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Question Number 211956    Answers: 1   Comments: 0

Question Number 211954    Answers: 1   Comments: 0

x,y are rational numbers where x≠0, y≠0, x≠y, then is it possible: x^5 +y^5 =2x^2 y^2 ?

$$\:{x},{y}\:{are}\:{rational}\:{numbers}\:{where} \\ $$$$\:{x}\neq\mathrm{0},\:{y}\neq\mathrm{0},\:{x}\neq{y},\:{then}\:{is}\:{it}\: \\ $$$$\:{possible}:\:\:{x}^{\mathrm{5}} +{y}^{\mathrm{5}} =\mathrm{2}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \:? \\ $$

Question Number 211953    Answers: 1   Comments: 2

Question Number 211949    Answers: 1   Comments: 0

Question Number 211946    Answers: 2   Comments: 0

Question Number 211944    Answers: 1   Comments: 0

f(x)=(((√(1+x))−(√(1−x)))/( (√(1+x))+(√(1−x)))) f^′ (x)=?

$$ \\ $$$${f}\left({x}\right)=\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\:\:\:\:{f}^{'} \left({x}\right)=? \\ $$$$ \\ $$$$ \\ $$

Question Number 211943    Answers: 2   Comments: 1

2^(m−1) =1+mn m, n ∈Z

$$\mathrm{2}^{{m}−\mathrm{1}} =\mathrm{1}+{mn} \\ $$$${m},\:{n}\:\in\mathbb{Z} \\ $$

Question Number 211932    Answers: 0   Comments: 4

determiner R1 R2 et R3 segment de longueur a est tangent aux cercles 1et 2. MN//EF; EF=a; OM=ON=((3a)/2). (length a is tangent to cirles C1 (radius R1)and circldC2(radius R2)).

$$\mathrm{determiner}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{1}\:\:\:\:\boldsymbol{\mathrm{R}}\mathrm{2}\:\mathrm{et}\:\boldsymbol{\mathrm{R}}\mathrm{3} \\ $$$$\mathrm{segment}\:\mathrm{de}\:\mathrm{longueur}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{tangent}}\:\boldsymbol{\mathrm{aux}} \\ $$$$\boldsymbol{\mathrm{cercles}}\:\mathrm{1}\boldsymbol{\mathrm{et}}\:\mathrm{2}. \\ $$$$\:\boldsymbol{\mathrm{MN}}//\boldsymbol{\mathrm{EF}};\:\:\boldsymbol{\mathrm{EF}}=\boldsymbol{\mathrm{a}};\:\:\mathrm{OM}=\mathrm{ON}=\frac{\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}. \\ $$$$\left(\mathrm{length}\:\boldsymbol{\mathrm{a}}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{cirles}\:\mathrm{C1}\:\left(\mathrm{radius}\:\mathrm{R1}\right)\mathrm{and}\:\mathrm{circldC2}\left(\mathrm{radius}\:\mathrm{R2}\right)\right). \\ $$

Question Number 211920    Answers: 1   Comments: 0

Question Number 211919    Answers: 2   Comments: 0

Question Number 211918    Answers: 0   Comments: 0

Question Number 211917    Answers: 2   Comments: 0

Question Number 211908    Answers: 1   Comments: 0

Question Number 211902    Answers: 1   Comments: 1

Question Number 211896    Answers: 2   Comments: 0

Question Number 211895    Answers: 2   Comments: 0

Question Number 211886    Answers: 2   Comments: 0

Question Number 211885    Answers: 1   Comments: 0

Question Number 211881    Answers: 1   Comments: 0

Question Number 211880    Answers: 2   Comments: 0

Question Number 211879    Answers: 3   Comments: 0

Question Number 211873    Answers: 1   Comments: 0

Question Number 211872    Answers: 0   Comments: 0

Question Number 211871    Answers: 1   Comments: 0

Question Number 211863    Answers: 0   Comments: 1

Question Number 211861    Answers: 1   Comments: 0

ab^(−) + ba^(−) = 4c a + b + 3c = ?

$$\overline {\mathrm{ab}}\:\:+\:\:\overline {\mathrm{ba}}\:\:=\:\:\mathrm{4c} \\ $$$$\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{3c}\:=\:? \\ $$

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