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<=> \(x^4\left(\sqrt{x+3}-2\right)\)\(+2018\left(x-1\right)=0\)
<=>\(x^4\left(\dfrac{x+3-4}{\sqrt{x+3}+2}\right)+2018\left(x-1\right)=0\)
<=>\(x^{\text{4}}\left(\dfrac{x-1}{\sqrt{x+3}+2}\right)+2018\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(\dfrac{x^4}{\sqrt{x+3}+2}+2018\right)=0\)
=>x-1=0 <=>x=1
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Ta có:
\(\sqrt{x^2-2018x+2018}+\sqrt{x^2-1009x+1009}=2x\)
\(\Leftrightarrow x-\sqrt{\left(2018x-2018\right)}+x-\sqrt{\left(1009x-1009\right)}=2x\)
\(\Leftrightarrow2x-\sqrt{\left(2018x-2018\right)}-\sqrt{\left(1009x-1009\right)}=2x\)
\(\Leftrightarrow\sqrt{\left(2018x\right)-2018}+\sqrt{\left(1009x-1009\right)}=0\)
\(\Leftrightarrow\sqrt{\left(2018x-2018\right)}=\sqrt{\left(1009x-1009\right)}=0\)
\(\Leftrightarrow2018x-2018=1009x-1009=0\Leftrightarrow x=1\)
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\(y\left(x+1\right)^2=-x^2+2018x-1\)
\(\Leftrightarrow y=\dfrac{-x^2+2018x-1}{\left(x+1\right)^2}=-1+\dfrac{2020x}{\left(x+1\right)^2}\)
\(\Rightarrow\dfrac{2020x}{\left(x+1\right)^2}\in Z\)
Mà x và \(x\left(x+2x\right)+1\) nguyên tố cùng nhau
\(\Rightarrow2020⋮\left(x+1\right)^2\)
Ta có 2020 chia hết cho đúng 2 số chính phương là 1 và 4
\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=1\\\left(x+1\right)^2=4\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow y\)
b.
Từ pt đầu:
\(x^2+xy-2y^2+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)+2\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+2y+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y-2\end{matrix}\right.\)
Thế xuống dưới ...
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ĐK: \(x\ge\frac{2017}{2018}\)
\(pt\Leftrightarrow2017\sqrt{2017x-2016}-2017+\sqrt{2018x-2017}-1=0\)
\(\Leftrightarrow2017\frac{2017\left(x-1\right)}{\sqrt{2017x-2016}+1}+\frac{2018\left(x-1\right)}{\sqrt{2018x-2017}+1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}\right)=0\)
Dễ thấy với \(x\ge\frac{2017}{2018}\Rightarrow\)\(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}>0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)
\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)
\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)
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ĐKXĐ: \(x\ge\dfrac{3}{2}\).
PT đã cho tương đương:
\(\dfrac{x-4}{\sqrt{2x-3}+\sqrt{x+1}}=x-4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\Leftrightarrow x=4\left(TMĐK\right)\\\sqrt{2x-3}+\sqrt{x+1}=1\left(1\right)\end{matrix}\right.\).
Ta có \(\left(1\right)\Leftrightarrow2x-3+x+1+2\sqrt{\left(2x-3\right)\left(x+1\right)}=1\)
\(\Leftrightarrow2\sqrt{\left(2x-3\right)\left(x+1\right)}=3-3x\).
Do đó 3 - 3x \(\ge0\Leftrightarrow x\le1\) (trái với đkxđ).
Suy ra (1) vô nghiệm.
Vậy ncpt là x = 4.