\(2014.\left|x-12\right|+\left(x-12\right)^2=2013.\left|12-x\right|\)
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PT
1
10 tháng 12 2017
\(Do:\left|x-12\right|=\left|12-x\right|\)
⇒2014.|x−12|+(x-12)2=2013.\(\left|x-12\right|\)
⇒2014.\(\left|x-12\right|\)+(x-12)2-2013.\(\left|x-12\right|\)=0
⇒(2014-2013).\(\left|x-12\right|\)+(x-12)2=0
⇒\(\left|x-12\right|+\left(x-12\right)^2\)=0
Do: \(\left|x-12\right|\ge0,\left(x-12\right)^2\ge0\)
⇒x-12=0
⇒x=12
PB
0
J
1
7 tháng 3 2020
Ta có:
\(2014\left|x-12\right|+\left(x-12\right)^2=2013\left|12-x\right|\)
\(\Rightarrow\left(x-12\right)^2=2013\left|12-x\right|-2014\left|x-12\right|\)
\(\Rightarrow\left(x-12\right)^2=-\left|x-12\right|\)
\(\Rightarrow x-12=0\Rightarrow x=12\)
8 tháng 2 2020
a) Ta có:
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=\left(x+11\right)\left(\frac{1}{15}+\frac{1}{16}\right)\)
Mà ta có:
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\ne\frac{1}{15}+\frac{1}{16}\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Ta có:
\(A=1+x+x^2+x^3+...+x^{100}\)
Đặt \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
23 tháng 10 2016
Đặt \(a=2x^2+x-2014\) , \(b=x^2-5x-2013\)
thì \(a^2+4b^2=4ab\Leftrightarrow a^2-4ab+4b^2=0\Leftrightarrow\left(a-2b\right)^2=0\)
Thay vào được \(\left[\left(2x^2+x-2014\right)-2\left(x^2-5x-2013\right)\right]^2=0\)
\(\Leftrightarrow11x+2012=0\Leftrightarrow x=-\frac{2012}{11}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 2 2019
a/ Đặt \(x^2+x+1=a\Rightarrow x^2+x+2=a+1\)
Pt trở thành \(a\left(a+1\right)-12=0\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a-3\right)\left(a+4\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+x+1=3\\x^2+x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2+x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x+2\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
2/ \(\dfrac{x+1}{2014}+1+\dfrac{x+2}{2013}+1=\dfrac{x+3}{2012}+1+\dfrac{x+4}{2011}+1\)
\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}=\dfrac{x+2015}{2012}+\dfrac{x+2015}{2011}\)
\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\right)=0\)
\(\Leftrightarrow x+2015=0\) (do \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))
\(\Rightarrow x=-2015\)
Ta có:
|12-x|=-|x-12|
Đ k: x-12>_0=>x>_12
=>2014.|x-12|+(x-12)^2=-2013.|x-12|
=>2014.(x-12)+(x-12)^2+2013.(x-12)=0
=>(x-12).(2014+x-12+2013)=0
=>(x-12).(x+2005)=0
=>x-12=0 và. x+2005=0
=>x=12 và x=-2005