mathrm{4}} \\ $$$$\:\:\:{ab}={c} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\::\:\:\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}}. \\ $$ Answered by MJS last…

Let-A-and-B-are-matrices-in-R-2017-2017-such-that-A-1-A-B-1-B-1-and-det-A-1-2017-Find-det-B-

Question Number 55857 by Joel578 last updated on 05/Mar/19 $$\mathrm{Let}\:{A}\:\mathrm{and}\:{B}\:\mathrm{are}\:\mathrm{matrices}\:\mathrm{in}\:\mathbb{R}^{\mathrm{2017}×\mathrm{2017}} \:\mathrm{such}\:\mathrm{that}\: \\ $$$${A}^{−\mathrm{1}} \:=\:\left({A}\:+\:{B}\right)^{−\mathrm{1}} \:−\:{B}^{−\mathrm{1}} \\ $$$$\mathrm{and}\: \\ $$$$\mathrm{det}\left({A}^{−\mathrm{1}} \right)\:=\:\mathrm{2017} \\ $$$$\mathrm{Find}\:\:\mathrm{det}\left({B}\right) \\ $$ Terms…

Question-55805

Question Number 55805 by bshahid010@gmail.com last updated on 04/Mar/19 Commented by MJS last updated on 04/Mar/19 $$\mathrm{b}=\mathrm{0}\:\Rightarrow\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}+\mathrm{1}}={a}^{\mathrm{2}} \\ $$$${b}={a}^{\mathrm{3}} \:\Rightarrow\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}+\mathrm{1}}={a}^{\mathrm{2}} \\…

Question-186867

Question Number 186867 by Mingma last updated on 11/Feb/23 Answered by aba last updated on 11/Feb/23 $${x}=−\mathrm{5} \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}\right)^{−\mathrm{4}} =\left(\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{4}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$ Commented by…