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}} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 145231 by mathdanisur last updated on 03/Jul/21 $$\mathrm{1}<{a}\leqslant{b}\:\:{then}\:{find} \\ $$$$\underset{\:\boldsymbol{{a}}} {\overset{\:\boldsymbol{{b}}} {\int}}\:{tan}^{-\mathrm{1}} \left(\frac{\mathrm{3}{x}}{\mathrm{1}-\mathrm{2}{x}^{\mathrm{2}} }\right){dx}=? \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 145224 by mathdanisur last updated on 03/Jul/21 $$\sqrt{\mathrm{2}\sqrt{\mathrm{3}}\:+\:\mathrm{2}}\:-\:\sqrt{\sqrt{\mathrm{3}}\:-\:\sqrt{\mathrm{2}}}\:=\:? \\ $$ Answered by Olaf_Thorendsen last updated on 03/Jul/21 $$\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{6}+\mathrm{2}\sqrt{\mathrm{6}}+\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$ \\ $$$${x}\:=\:\sqrt{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}}−\sqrt{\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}}…
Question Number 14157 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 29/May/17 Commented by ajfour last updated on 30/May/17 $${kindly}\:{view}\:{my}\:{solution}\:{in} \\ $$$${Q}.\mathrm{14306}\:{solved}\:{by}\:{coordinate} \\ $$$${geometry}.. \\ $$ Commented by…