làm ơn giúp mình với mình cần gấp:(((

ktvn100

2 năm trước

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`|x - 2019|^2020 + |x - 2020|^2021 = 1`

`<=> |x - 2019|^2020 = 1 - |x - 2020|^2021`

Vì `|x - 2019|^2020 >= 0 AA x`

`<=> 1 - |x - 2020|^2021 >= 0 AA x`

`<=> - |x - 2020|^2021 >= -1 AA x`

`<=> |x - 2020|^2021 <= 1 AA x`
mà `|x - 2020|^2021 >= 0 forall x`

`<=> 0 <= |x - 2020|^2021 <= 1`

`<=> |x - 2020|^2021 in {0; 1}`

`-)` Nếu `|x - 2020|^2021 = 0`

`<=> |x - 2020| = 0`

`<=> x - 2020 = 0`

`<=> x = 2020`

Thay `x = 2020` vào phương trình `|x - 2019|^2020 + |x - 2020|^2021 = 1`, ta có:

`|2020 - 2019|^2020 + |2020 - 2020|^2021 = 1`

`<=> |1|^2020 + |0|^2021 = 1`

`<=> 1 + 0 = 1`(luôn đúng)

`<=> x = 2020`(chọn)

`-)` Nếu `|x - 2020|^2021 = 1`

`<=> |x - 2020| = 1`

`<=>` $\left[ \begin{array}{l}x- 2020=1\\x- 2020 = -1\end{array} \right.$

`<=>` $\left[ \begin{array}{l}x = 2021\\x = 2019\end{array} \right.$

`+)` Thay `x = 2021` vào phương trình `|x - 2019|^2020 + |x - 2020|^2021 = 1`, ta có:

`|2021 - 2019|^2020 + |2021 - 2020|^2021 = 1`

`<=> |2|^2020 + |1|^2021 = 1`

`<=> 2^2020 + 1 = 1`(Vô lí)

`<=> x = 2021`(Loại)

`+)` Thay `x = 2019` vào phương trình `|x - 2019|^2020 + |x - 2020|^2021 = 1`, ta có:

`|2019 - 2019|^2020 + |2019 - 2020|^2021 = 1`

`<=> |0|^2020 + |-1|^2021 = 1`

`<=> 0 + 1 = 1`(Luôn đúng)

`<=> x = 2019`(Chọn)

Vậy `x in {2019; 2021}`

Vote (0) Phản hồi (0) 2 năm trước

khanhlyylhnahk

Đáp án:

Vậy $x \in {2019; 2021}$

Giải thích các bước giải:

${| x - 2019 |}^{2020} + {| x - 2020 |}^{2021} = 1$

$\Leftrightarrow {| x - 2019 |}^{2020} = 1 - {| x - 2020 |}^{2021}$

Vì ${| x - 2019 |}^{2020} \geq 0 \forall x$

$\Leftrightarrow 1 - {| x - 2020 |}^{2021} \geq 0 \forall x$

$\Leftrightarrow - {| x - 2020 |}^{2021} \geq - 1 \forall x$

$\Leftrightarrow {| x - 2020 |}^{2021} \leq 1 \forall x$
mà ${| x - 2020 |}^{2021} \geq 0 \forall x$

$\Leftrightarrow 0 \leq {| x - 2020 |}^{2021} \leq 1$

$\Leftrightarrow {| x - 2020 |}^{2021} \in {0; 1}$

Nếu ${| x - 2020 |}^{2021} = 0$

$\Leftrightarrow | x - 2020 | = 0$

$\Leftrightarrow x - 2020 = 0$

$\Leftrightarrow x = 2020$

Thay $x = 2020$ vào phương trình

${| x - 2019 |}^{2020} + {| x - 2020 |}^{2021} = 1$

${| 2020 - 2019 |}^{2020} + {| 2020 - 2020 |}^{2021} = 1$

$\Leftrightarrow {| 1 |}^{2020} + {| 0 |}^{2021} = 1$

$\Leftrightarrow x = 2020$ (chọn)

Nếu ${| x - 2020 |}^{2021} = 1$

$\Leftrightarrow | x - 2020 | = 1$

$\Leftrightarrow$ $[\begin{array}{l} x - 2020 = 1 \\ x - 2020 = - 1 \end{array}$

$\Leftrightarrow$ $[\begin{array}{l} x = 2021 \\ x = 2019 \end{array}$

Thay $x = 2021$ vào phương trình

${| x - 2019 |}^{2020} + {| x - 2020 |}^{2021} = 1$ , ta có:

${| 2021 - 2019 |}^{2020} + {| 2021 - 2020 |}^{2021} = 1$

$\Leftrightarrow {| 2 |}^{2020} + {| 1 |}^{2021} = 1$

$\Leftrightarrow 2^{2020} + 1 = 1$ (Vô lí)

$\Leftrightarrow x = 2021$ (Loại)

Thay $x = 2019$ vào phương trình

${| x - 2019 |}^{2020} + {| x - 2020 |}^{2021} = 1$ , ta có:

${| 2019 - 2019 |}^{2020} + {| 2019 - 2020 |}^{2021} = 1$

$\Leftrightarrow {| 0 |}^{2020} + {| - 1 |}^{2021} = 1$

$\Leftrightarrow x = 2019$

Vậy $x \in {2019; 2021}$

Vote (0) Phản hồi (0) 2 năm trước

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