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x-x-2-1-x-10-3-x-x-x-2-2-x-9-3-x-x-x-2-3-x-8-3-x-x-x-2-4-x-7-3-x-x-x-2-5-x-6-3-x-1-25-x-x-x-1-25-Z-A-




Question Number 164970 by Zaynal last updated on 24/Jan/22
  ⌊x(x^(2+1) /(x/(10−3^x ))) + x(x^(2+2) /(x/(9−3^x ))) + x(x^(2+3) /(x/(8−3^x ))) +x (x^(2+4) /(x/(7−3^x ))) + x(x^(2+5) /(x/(6−3^x ))) ≥ (1/((25)/x^x^(x ((1/(25))))  ))⌋     {Z.A}
$$\:\:\lfloor\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{1}} }{\frac{\boldsymbol{{x}}}{\mathrm{10}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{2}} }{\frac{\boldsymbol{{x}}}{\mathrm{9}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{3}} }{\frac{\boldsymbol{{x}}}{\mathrm{8}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\boldsymbol{{x}}\:\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{4}} }{\frac{\boldsymbol{{x}}}{\mathrm{7}−\mathrm{3}^{\boldsymbol{{x}}} }}\:+\:\boldsymbol{{x}}\frac{\boldsymbol{{x}}^{\mathrm{2}+\mathrm{5}} }{\frac{\boldsymbol{{x}}}{\mathrm{6}−\mathrm{3}^{\boldsymbol{{x}}} }}\:\geqslant\:\frac{\mathrm{1}}{\frac{\mathrm{25}}{\boldsymbol{{x}}^{\boldsymbol{{x}}^{\boldsymbol{{x}}\:\left(\frac{\mathrm{1}}{\mathrm{25}}\right)} } }}\rfloor \\ $$$$\:\:\:\left\{\mathrm{Z}.\mathrm{A}\right\} \\ $$$$\: \\ $$

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