giải pt: \(\sqrt{x-2019}+\sqrt{2021-x}\)=(x-2020)2+2
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9 tháng 5 2021
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 ...
3 tháng 11 2018
Ta có :
\(\dfrac{1}{\sqrt{x+1}+\sqrt{x+2}}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{\left(\sqrt{x+1}+\sqrt{x+2}\right)\left(\sqrt{x+1}-\sqrt{x+2}\right)}=\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}=-\sqrt{x+1}+\sqrt{x+2}\)
Tương tự :
\(\dfrac{1}{\sqrt{x+2}+\sqrt{x+3}}=-\sqrt{x+2}+\sqrt{x+3}\)
\(\dfrac{1}{\sqrt{x+3}+\sqrt{x+4}}=-\sqrt{x+3}+\sqrt{x+4}\)
....
\(\dfrac{1}{\sqrt{x+2019}+\sqrt{x+2010}}=-\sqrt{x+2019}+\sqrt{x+2010}\)
Từ những ý trên , pt trở thành :
\(-\sqrt{x+1}+\sqrt{x+2}-\sqrt{x+2}+\sqrt{x+3}-\sqrt{x+3}+\sqrt{x+4}-.....-\sqrt{x+2019}+\sqrt{x+2020}=11\)
\(\Leftrightarrow\sqrt{x+2020}-\sqrt{x+1}=11\)
\(\Leftrightarrow x+2020-2\sqrt{\left(x+2020\right)\left(x+1\right)}+x+1=121\)
\(\Leftrightarrow2x+1900=2\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x+950=\sqrt{\left(x+1\right)\left(x+2020\right)}\)
\(\Leftrightarrow x^2+1900x+902500=x^2+2021x+2020\)
\(\Leftrightarrow121x-900480=0\)
\(\Leftrightarrow x=\dfrac{900480}{121}\)
HT
1
27 tháng 3 2022
Ta có: VT = \(\dfrac{x+1}{2021}\)+1 - (\(\dfrac{x+2}{2020}\)+1) = \(\dfrac{x+3}{2019}\)+1=VP
=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}-\dfrac{x+2022}{2019}=0\)
=>\(\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}\right)=0\)
=>x +2022 = 0=> x =-2022
LP
1
YY
1 tháng 6 2018
\(x+y+z-6046=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
\(\left(x-2019\right)+\left(x-2020\right)+\left(x-2021\right)+1+4+9\)\(=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)
đặt :\(\hept{\begin{cases}\sqrt{x-2019}=a\\\sqrt{y-2020}=b\\\sqrt{z-2021}=c\end{cases}\left(đk:a,b,c\ge0\right)}\)
PT <=> \(a^2+b^2+c^2+1+4+9=2a+4b+6c\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-6\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}\left(tm\right)}}\)
\(\Rightarrow\hept{\begin{cases}x=2020\\y=2024\\z=2030\end{cases}}\)
14 tháng 7 2021
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
ĐKXĐ: \(2019\le x\le2020\)
\(VT=\sqrt{x-2019}+\sqrt{2021-x}\le\sqrt{2\left(x-2019+2021-x\right)}=2\)
\(VP=\left(x-2020\right)^2+2\ge2\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}x-2019=2021-x\\x-2020=0\end{matrix}\right.\) \(\Leftrightarrow x=2020\)