5) x+1/x-1 + x-1/x+1 = 4/(x+1)(x-1) 6) x-1/x+1 – x+1/x-1 = 2/x^2-1

Question

5) x+1/x-1 + x-1/x+1 = 4/(x+1)(x-1)
6) x-1/x+1 – x+1/x-1 = 2/x^2-1

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Maria 3 phút 2021-10-17T04:32:34+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-10-17T04:33:49+00:00

    Đáp án:

    `\text{Em tham khảo!}`

    Giải thích các bước giải:

    `5)(x+1)/(x-1)+(x-1)/(x+1)=4/((x-1)(x+1))(x ne +-1)`

    `<=>(x+1)^2+(x-1)^2=4`

    `<=>x^2+2x+1+x^2-2x+1=4`

    `<=>2x^2+2=4`

    `<=>2x^2=2`

    `<=>x^2=1`

    `<=>x=+-1(loại)`

    `6)(x-1)/(x+1)-(x+1)/(x-1)=2/(x^2-1)(x ne +-1)`

    `<=>(x-1)^2-(x+1)^2=2`

    `<=>x^2-2x+1-x^2-2x-1=2`

    `<=>-4x=2`

    `<=>x=-1/2`

    0
    2021-10-17T04:33:50+00:00

    Đáp án:

     

    Giải thích các bước giải:

    `5)` `(x+1)/(x-1)+(x-1)/(x+1)=4/((x+1)(x-1))(ĐK:x\ne1;x\ne-1)`

    `↔((x+1)^2+(x-1)^2)/(x^2-1)=4/(x^2-1)`

    `→x^2+2x+1+x^2-2x+1=4`

    `↔2x^2-2=0`

    `↔2(x^2-1)=0`

    `↔2(x-1)(x+1)=0`

    `↔` \(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\) 

    `↔` \(\left[ \begin{array}{l}x=1(l)\\x=-1(l)\end{array} \right.\) 

    Vậy phương trình vô nghiệm

    `6)` `(x-1)/(x+1)-(x+1)/(x-1)=2/(x^2-1)(ĐK”x\ne1;x\ne-1)`

    `↔((x-1)^2-(x+1)^2)/(x^2-1)=2/(x^2-1)`

    `→x^2-2x+1-x^2-2x-1=2`

    `↔-4x=2`

    `↔x=-1/2` (thoả mãn ĐKXĐ)

    Vậy `S={-1/2}`

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