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ĐKXĐ : \(\left\{{}\begin{matrix}x>2019\\y>2020\\z>2021\end{matrix}\right.\)
Đặt \(\sqrt{x-2019}=a,......\)
Ta được PT : \(\dfrac{1-a}{a^2}+\dfrac{1-b}{b^2}+\dfrac{1-c}{c^2}+\dfrac{3}{4}=0\)
\(\Leftrightarrow\dfrac{1}{a^2}-\dfrac{1}{a}+\dfrac{1}{4}+\dfrac{1}{b^2}-\dfrac{1}{b}+\dfrac{1}{4}+\dfrac{1}{c^2}-\dfrac{1}{c}+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)
- Thấy : \(\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2\ge0,......\)
\(\Rightarrow\left(\dfrac{1}{a}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2+\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2\ge0\)
- Dấu " = " xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{2}\\\dfrac{1}{b}=\dfrac{1}{2}\\\dfrac{1}{c}=\dfrac{1}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=2\end{matrix}\right.\)
- Thay lại a. b. c ta được : \(\left\{{}\begin{matrix}\sqrt{x-2019}=2\\\sqrt{y-2020}=2\\\sqrt{z-2021}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2019=4\\y-2020=4\\z-2021=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2023\\y=2024\\z=2025\end{matrix}\right.\) ( TM )
Vậy ...
\(DK:x\ge\frac{2019}{2020}\)
\(\Leftrightarrow\left(2020x-2019-2\sqrt{2020x-2019}+1\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2020x-2019}-1\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2020x-2019}-1=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow x=1\left(n\right)\)
Vay nghiem cua PT la \(x=1\)
ĐKXĐ:...
\(\Leftrightarrow x^2-2x+1+2020x-2019-2\sqrt{2020x-2019}+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2020x-2019}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\sqrt{2020x-2019}-1=0\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(2018x^2-\left(m-2019\right)x-2020=0\)
Ta có \(\Delta=b^2-4ac\)
\(=\left[-\left(m-2019\right)\right]^2-4.2018.\left(-2020\right)\)
\(=\left(m-2019\right)^2+4.2018.2020>0\)( vì \(\left(m-2019\right)^2\ge0\forall x\))
Phương trình có 2 nghiệm \(x_1,x_2\) Áp dụng hệ thức Vi-ét ta có
\(\hept{\begin{cases}x_1+x_2=\frac{m-2019}{2018}\left(1\right)\\x_1.x_2=\frac{-2020}{2018}\left(2\right)\end{cases}}\)
Ta có \(\sqrt{x_1^2+2019}-x_2=\sqrt{x_2^2+2019}-x_2\)
\(\Leftrightarrow\sqrt{x_1^2+2019}-x_2+x_2=\sqrt{x_2^2+2019}\)
\(\Leftrightarrow\sqrt{x_1^2+2019}+0=\sqrt{x_2^2+2019}\)
\(\Leftrightarrow x_1^2+2019=x_2^2+2019\)
\(\Leftrightarrow x_1^2-x_2^2=0\)
\(\Leftrightarrow\left(x_1-x_2\right).\left(x_1+x_2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right).\frac{m-2019}{2018}=0\Rightarrow x_1-x_2=0\left(3\right)\)
Thay (3) vào (!) ta có \(\hept{\begin{cases}x_1+x_2=\frac{m-2019}{2018}\\x_1-x_2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x_1=\frac{m-2019}{2018}\\x_1-x_2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x_1=\frac{m-2019}{4036}\\x_2=\frac{m-2019}{4036}\end{cases}}\)
\(\Rightarrow x_1.x_2=\frac{-2020}{2018}=\frac{-1010}{1009}\)
\(\Leftrightarrow\frac{m-2019}{4036}.\frac{m-2019}{4036}=\frac{-1010}{1009}\)
\(\Leftrightarrow\frac{\left(m-2019\right)^2}{4036^2}=\frac{-1010}{1009}\)
\(\Leftrightarrow\left(m-2019\right)^2=\frac{4036^2.\left(-1010\right)}{1009}\)
\(\Leftrightarrow\left(m-2019\right)^2=-16305440\left(VL\right)\)
Vậy không có m để thỏa mãn bài toán
Đặt t=\(\sqrt{2019-x^{ }2}\)>0, nên \(t^2\)=2019-\(x^2\) hay \(x^2\)=2019-\(t^2\).
từ đề bài ta có: 2019-\(t^2\)-\(t^2\)-2017t=0
hay 2\(t^2\)+2017t-2019=0, nên t=1 và t=-2019/2<0 loại
t=1, nên \(x^2\)=2018, nên x=2018 hoặc x=-2018 thỏa điều kiện 2019-\(x^2\)>=0