\(Do:\left|x-12\right|=\left|12-x\right|\)

⇒2014.+(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

Ta có:

\(2014\left|x-12\right|+\left(x-12\right)^2=2013\left|12-x\right|\)

\(\Rightarrow\left(x-12\right)^2=2013\left|12-x\right|-2014\left|x-12\right|\)

\(\Rightarrow\left(x-12\right)^2=-\left|x-12\right|\)

\(\Rightarrow x-12=0\Rightarrow x=12\)

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 \(B=x+x^2+x^3+...+x^{100}\)

\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)

\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)

\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)

\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)

\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)

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 \(\left[\left(2x^2+x-2014\right)-2\left(x^2-5x-2013\right)\right]^2=0\)

\(\Leftrightarrow11x+2012=0\Leftrightarrow x=-\frac{2012}{11}\)

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 \(\dfrac{1}{2014}+\dfrac{1}{2013}-\dfrac{1}{2012}-\dfrac{1}{2011}\ne0\))

\(\Rightarrow x=-2015\)

Chứng minh gì bạn?