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Nguyệt Ánh 10 months 2021-04-22T05:51:57+00:00 1 Answers 11 views 0

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    2021-04-22T05:53:25+00:00

    Giải thích các bước giải:

    Ta có:

    $\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac1{x+y+z}=\dfrac{(y+z+1)+(x+z+2)+(x+y-3)}{x+y+z}=\dfrac{2(x+y+z)}{x+y+z}=2$

    $\to x+y+z=2, y+z+1=2x, x+z+2=2y, x+y-3=2z$

    Ta có $y+z+1=2x\to y+z=2x-1$

    Mà $x+y+z=2\to x+(2x-1)=2\to 3x-1=2\to 3x=3\to x=1$

    Ta có: $x+z+2=2y$

    $\to (x+y+z)+2=3y$

    $\to 2+2=3y$

    $\to y=\dfrac43$

    Ta có $x+y-3=2z$

    $\to x+y+z-3=3z$

    $\to 2-3=3z$

    $\to z=-\dfrac13$

    $\to A=2016\cdot 1+(\dfrac43)^{2017}+(-\dfrac13)^{2017}$

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