Find x such that :
\(\left|\left|6x-2\right|-5\right|=2016x-2017\)
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\(\left|\left|6x-2\right|-5\right|=2016x-2017\)
Xét trường hợp 1: \(\left|6x-2\right|-5=2016x-2017\)
\(\Rightarrow\left|6x-2\right|=2016x-2017+5\)
\(\Rightarrow\left|6x-2\right|=2016x-2012\)
\(\Rightarrow\left[{}\begin{matrix}6x-2=2016x-2012\\6x-2=-\left(2016x-2012\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}6x-2-2016x+2012=0\\6x-2+2016x-2012=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2010x+2010=0\\2022x-2014=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2010x=-2010\\2022x=2014\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2010:-2010\\x=2014:2022\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1007}{1011}\end{matrix}\right.\)
Xét trường hợp 2: \(\left|6x-2\right|-5=-\left(2016x-2017\right)\)
\(\Rightarrow\left|6x-2\right|=-\left(2016x-2017\right)+5\)
\(\Rightarrow\left|6x-2\right|=-2016x+2017+5\)
\(\Rightarrow\left|6x-2\right|=-2016x+2022\)
\(\Rightarrow\left|6x-2\right|=-\left(2016x-2022\right)\)
\(\Rightarrow\left[{}\begin{matrix}6x-2=-\left(2016x-2022\right)\\6x-2=-\left[-\left(2016x-2022\right)\right]\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}6x-2=-\left(2016x-2022\right)\\6x-2=2016x-2022\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}6x-2-\left[-\left(2016x-2022\right)\right]=0\\6x-2-\left(2016x-2022\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}6x-2+2016x-2022=0\\6x-2-2016x+2022=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2022x-2024=0\\-2010x+2020=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2022x=2024\\-2010x=-2020\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2024:2022\\x=\left(-2020\right):\left(-2010\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1012}{1011}\\x=\dfrac{202}{201}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\dfrac{1007}{1011}\) hoặc \(x=\dfrac{1012}{1011}\) hoặc \(x=\dfrac{202}{201}\)
a) \(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)
\(\Leftrightarrow x-\sqrt{3}=\pm\frac{\sqrt{3}}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\sqrt{3}=-\frac{\sqrt{3}}{2}\\x-\sqrt{3}=\frac{\sqrt{3}}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{2}\\\frac{3\sqrt{3}}{2}\end{cases}}\)
Nghiệm cuối cùng là : \(x_1=\frac{\sqrt{3}}{2};x_2=\frac{3\sqrt{3}}{2}\)
b) || 6x - 2 | - 5 | = 2016. x -2017
<=> || 6x - 2 | -5 | -2016x = -2017
<=> \(\orbr{\begin{cases}\left|6x-2\right|-5-2016.x=-2017,\left|6x-2\right|-5\ge0\\-\left(\left|6x-2\right|-5\right)-2016x=-2017,\left|6x-2\right|-5< 0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=1,x\in\left[-\infty,-\frac{1}{2}\right];\left[\frac{7}{6};+\infty\right]\\x=\frac{1012}{1011},x\in\left[-\frac{1}{2},\frac{7}{6}\right]\end{cases}}\)
<=>\(\orbr{\begin{cases}x\in\varnothing\\x=\frac{1012}{1011}\end{cases}}\)
Vậy x = \(\frac{1012}{1011}\)
Tử số bằng mẫu số
K-2016=1
K=2017
Muốn biết tại sao tử= mẫu thì tích nha
\(K-2016=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)
\(K-2016=\frac{1\times2017+2\times2016+3\times2015+...+2017\times1}{2017\times1+2016\times2+2015\times3+...+2017\times1}\)
\(K-2016=1\)
\(\Rightarrow K=1+2016\)
\(\Rightarrow K=2017\)
a, Ta có:
\(\orbr{\begin{cases}x-\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{4}}\\x-\sqrt{\frac{3}{4}}=-\sqrt{\frac{3}{4}}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\sqrt{\frac{3}{4}}\\x=0\end{cases}}}\)
Câu 1:
\(\Leftrightarrow x^2-2x+1-2\left|x-1\right|-3=0\)
\(\Leftrightarrow\left(\left|x-1\right|\right)^2-2\left| x-1\right|-3=0\)
\(\Leftrightarrow\left(\left|x-1\right|-3\right)\left(\left|x-1\right|+1\right)=0\)
=>x-1=3 hoặc x-1=-3
=>x=4 hoặc x=-2