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Question Number 140816 Answers: 0 Comments: 0
Question Number 140815 Answers: 0 Comments: 0
Question Number 140813 Answers: 0 Comments: 0
Question Number 142208 Answers: 1 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{which}\:\mathrm{is} \\ $$$$\mathrm{orthogonal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{circles}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{7}{x}−{y}=\mathrm{0} \\ $$$$\mathrm{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}{y}+\mathrm{5}=\mathrm{0}\:\mathrm{and}\:\mathrm{which}\:\mathrm{passes} \\ $$$$\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{3},\mathrm{0}\right). \\ $$
Question Number 142207 Answers: 2 Comments: 0
Question Number 142205 Answers: 0 Comments: 5
Question Number 142204 Answers: 1 Comments: 0
Question Number 140810 Answers: 0 Comments: 0
$${Find}\:{the}\:{natural}\:{value}\:{of}\:\boldsymbol{{x}}\:{that} \\ $$$${satisfies}\:{the}\:{equation}: \\ $$$$\frac{\mathrm{1}}{\mathrm{6}\centerdot\mathrm{7}}\:+\:\frac{\mathrm{1}}{\mathrm{7}\centerdot\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{8}\centerdot\mathrm{9}}\:+...+\:\frac{\mathrm{1}}{{x}\centerdot\left({x}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{12}} \\ $$
Question Number 140809 Answers: 1 Comments: 0
$$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:=\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right) \\ $$
Question Number 140805 Answers: 1 Comments: 0
$${find}\:{x}\:\mathrm{2cosh}\:\mathrm{2}{x}+\mathrm{10sinh}\:\mathrm{2}{x}=\mathrm{5} \\ $$
Question Number 140803 Answers: 1 Comments: 0
Question Number 140798 Answers: 0 Comments: 1
$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} \begin{pmatrix}{\mathrm{4}{n}}\\{{n}}\end{pmatrix}} \\ $$
Question Number 140793 Answers: 0 Comments: 0
$$\int\sqrt{\frac{{x}+\mathrm{2}}{{e}^{{x}} }}{dx}=...? \\ $$
Question Number 140786 Answers: 0 Comments: 0
Question Number 140785 Answers: 0 Comments: 2
$${P}-{intersection}\:{point}\:{of}\:{bimedians}\:{in} \\ $$$${ABCD}-{convexe}\:{quadrilateral}\:{with} \\ $$$${a};{b};{c};{d}-{sides},\:{e};{f}-{diagonals}, \\ $$$${E}-{point}\:{in}\:{plane},\:{x};{y};{z};{t}-{distances} \\ $$$${from},\:{E}-{to}\:{A};{B};{C};{D}.\:{Prove}\:{that}... \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)+\mathrm{4}{PE}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{t}^{\mathrm{2}} \\ $$
Question Number 140784 Answers: 0 Comments: 0
$${Let}\:{the}\:{i}-{j}\:{plane}\:{be}\:{the}\:{complex} \\ $$$$\:{plane},\:{with}\:{basic}\:{operations} \\ $$$$\:\:{ij}=−{i} \\ $$$$\:\:{ji}=−{j} \\ $$$$\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{j}^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}={r}+{xi}+{yj}\:\:\:\:{w}={s}+{pi}+{qj} \\ $$$${zw}={rs}+{pri}+{qrj}+{sxi}−{px}−{qxi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{syj}−{pyj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{qx}\right){i} \\ $$$$\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{py}\right){j} \\ $$$${wz}={rs}+{sxi}+{syj}+{pri}−{px}−{pyi} \\ $$$$\:\:\:\:\:\:\:\:\:\:+{qrj}−{qxj}−{qy} \\ $$$$\:\:\:=\:\left({rs}−{px}−{qy}\right)+\left({pr}+{sx}−{py}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left({qr}+{sy}−{qx}\right){j} \\ $$$$\left({little}\:{difference}..