Question Number 224028 by behi834171 last updated on 15/Aug/25 Answered by Rasheed.Sindhi last updated on 15/Aug/25 $$\frac{\left({x}+{c}^{\mathrm{2}} \right)\left({x}−{ab}\right)}{\left({x}+{a}^{\mathrm{2}} \right)\left({x}+{b}^{\mathrm{2}} \right)}=\frac{{c}}{{a}+{b}} \\ $$$$\left({a}+{b}\right)\left({x}^{\mathrm{2}} +\left(−{ab}+{c}^{\mathrm{2}} \right){x}−{abc}^{\mathrm{2}} \right)…
Question Number 224029 by behi834171 last updated on 15/Aug/25 Answered by Ghisom last updated on 15/Aug/25 $$\mathrm{you}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$ Answered by fantastic last updated on…
Question Number 224030 by behi834171 last updated on 15/Aug/25 Answered by Rasheed.Sindhi last updated on 15/Aug/25 $${x}^{\mathrm{3}} \geqslant\mathrm{18}\Rightarrow{x}\neq\mathrm{1},\mathrm{2} \\ $$$$\mathrm{3}^{\mathrm{3}} =\mathrm{3}^{\mathrm{3}} +\mathrm{17}\:× \\ $$$$\mathrm{3}^{\mathrm{4}} =\mathrm{4}^{\mathrm{3}}…
Question Number 224041 by Tawa11 last updated on 15/Aug/25 Answered by mr W last updated on 19/Aug/25 $${say}\:{f}\left({x}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{kx}^{{k}} \\ $$$${since}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:…,\:{a}_{{n}} \:{are}\:{the}\:{roots}\:{of}…
Question Number 224042 by Tawa11 last updated on 15/Aug/25 Answered by mr W last updated on 16/Aug/25 $${let}\:{a}_{{n}} ={b}_{{n}} +{k} \\ $$$${b}_{{n}+\mathrm{2}} +{k}=\mathrm{2}\left({b}_{{n}+\mathrm{1}} +{k}\right)+{b}_{{n}} +{k}−\mathrm{1}…
Question Number 224034 by MirHasibulHossain last updated on 15/Aug/25 $$\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}} }=\mathrm{18}\sqrt{\mathrm{3}}\:.\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$ Answered by BaliramKumar last updated on 15/Aug/25 $${x}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{11}\sqrt{\mathrm{2}}\:+\mathrm{9}\sqrt{\mathrm{3}}\:\:} \\ $$ Answered…
Question Number 224015 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\sqrt{\mathrm{a}}\:+\:\sqrt{\mathrm{b}}\:+\:\sqrt{\mathrm{c}}\:\leqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\sqrt{\mathrm{abc}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224016 by hardmath last updated on 14/Aug/25 $$\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{2}=\mathrm{abc} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{a}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{b}}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}+\mathrm{c}}}\:\leqslant\:\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 224017 by hardmath last updated on 14/Aug/25 $$\mathrm{x},\mathrm{y},\mathrm{z}>\mathrm{0} \\ $$$$\mathrm{xy}+\mathrm{yz}+\mathrm{zx}+\mathrm{2xyz}=\mathrm{1} \\ $$$$\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }\:\leqslant\:\frac{\mathrm{3}\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 224018 by hardmath last updated on 14/Aug/25 $$\mathrm{x}\:\neq\:\mathrm{y} \\ $$$$\lambda\:\geqslant\:\mathrm{1} \\ $$$$\begin{cases}{\mathrm{x}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{y}\:−\:\lambda\right)^{\mathrm{2}} }\\{\mathrm{y}\:+\:\lambda^{\mathrm{2}} \:=\:\left(\mathrm{x}\:−\:\lambda\right)^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:−\:\mathrm{1}}\right)^{\mathrm{2025}} =\:\:? \\…