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\(DK:x\ge\frac{2019}{2020}\)
\(\Leftrightarrow\left(2020x-2019-2\sqrt{2020x-2019}+1\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{2020x-2019}-1\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\sqrt{2020x-2019}-1=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow x=1\left(n\right)\)
Vay nghiem cua PT la \(x=1\)
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 \(x\ge\frac{2017}{2018}\Rightarrow\)\(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}>0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
ĐKXĐ:...
\(\Leftrightarrow x^2-2x+1+2020x-2019-2\sqrt{2020x-2019}+1=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2020x-2019}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\\sqrt{2020x-2019}-1=0\end{matrix}\right.\)
\(\Rightarrow x=1\)
\(\Delta=2019^2-4m+4\)
\(x_1^2-x_1x_2-2018x_1x_2+2018x_2^2=0\)
\(\Leftrightarrow x_1\left(x_1-x_2\right)-2018x_2\left(x_1-x_2\right)=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1-2018x_2\right)=0\)
TH1: \(x_1=x_2\Rightarrow\Delta=0\Rightarrow2019^2-4m+4=0\Rightarrow m=\frac{2019^2+4}{4}\)
TH2: \(x_1=2018x_2\) kết hợp Viet ta có hệ:
\(\left\{{}\begin{matrix}x_1+x_2=2019\\x_1=2018x_2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=2018\\x_2=1\end{matrix}\right.\)
\(x_1x_2=m-1\Rightarrow m-1=2018\Rightarrow m=2019\)
Lời giải:
Đặt mẫu số của $B$ là $M$.
Từ \(2018x^3=2019y^3=2020z^3\)
\(\Rightarrow \sqrt[3]{2018}x=\sqrt[3]{2019}y=\sqrt[3]{2020}z=\frac{\sqrt[3]{2018}}{\frac{1}{x}}=\frac{\sqrt[3]{2019}}{\frac{1}{y}}=\frac{\sqrt[3]{2020}}{\frac{1}{z}}=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\)
\(=\frac{\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020}}{8}=\frac{M}{8}\)
\(\Rightarrow \left\{\begin{matrix} x=\frac{M}{8\sqrt[3]{2018}}\\ y=\frac{M}{8\sqrt[3]{2019}}\\ z=\frac{M}{8\sqrt[3]{2020}}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 2018x^2=\frac{\sqrt[3]{2018}M^2}{64}\\ 2019y^2=\frac{\sqrt[3]{2019}M^2}{64}\\ 2020z^2=\frac{\sqrt[3]{2020}M^2}{64}\end{matrix}\right.\)
\(\Rightarrow 2018x^2+2019y^2+2020z^2=\frac{M^2(\sqrt[3]{2018}+\sqrt[3]{2019}+\sqrt[3]{2020})}{64}=\frac{M^3}{64}\)
\(\Rightarrow B=\frac{\sqrt[3]{\frac{M^3}{64}}}{M}=\frac{M}{4M}=\frac{1}{4}\)
\(\text{Δ}=\left(-1\right)^2-4\cdot\left(3m-11\right)=1-12m+44=-12m+45\)
Để phương trình có hai nghiệm phân biệt thì -12m+45>0
hay m<45/12
Theo Vi-et, ta đc: \(x_1+x_2=1;x_1x_2=3m-11\)
Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}x_1+x_2=1\\2017x_1+2018x_2=2019\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=2\end{matrix}\right.\)
Ta có: \(x_1x_2=3m-11\)
=>3m-11=-2
=>m=3
\(2018x^2-2017-1=0\)
\(2018x^2-2018=0\)
\(2018\left(x^2-1\right)=0\)
\(\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
đề phải là : 2018x^2-2017x-1 = 0
pt <=> (2018x^2-2018x)+(x-1) = 0
<=> 2018.x.(x-1)+(x-1) = 0
<=> (x-1).(2018x+1) = 0
<=> x-1=0 hoặc 2018x+1=0
<=> x=1 hoặc x=-1/2018
Vậy ............
Tk mk nha