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

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

\(\frac{x}{2016}+\frac{x-1}{2015}+\frac{x-2}{2014}+\frac{x-3}{2013}=4\)

\(\Leftrightarrow\left(\frac{x}{2016}-1\right)+\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)+\left(\frac{x-3}{2013}-1\right)=0\)

\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2015}+\frac{x-2016}{2014}+\frac{x-2016}{2013}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}+\frac{1}{2013}\right)=0\)

Dễ thấy cái vế sau > 0 nên x=2016

Câu b có cách nào hay hơn bằng cách phá ko ta,hóng quá:)

\(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)

\(\Leftrightarrow8x^3+12x^2+6x+1+27x^3-27x^2+9x-1=125x^3\)

\(\Leftrightarrow35x^3-15x^2+15x=125x^3\)

\(\Leftrightarrow90x^3+15x^2-15x=0\)

\(\Leftrightarrow x\left(90x^2+15x-15\right)=0\)

\(\Leftrightarrow x\left(3x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow x=0;x=-\frac{1}{2};x=\frac{1}{3}\)

Câu c có cách giải rất hay đó nha :) 

\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)

\(\Leftrightarrow\left(2x-5\right)^3+\left(3x-3\right)^3+\left(8-5x\right)^3=0\)

Đặt \(2x-5=a;3x-3=b;8-5x=c\Rightarrow a+b+c=0\)

\(\Rightarrow a+b=-c\)

\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab\left(-c\right)=3abc\)

Khi đó:

\(\left(2x-5\right)^3+27\left(x-1\right)^3+\left(8-5x\right)^3=0\)

\(\Leftrightarrow3\left(2x-5\right)\left(3x-3\right)\left(8-5x\right)=0\)

\(\Leftrightarrow x=\frac{5}{2};x=1;x=\frac{8}{5}\)

Đặt 2x²+x-2015=a; x²-5x-2016=b

phương trình tương đương a²+4b²=4ab

=> a²-4ab+4b²=0

=> (a-2b)²=0

=> a=2b

vậy 2x²+x-2015=2*(x²-5x-2016)

=> x=\(\frac{-2017}{11}\)

Ta có :

\(\left(x^2-2014\right)\left(x^2-2015\right)\left(x^2-2016\right)\)\(=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x^2-2014=0\\x^2-2015=0\\x^2-2016=0\end{cases}}\)

Giải (1) :

    \(x^2-2014=0\)

     \(\hept{\begin{cases}x=\sqrt{2014}\\x=-\sqrt{2014}\end{cases}}\)

Giải (2) :

     \(x^2-2015=0\)

        \(\hept{\begin{cases}x=\sqrt{2015}\\x=-\sqrt{2015}\end{cases}}\)

Giải (3) :

   \(x^2-2016=0\)

    \(\hept{\begin{cases}x=\sqrt{2016}\\x=-\sqrt{2016}\end{cases}}\)

Vậy nghiệm nhỏ nhất của phương trình là \(x=-\sqrt{2016}\)

Chú ý : \(x^2-2014=0\)(1)

            \(x^2-2015=0\)(2)

            \(x^2-2016=0\)(3)

nếu x<2017 thì x-2017<2017

vì tổng của các giá trị tuyệt đối không thể là số âm nên x<2017 loại.

xét \(x\ge2017\), ta có:\(\left|x-2014\right|=x-2014\\ \left|2x-2015\right|=2x-2015\\\left|3x-2016\right|=3x-2016\)

khi đó:

\(x-2014+2x-2015+3x-2016=x-2017\\ \Leftrightarrow6x=4028\\ \Leftrightarrow x=\dfrac{2014}{3}\left(loại\right)\)

vậy phương trình đã cho vô nghiệm.

\(\left|x-2014\right|+\left|2x-2015\right|+\left|3x-2016\right|=x-2017\)

Do \(\left|x-2014\right|+\left|2x-2015\right|+\left|3x-2016\right|\ge0\forall x\)

\(\Rightarrow x-2017\ge0\\ \Leftrightarrow x\ge2017\)

\(\Rightarrow\left\{{}\begin{matrix}x-2014\ge3>0\\2x-2015\ge2019>0\\3x-2016\ge4035>0\end{matrix}\right.\)

\(pt\Leftrightarrow\left|x-2014\right|+\left|2x-2015\right|+\left|3x-2016\right|=x-2017\\ \Leftrightarrow x-2014+2x-2015+3x-2016=x-2017\\ \Leftrightarrow6x-6045=x-2017\\ \Leftrightarrow6x-x=-2017+6045\\ \Leftrightarrow5x=4028\\ \Leftrightarrow x=\dfrac{4028}{5}\\ \)

Vậy pt có nghiệm \(x=\dfrac{4028}{5}\)

a:

ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)

 \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)

=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)

=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)

=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)

b:

ĐKXĐ: x<>-3

 \(y=\left(x+3\right)+\dfrac{4}{x+3}\)

=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)

\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)

=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)

y'=0

=>\(\left(x+3\right)^2-4=0\)

=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)

=>(x+5)(x+1)=0

=>x=-5 hoặc x=-1

c:

ĐKXĐ: x<>-2

 \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)

=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)

=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)

=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)

\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)

d: 

ĐKXĐ: x<>2

\(y=x-2+\dfrac{9}{x-2}\)

=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)

\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)

=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)

y'=0

=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)

=>\(\left(x-2\right)^2-9=0\)

=>(x-2-3)(x-2+3)=0

=>(x-5)(x+1)=0

=>x=5 hoặc x=-1

tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi 

\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)

\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)

\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)

\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)