Question

про числа A и B известно что A + 1 / B равно 7 и B + 1 / A равно 8 найдите значение выражения AB + 1 / A B

Answer 1

Составим систему и определим значения $A$ и $B:$

$left { {bigg{A + dfrac{1}{B} = 7 (1)} atop bigg{B + dfrac{1}{A} = 8 (2)}} right. \(1) A = 7 - dfrac{1}{B} = dfrac{7B - 1}{B}\(2) B + dfrac{1}{dfrac{7B - 1}{B}} = 8\B + dfrac{B}{7B - 1} = 8\dfrac{B(7B-1) + B}{7B-1} = 8\dfrac{7B^{2}}{7B - 1} = 8\ 7B^{2} = 8(7B - 1)\7B^{2} = 56B - 8$

$7B^{2} - 56B + 8 = 0\a = 7; b = -56; c = -8\D = b^{2} - 4ac = (-56)^{2} - 4 cdotp 7 cdotp 8 = 3136 - 224 = 2912\B_{1,2} = dfrac{-b pmsqrt{D}}{2a} = dfrac{-(-56) pmsqrt{2912}}{2 cdotp 7} = dfrac{56 pm 4sqrt{182}}{14} = dfrac{2(28 pm 2sqrt{182})}{14} =\\= dfrac{28 pm 2sqrt{182}}{7} = left[begin{array}{ccc}B_{1} = dfrac{28 + 2sqrt{182}}{7}\B_{2} = dfrac{28 - 2sqrt{182}}{7}end{array}right$

$(1) A_{1} = dfrac{7 cdotp dfrac{28 + 2sqrt{182}}{7} - 1}{dfrac{28 + 2sqrt{182}}{7}} = dfrac{(28 + 2sqrt{182} - 1) cdotp 7}{28 + 2sqrt{182}}} = dfrac{(27 + 2sqrt{182}) cdotp 7}{28 + 2sqrt{182}}} =\\= dfrac{189 + 14sqrt{182}}{28 + 2sqrt{182}}} = dfrac{(189 + 14sqrt{182})(28 - 2sqrt{182})}{(28 + 2sqrt{182})(28 - 2sqrt{182})}} = dfrac{7(27 + 2sqrt{182}) cdotp 2(14 - sqrt{182})}{784 - 4 cdotp 182} =$

$= dfrac{7(27 + 2sqrt{182}) cdotp 2(14 - sqrt{182})}{56} = dfrac{(27 + 2sqrt{182})(14 - sqrt{182})}{4} = \\= dfrac{378 - 27sqrt{182} + 28sqrt{182} - 364}{4} = dfrac{14 + sqrt{182}}{4}$

$(1) A_{2} = dfrac{7 cdotp dfrac{28 - 2sqrt{182}}{7} - 1}{dfrac{28 - 2sqrt{182}}{7}} = dfrac{(28 - 2sqrt{182} - 1) cdotp 7}{28 - 2sqrt{182}}} = dfrac{(27 - 2sqrt{182}) cdotp 7}{28 - 2sqrt{182}}} =\\= dfrac{189 - 14sqrt{182}}{28 - 2sqrt{182}}} = dfrac{(189 - 14sqrt{182})(28 + 2sqrt{182})}{(28 - 2sqrt{182})(28 + 2sqrt{182})}} = dfrac{7(27 - 2sqrt{182}) cdotp 2(14 + sqrt{182})}{784 - 4 cdotp 182} =$

$= dfrac{7(27 - 2sqrt{182}) cdotp 2(14 + sqrt{182})}{56} = dfrac{(27 - 2sqrt{182})(14 + sqrt{182})}{4} = \\= dfrac{378 + 27sqrt{182} - 28sqrt{182} - 364}{4} = dfrac{14 - sqrt{182}}{4}$

Высчитываем выражение $AB + dfrac{1}{AB}$, подставляя значения букв $A$ и $B:$

$1) A_{1}B_{1} + dfrac{1}{A_{1}B_{1}} = dfrac{14 + sqrt{182}}{4} cdotp dfrac{28 + 2sqrt{182}}{7} + dfrac{1}{dfrac{14 + sqrt{182}}{4} cdotp dfrac{28 + 2sqrt{182}}{7}} = \\= dfrac{(14 + sqrt{182}) cdotp 2(14 + sqrt{182})}{4 cdotp 7} + dfrac{1}{dfrac{(14 + sqrt{182}) cdotp 2(14 + sqrt{182})}{4 cdotp 7}} =\= dfrac{(14 + sqrt{182})^{2}}{14} + dfrac{1}{dfrac{(14 + sqrt{182})^{2}}{14}} = dfrac{378 + 28sqrt{182}}{14} + dfrac{1}{dfrac{378 + 28sqrt{182}}{14}} =$

$= dfrac{14(27 + 2sqrt{182})}{14} + dfrac{1}{dfrac{14(27 + 2sqrt{182})}{14}} = 27 + 2sqrt{182} + dfrac{1}{27 + 2sqrt{182}} =\\= dfrac{(27 + 2sqrt{182})^{2} + 1}{27 + 2sqrt{182}} = dfrac{1457 + 108sqrt{182} + 1}{27 + 2sqrt{182}} = dfrac{1458 + 108sqrt{182}}{27 + 2sqrt{182}} = dfrac{54(27 + 2sqrt{182})}{27 + 2sqrt{182}} =\\= 54$

$2) A_{2}B_{2} + dfrac{1}{A_{2}B_{2}} = dfrac{14 - sqrt{182}}{4} cdotp dfrac{28 - 2sqrt{182}}{7} + dfrac{1}{dfrac{14 - sqrt{182}}{4} cdotp dfrac{28 - 2sqrt{182}}{7}} = \\= dfrac{(14 - sqrt{182}) cdotp 2(14 - sqrt{182})}{4 cdotp 7} + dfrac{1}{dfrac{(14 - sqrt{182}) cdotp 2(14 - sqrt{182})}{4 cdotp 7}} =\= dfrac{(14 - sqrt{182})^{2}}{14} + dfrac{1}{dfrac{(14 - sqrt{182})^{2}}{14}} = dfrac{378 - 28sqrt{182}}{14} + dfrac{1}{dfrac{378 - 28sqrt{182}}{14}} =$

$= dfrac{14(27 - 2sqrt{182})}{14} + dfrac{1}{dfrac{14(27 - 2sqrt{182})}{14}} = 27 - 2sqrt{182} + dfrac{1}{27 - 2sqrt{182}} =\\= dfrac{(27 - 2sqrt{182})^{2} + 1}{27 - 2sqrt{182}} = dfrac{1457 - 108sqrt{182} + 1}{27 - 2sqrt{182}} = dfrac{1458 - 108sqrt{182}}{27 - 2sqrt{182}} = dfrac{54(27 - 2sqrt{182})}{27 - 2sqrt{182}} =\\= 54$

Ответ: 54.