\right) \\ $$$$\Rightarrow\:\:{zw}−{wz}=\:\left({py}−{qx}\right)\left({i}−{j}\right) \\ $$$$\:\:\:=\:\mathrm{0}\:\:{if}\:\:\:{py}−{qx}=\:\mathrm{0} \\ $$$${z}^{\mathrm{2}} =\:\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)+\left(\mathrm{2}{rx}−{xy}\right){i} \\ $$$$\:\:\:\:\:\:\:\:\:+\left(\mathrm{2}{ry}−{xy}\right){j} \\ $$$${And}\:{if}\:\:{y}=\mathrm{0}\:\:\Rightarrow\:\:{z}={r}+{xi} \\ $$$${z}^{\mathrm{2}} =\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{rxi} \\ $$$$\:\:{either}\:{way}! \\ $$$$\left({z}^{\mathrm{2}} \right){z}={r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+\mathrm{2}{r}^{\mathrm{2}} {xi} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\:\left({z}^{\mathrm{2}} \right){z}=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$${z}\left({z}^{\mathrm{2}} \right)=\:{r}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)+{x}\left({r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{r}^{\mathrm{2}} {xi}−\mathrm{2}{rx}^{\mathrm{2}} \\ $$$$\:\:\:\:=\:{r}\left({r}^{\mathrm{2}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{x}\left(\mathrm{3}{r}^{\mathrm{2}} −{x}^{\mathrm{2}} \right){i} \\ $$$$\:{so}\:\:\:{z}\left({z}^{\mathrm{2}} \right)=\left({z}^{\mathrm{2}} \right){z}\:=\:{z}^{\mathrm{3}} \:\:\left({so}\:{far}\:{so}\:{good}\right) \\ $$$$..... \\ $$$$ \\ $$
Question Number 142200 Answers: 2 Comments: 0
$$\:{Let}\:{p}\:\&\:{q}\:{real}\:{positive}\:{number} \\ $$$$\:\:{what}\:{the}\:{minimum}\:{of}\:\left(\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }\:+\frac{{q}}{{p}}\right)^{\mathrm{3}} . \\ $$
Question Number 140779 Answers: 1 Comments: 1
$$\begin{cases}{\mathrm{7}{x}\equiv\mathrm{3}\left({mod}\mathrm{5}\right)}\\{\mathrm{5}{x}\equiv\mathrm{3}\left({mod}\mathrm{9}\right)}\end{cases} \\ $$$${solve}\:{for}\:{x} \\ $$
Question Number 140774 Answers: 3 Comments: 3
$${Prove}\:{that} \\ $$$$\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{5}}{\mathrm{3}\sqrt{\mathrm{6}}}\right)^{\mathrm{1}/\mathrm{3}} =\:\sqrt{\mathrm{2}} \\ $$
Question Number 140789 Answers: 2 Comments: 0
$$\:\:\:\:\:\:\:\:.......{nice}.....{math}.... \\ $$$$\:\:\:\:\:\:{calculate}::\:\Theta:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{}\:=?? \\ $$$$ \\ $$
Question Number 140768 Answers: 3 Comments: 0
$$\:\int_{−\infty} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{4}} +\mathrm{16}}\:\mathrm{dx}\:=? \\ $$
Question Number 140767 Answers: 1 Comments: 0
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\::\:\mathrm{2cot}\:^{\mathrm{2}} \mathrm{x}\:+\:\mathrm{csc}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2}\:=\:\mathrm{0}\: \\ $$
Question Number 140751 Answers: 1 Comments: 0
Question Number 140750 Answers: 1 Comments: 0
$$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\mathrm{cos}^{\mathrm{3}} \mathrm{k}=? \\ $$
Question Number 140749 Answers: 1 Comments: 0
$$\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{k}}} \left(\mathrm{1}−\mathrm{5}^{−\frac{\mathrm{1}}{\mathrm{k}\left(\mathrm{k}+\mathrm{1}\right)}} \right)=? \\ $$
Question Number 140748 Answers: 3 Comments: 0
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