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AlgebraQuestion and Answers: Page 29

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

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

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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)).

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

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ab^(−) + ba^(−) = 4c a + b + 3c = ?

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

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Compare: 5^(100) , 6^(91) , 7^(90) , 8^(85)

$$\mathrm{Compare}: \\ $$$$ \\ $$$$\mathrm{5}^{\mathrm{100}} \:,\:\mathrm{6}^{\mathrm{91}} \:,\:\mathrm{7}^{\mathrm{90}} \:,\:\mathrm{8}^{\mathrm{85}} \\ $$

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