Menu Close

Category: Algebra

Question-211509

Question Number 211509 by RojaTaniya last updated on 11/Sep/24 Answered by A5T last updated on 11/Sep/24 $$\frac{\mathrm{6}{a}+{b}}{\mathrm{6}{c}+{d}}={k}=\frac{\mathrm{5}{a}+{b}}{\mathrm{5}{c}+{d}} \\ $$$$\Rightarrow\mathrm{6}{a}+{b}=\mathrm{6}{ck}+{dk}…\left({i}\right);\mathrm{5}{a}+{b}=\mathrm{5}{ck}+{dk}…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right)\Rightarrow{a}={ck};\mathrm{6}\left({ii}\right)−\mathrm{5}\left({i}\right)\Rightarrow{b}={dk} \\ $$$$\mathrm{7}{a}+{b}=\mathrm{8}\left(\mathrm{7}{c}+{d}\right)\Rightarrow\mathrm{7}{ck}+{dk}=\mathrm{8}\left(\mathrm{7}{c}+{d}\right) \\ $$$$\Rightarrow{k}=\frac{\mathrm{8}\left(\mathrm{7}{c}+{d}\right)}{\mathrm{7}{c}+{d}}=\mathrm{8}…

determiner-les-valeurs-de-pet-q-sachant-que-2-et-3-sont-les-2-et-3-sont-les-racines-de-l-equation-2pqz-2-5z-4-p-q-0-

Question Number 211520 by a.lgnaoui last updated on 12/Sep/24 $$\mathrm{determiner}\:\mathrm{les}\:\mathrm{valeurs}\:\:\mathrm{de}\:\boldsymbol{\mathrm{p}}\mathrm{et}\:\boldsymbol{\mathrm{q}}\:\mathrm{sachant}\:\mathrm{que}\:−\mathrm{2}\:\mathrm{et}\:\mathrm{3}\:\mathrm{sont}\:\mathrm{les}\: \\ $$$$−\mathrm{2}\:\mathrm{et}\:\mathrm{3}\:\:\mathrm{sont}\:\mathrm{les}\:\mathrm{racines}\:\mathrm{de}\:\mathrm{l}\:\mathrm{equation}: \\ $$$$\mathrm{2}\boldsymbol{\mathrm{pqz}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{\mathrm{z}}−\mathrm{4}\left(\boldsymbol{\mathrm{p}}+\boldsymbol{\mathrm{q}}\right)=\mathrm{0} \\ $$$$ \\ $$ Commented by MrGaster last updated on…

If-a-b-c-b-c-a-c-a-b-1-and-a-b-c-0-then-prove-that-1-a-1-b-1-c-

Question Number 211485 by MATHEMATICSAM last updated on 10/Sep/24 $$\mathrm{If}\:\frac{{a}\:−\:{b}}{{c}}\:+\:\frac{{b}\:−\:{c}}{{a}}\:+\:\frac{{c}\:+\:{a}}{{b}}\:=\:\mathrm{1}\:\mathrm{and}\: \\ $$$${a}\:−\:{b}\:+\:{c}\:\neq\:\mathrm{0}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}}\:=\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:. \\ $$ Answered by Frix last updated on 10/Sep/24 $$\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}+{a}}{{b}}=\mathrm{1} \\…

Question-211398

Question Number 211398 by Spillover last updated on 08/Sep/24 Answered by Frix last updated on 08/Sep/24 $$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}\:\overset{{t}=\frac{{x}+\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}}{\:\sqrt{\mathrm{2}}}} {=} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\int\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2tan}^{−\mathrm{1}} \:\sqrt{\mathrm{2}}{t}\:= \\…

Question-211402

Question Number 211402 by efronzo1 last updated on 08/Sep/24 Answered by A5T last updated on 08/Sep/24 $${Q}\mathrm{211029},\:{we}\:{can}\:{find}\:{a}\:{polynomial}\:{f}\left({x}\right) \\ $$$${f}\left({x}\right)={x}^{{n}} \underset{−} {+}\mathrm{1} \\ $$$${f}\left(\mathrm{3}\right)=\mathrm{28}\Rightarrow{f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{1} \\…

Question-211393

Question Number 211393 by BaliramKumar last updated on 08/Sep/24 Commented by BaliramKumar last updated on 08/Sep/24 $${a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:=\:\mathrm{25}\:\:\:\:\:\:\:\:\:\:\:{a}+{b}\:=\:\mathrm{5} \\ $$$$\left({a}−{b}\right)\left({a}+{b}\right)\:=\:\mathrm{25} \\ $$$$\left({a}−{b}\right)\mathrm{5}\:=\:\mathrm{25} \\ $$$${a}−{b}\:=\:\mathrm{5}………\left({i}\right)…