Find x such that :
\(\left|\left|6x-2\right|-5\right|=2016x-2017\)
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\(\left|\left|6x-2\right|-5\right|=2016x-2017\)
\(\left|\left|6x-2\right|-5\right|=2016x-2017\)
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 \(x=\dfrac{202}{201}\)
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Tìm x biết
a,\(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)
b,||6x-2|-5|=2016x-2017
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 = \(\frac{1012}{1011}\)
Find K such that:
K - 2016 = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)
Tử số bằng mẫu số
K-2016=1
K=2017
Muốn biết tại sao tử= mẫu thì tích nha
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\(K-2016=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017\times1+2016\times2+2015\times3+...+2\times2016+1\times2017}\)
\(K-2016=\frac{1\times2017+2\times2016+3\times2015+...+2017\times1}{2017\times1+2016\times2+2015\times3+...+2017\times1}\)
\(K-2016=1\)
\(\Rightarrow K=1+2016\)
\(\Rightarrow K=2017\)
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Xem thêm câu trả lời
Tìm x biết :
a)\(^{\left(x-\sqrt{\frac{3}{4}}\right)^2=\frac{3}{4}}\)
b)\(||6x-2|-5|=2016x-2017\)
a, Ta có:
\(\orbr{\begin{cases}x-\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{4}}\\x-\sqrt{\frac{3}{4}}=-\sqrt{\frac{3}{4}}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\sqrt{\frac{3}{4}}\\x=0\end{cases}}}\)
mình xin lỗi , mình ghi sai đề
a)\(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)
Find the value of such that
\(\frac{x-2}{\left(a+3\right)\left(5-a\right)}=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(5-a\right)}\) (\(\left(a\ne-3;a\ne5\right)\)
=> 2(x-2) =5-a+a+3
=>2x =4+8
=> x =6
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1. Find the positive value of x such that:
\(x^2-2-2x-2\left|x-1\right|=0\)
2. Find the remainder of the division:
\(\left(x^3-13+5x-3x^2\right):\left(x-3\right)\)
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
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1.If 2x-y5 then the value of Mleft(x+2y-3right)^2-left(6x+2yright)left(x+2y-3right)+9x^2+6xy+y^22.The free coefficient in the following poly nomaial: left(2x-2right)left(x+1right)left(7-x^2right)is:3.The greatest integer number x such that frac{2x-1}{x-3}-1 0 is:4.How many of the integer n such that satisfy the inequality left(n-3right)^2-left(n-4right)left(n+4right) 43 are less than 3?5.The opposite fraction of frac{x-2}{7-x} is:
Đọc tiếp
1.If 2x-y=5 then the value of M=\(\left(x+2y-3\right)^2-\left(6x+2y\right)\left(x+2y-3\right)+9x^2+6xy\)
\(+y^2\)
2.The free coefficient in the following poly nomaial: \(\left(2x-2\right)\left(x+1\right)\left(7-x^2\right)is:\)
3.The greatest integer number x such that \(\frac{2x-1}{x-3}-1< 0\) is:
4.How many of the integer n such that satisfy the inequality \(\left(n-3\right)^2-\left(n-4\right)\left(n+4\right)< =43\) are less than 3?
5.The opposite fraction of \(\frac{x-2}{7-x}\) is:
\(CMR:P\left(x\right)=x^{2016}+2.x^{2015}+....+2016x+2017\)không có nghiệm nguyên
PHÂN TÍCH:
a)left(x^2+8x+12right)left(x^2+12x+32right)+16
b)left(x^2+6x+8right)left(x^2+8x+15right)-24
c)left(x^2-6x+5right)left(x^2-10x+21right)-20
d)left(x^2+x-2right)left(x^2+9x+18right)-28
e)left(x^2-11x+28right)left(x^2-7x+10right)-72
f) left(x^2+5x+6right)left(x^2-15x+56right)-144
g)left(x^2-xright)^2+3left(x^2-xright)+2
h)left(x^2+5xright)^2+10x^2+50x+24
i)x^4+2016x^2+2015x+2016
Đọc tiếp
PHÂN TÍCH:
a)\(\left(x^2+8x+12\right)\left(x^2+12x+32\right)+16\)
b)\(\left(x^2+6x+8\right)\left(x^2+8x+15\right)-24\)
c)\(\left(x^2-6x+5\right)\left(x^2-10x+21\right)-20\)
d)\(\left(x^2+x-2\right)\left(x^2+9x+18\right)-28\)
e)\(\left(x^2-11x+28\right)\left(x^2-7x+10\right)-72\)
f) \(\left(x^2+5x+6\right)\left(x^2-15x+56\right)-144\)
g)\(\left(x^2-x\right)^2+3\left(x^2-x\right)+2\)
h)\(\left(x^2+5x\right)^2+10x^2+50x+24\)
i)\(x^4+2016x^2+2015x+2016\)
một lượt tối đa 2 câu làm vậy có thánh nào dmas beensg tới
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