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Author: Tinku Tara

Q-210956-im-read-leithold-book-again-in-this-book-1-define-ln-x-1-x-dx-x-x-gt-0-2-define-ln-e-1-1-e-dx-x-3-define-exp-x-y-ln-y-x-d-ln-u-du-1-u-d-ln-u-dx

Question Number 210967 by mahdipoor last updated on 24/Aug/24 $${Q}.\mathrm{210956} \\ $$$${im}\:{read}\:{leithold}\:{book}\:{again}\:,\:{in}\:{this}\:{book}\:: \\ $$$$\left.\mathrm{1}\right\}{define}\::\:{ln}\left({x}\right)=\int_{\mathrm{1}} ^{\:{x}} {dx}/{x}\:\:\:\:\:\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right\}{define}\::\:{ln}\left({e}\right)=\mathrm{1}=\int_{\mathrm{1}} ^{\:{e}} {dx}/{x} \\ $$$$\left.\mathrm{3}\right\}{define}\::\:{exp}\left({x}\right)={y}\:\Leftrightarrow\:{ln}\left({y}\right)={x} \\ $$$$\frac{{d}\left({ln}\left({u}\right)\right)}{{du}}=\frac{\mathrm{1}}{{u}}\:\Rightarrow\:\frac{{d}\left({ln}\left({u}\right)\right)}{{dx}}=\frac{{du}/{dx}}{{u}}\:\Rightarrow \\…

Question-210961

Question Number 210961 by RojaTaniya last updated on 24/Aug/24 Answered by A5T last updated on 24/Aug/24 $$\frac{{a}−\mathrm{1}+\mathrm{2}}{{a}−\mathrm{1}}+\frac{{b}−\mathrm{1}+\mathrm{2}}{{b}−\mathrm{1}}+\frac{{c}−\mathrm{1}+\mathrm{2}}{{c}−\mathrm{1}}=\mathrm{10} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}−\mathrm{1}}+\frac{\mathrm{1}}{{b}−\mathrm{1}}+\frac{\mathrm{1}}{{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=\mathrm{0};{ab}+{bc}+{ca}=\frac{{m}−\mathrm{1}}{{m}};{abc}=\frac{−{m}−\mathrm{1}}{{m}} \\ $$$$\Rightarrow\frac{{ab}+{bc}+{ca}−\mathrm{2}\left({a}+{b}+{c}\right)+\mathrm{3}}{{abc}−{ab}−{bc}−{ac}+{a}+{b}+{c}−\mathrm{1}}=\frac{\mathrm{7}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\frac{{m}−\mathrm{1}+\mathrm{3}{m}}{{m}}}{\frac{−{m}−\mathrm{1}+\mathrm{1}−{m}−{m}}{{m}}}=\frac{\mathrm{7}}{\mathrm{2}}\Rightarrow\frac{\mathrm{4}{m}−\mathrm{1}}{−\mathrm{3}{m}}=\frac{\mathrm{7}}{\mathrm{2}}…

Question-210948

Question Number 210948 by cherokeesay last updated on 23/Aug/24 Commented by cherokeesay last updated on 23/Aug/24 $$\mathrm{If}\:\mathrm{only}\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{accompanied}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{diagam}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{perfect}. \\ $$$${thanks}. \\ $$ Answered by…