Question Number 145314 by loveineq last updated on 04/Jul/21 $$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)\:=\:\left({a}+{b}\right)^{\mathrm{2}} \:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\left({a}+{b}\right)\sqrt{\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} }\:\leqslant\:\left({a}+\mathrm{1}\right)^{\mathrm{2}} +\left({b}+\mathrm{1}\right)^{\mathrm{2}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[\left({a}+\mathrm{1}\right)^{\mathrm{3}} +\left({b}+\mathrm{1}\right)^{\mathrm{3}} \right] \\ $$ Commented by justtry last…
Question Number 14244 by Ruth1 last updated on 30/May/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 30/May/17 $$\circledcirc\mathrm{10} \\ $$$$\frac{{jx}}{\mathrm{1}+{jy}}=\frac{\mathrm{3}{x}+{j}\mathrm{4}}{{x}+\mathrm{3}{y}}\Rightarrow{x}^{\mathrm{2}} {j}+\mathrm{3}{xyj}=\mathrm{3}{x}+{j}\mathrm{4}+\mathrm{3}{xyj}+\mathrm{4}{yj}^{\mathrm{2}} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −\mathrm{4}\right)−\left(\mathrm{3}{x}−\mathrm{4}{y}\right)=\mathrm{0}\Rightarrow \\ $$$$\begin{cases}{{x}^{\mathrm{2}}…
Question Number 145304 by imjagoll last updated on 04/Jul/21 $$\:\mathrm{1}+\frac{\mathrm{3x}}{\mathrm{1}!}\:+\frac{\mathrm{5x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{7x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{\mathrm{9x}^{\mathrm{4}} }{\mathrm{4}!}+…+\infty=? \\ $$ Answered by qaz last updated on 04/Jul/21 $$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)\mathrm{x}^{\mathrm{n}}…
Question Number 79757 by mathocean1 last updated on 27/Jan/20 $$\mathrm{And}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{4}{y}=\mathrm{0}. \\ $$$$\left({T}\right)\:\mathrm{is}\:\mathrm{his}\:\mathrm{his}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{at}\:\mathrm{M}\left({x}_{\mathrm{0}} ;{y}_{\mathrm{0}} \right) \\ $$$${passing}\:{by}\:{D}\left(\mathrm{2};\mathrm{1}\right). \\ $$$$ \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{y}\:\mathrm{verify}\:\mathrm{y}_{\mathrm{0}} ^{\mathrm{2}}…
Question Number 14211 by prakash jain last updated on 29/May/17 $${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{4} \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{9} \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{16} \\ $$ Commented by…
Question Number 145280 by mathdanisur last updated on 03/Jul/21 $${if}\:\:\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{x}}+\mathrm{2}\:\:{find}\:\:\frac{\boldsymbol{{x}}^{\mathrm{3}} +\mathrm{1}}{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by puissant last updated on 03/Jul/21 $$\mathrm{x}^{\mathrm{3}} =\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\:\Rightarrow\:\mathrm{x}^{\mathrm{3}}…
Question Number 145279 by mathdanisur last updated on 03/Jul/21 $$\mathrm{2}^{\boldsymbol{{a}}!} \:+\:\mathrm{2}^{\boldsymbol{{b}}!} \:+\:\mathrm{2}^{\boldsymbol{{c}}!} \:=\:\boldsymbol{{x}} \\ $$$${Find}\:{natural}\:{numbers}\:\boldsymbol{{a}};\boldsymbol{{b}};\boldsymbol{{c}}\:{such} \\ $$$${that}\:{the}\:{number}\:“\boldsymbol{{x}}''\:{is}\:{a}\:{cube}\:{of} \\ $$$${any}\:{number}. \\ $$ Answered by SinNombre last…
Question Number 145259 by mathdanisur last updated on 03/Jul/21 $$\int\:\frac{\left(\mathrm{3}\sqrt{{x}}+\mathrm{2}\right)^{\mathrm{5}} }{\:\sqrt{{x}}}\:{dx}\:=\:? \\ $$ Answered by Dwaipayan Shikari last updated on 03/Jul/21 $${put}\:\sqrt{{x}}={t}\Rightarrow\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}={dt} \\ $$$$=\mathrm{2}.\mathrm{3}^{\mathrm{5}} \int\left({t}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{5}}…
Question Number 145240 by loveineq last updated on 03/Jul/21 $$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}\:\mathrm{and}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{cyc}} {\sum}{a}^{\mathrm{3}} +\underset{{cyc}} {\sum}\left({a}+{b}\right)^{\mathrm{3}} \:\leqslant\:\mathrm{27} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\left({b}+{c}\right)^{\mathrm{3}} +\left({c}+{a}\right)^{\mathrm{3}} \:\geqslant\:\frac{\mathrm{1}}{\mathrm{2}}\left[{c}^{\mathrm{3}}…
Question Number 145238 by liberty last updated on 03/Jul/21 Answered by gsk2684 last updated on 03/Jul/21 $$\mathrm{x}=\mathrm{1}\:\mathrm{or}\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$ Answered by Rasheed.Sindhi last updated on…