1- √x-2019 trên x-2019 +1- √y-2020 tên y-2020 +1- √z-2021 trên x-2021` +3 phần 4 =0 giúp mik với đề :giải pt 25/07/2021 Bởi Eden 1- √x-2019 trên x-2019 +1- √y-2020 tên y-2020 +1- √z-2021 trên x-2021` +3 phần 4 =0 giúp mik với đề :giải pt
Đáp án: `ĐK : x > 2019 ; y > 2020 ; z > 2021` `(1 – \sqrt{x – 2019})/(x – 2019) + (1 – \sqrt{y – 2020})/(y – 2020) + (1 – \sqrt{z – 2021})/(z – 2021) + 3/4 =0` `↔ 1/(x – 2019) – 2 . 1/(\sqrt{x – 2019}) . 1/2 + 1/4 + 1/(y – 2020) – 2 . 1/(\sqrt{y – 2020}) . 1/2 + 1/4 + 1/(z – 2021) – 2. 1/(\sqrt{z – 2021}) . 1/2 + 1/4 = 0` `↔ (1/(\sqrt{x – 2019}) – 1/2)^2 + (1/(\sqrt{y – 2020}) – 1/2)^2 + (1/(\sqrt{z – 2021}) – 1/2)^2 = 0` `↔ 1/(\sqrt{x – 2019}) – 1/2 = 1/(\sqrt{y – 2020}) – 1/2 = 1/(\sqrt{z – 2021}) – 1/2 = 0` `↔ 1/(\sqrt{x – 2019}) = 1/(\sqrt{y – 2020}) = 1/(\sqrt{z – 2021}) = 1/2` `↔ \sqrt{x – 2019} = \sqrt{y – 2020} = \sqrt{z – 2021} = 2` `↔ x – 2019 = y – 2020 = z – 2021 = 4` `↔ x = 2023 ; y = 2024 ; z = 2025` Giải thích các bước giải: Bình luận
Đáp án:
`ĐK : x > 2019 ; y > 2020 ; z > 2021`
`(1 – \sqrt{x – 2019})/(x – 2019) + (1 – \sqrt{y – 2020})/(y – 2020) + (1 – \sqrt{z – 2021})/(z – 2021) + 3/4 =0`
`↔ 1/(x – 2019) – 2 . 1/(\sqrt{x – 2019}) . 1/2 + 1/4 + 1/(y – 2020) – 2 . 1/(\sqrt{y – 2020}) . 1/2 + 1/4 + 1/(z – 2021) – 2. 1/(\sqrt{z – 2021}) . 1/2 + 1/4 = 0`
`↔ (1/(\sqrt{x – 2019}) – 1/2)^2 + (1/(\sqrt{y – 2020}) – 1/2)^2 + (1/(\sqrt{z – 2021}) – 1/2)^2 = 0`
`↔ 1/(\sqrt{x – 2019}) – 1/2 = 1/(\sqrt{y – 2020}) – 1/2 = 1/(\sqrt{z – 2021}) – 1/2 = 0`
`↔ 1/(\sqrt{x – 2019}) = 1/(\sqrt{y – 2020}) = 1/(\sqrt{z – 2021}) = 1/2`
`↔ \sqrt{x – 2019} = \sqrt{y – 2020} = \sqrt{z – 2021} = 2`
`↔ x – 2019 = y – 2020 = z – 2021 = 4`
`↔ x = 2023 ; y = 2024 ; z = 2025`
Giải thích các bước giải: