Question Number 126930 by bramlexs22 last updated on 25/Dec/20 $$\:\:\sqrt{\mathrm{1997}×\mathrm{1996}×\mathrm{1995}×\mathrm{1994}+\mathrm{1}}\:=? \\ $$ Commented by bramlexs22 last updated on 25/Dec/20 $$\Rightarrow{let}\:\mathrm{1994}={a}\:\rightarrow{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)\left({a}+\mathrm{3}\right)+\mathrm{1}= \\ $$$$\Leftrightarrow\left({a}^{\mathrm{2}} +{a}\right)\left({a}^{\mathrm{2}} +\mathrm{5}{a}+\mathrm{6}\right)+\mathrm{1}\:= \\…
Question Number 126931 by arash sharifi last updated on 25/Dec/20 $${y}=\left(\mathrm{1}−{x}\right)^{{cosx}} \\ $$ Answered by mahdipoor last updated on 25/Dec/20 $${salam}\:{arash}\:{jan},{soalat}\:{ro}\:{khamel}\:{benevis} \\ $$ Terms of…
Question Number 192463 by Spillover last updated on 18/May/23 Answered by AST last updated on 18/May/23 $$\mathrm{420}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}×\mathrm{7} \\ $$$$\mathrm{264}=\mathrm{2}^{\mathrm{3}} ×\mathrm{3}×\mathrm{11} \\ $$$${Third}\:{number}=\mathrm{2}^{{a}} \mathrm{3}^{{b}} \mathrm{5}^{{c}}…
Question Number 192440 by cortano12 last updated on 18/May/23 $$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\…
Question Number 192437 by MATHEMATICSAM last updated on 18/May/23 $$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} \:+\:{b}^{\mathrm{5}} \:+\:{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{5}} } \\ $$ Answered by Frix last…
Question Number 192425 by mehdee42 last updated on 17/May/23 $${Question} \\ $$$${if}\:\:“{k}''\:{is}\:{odd}\:\:\&\:{A}=\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +…+{n}^{{k}\:\:} \:\&\:\:{B}=\mathrm{1}+\mathrm{2}+…+{n} \\ $$$${prove}\:{that}\:\::\:\:{B}\:\mid\:{A}\: \\ $$ Answered by MM42 last updated on…
Question Number 126887 by mnjuly1970 last updated on 25/Dec/20 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{calculus}… \\ $$$${please}\:\:{solve}:\:\left(\:{with}\:{explanation}\right) \\ $$$$\:\:\:\:\left\{_{{x}^{\mathrm{2}} +{y}=\mathrm{3}} ^{{x}+{y}^{\mathrm{2}} =\mathrm{5}} \right. \\ $$$$ \\ $$ Commented by Lordose…
Question Number 126882 by Mathgreat last updated on 25/Dec/20 $$\boldsymbol{{x}}^{\boldsymbol{{x}}} =\mathrm{3} \\ $$ Answered by Lordose last updated on 25/Dec/20 $$\mathrm{x}^{\mathrm{x}} \:=\:\mathrm{3}\:\left\{\mathrm{x}^{\mathrm{x}} \:=\:\mathrm{e}^{\mathrm{xln}\left(\mathrm{x}\right)} \right\} \\…
Question Number 61343 by bhanukumarb2@gmail.com last updated on 01/Jun/19 Answered by MJS last updated on 01/Jun/19 $${x}={a}−\frac{\mathrm{1}}{{a}−\frac{\mathrm{1}}{{a}−…}} \\ $$$${x}={a}−\frac{\mathrm{1}}{{x}} \\ $$$${x}^{\mathrm{2}} −{ax}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{a}}{\mathrm{2}}\pm\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}…
Question Number 192409 by Abdullahrussell last updated on 17/May/23 Answered by Frix last updated on 17/May/23 $$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{2}} −\mathrm{3}{xyz} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}}…