a: (x-2)(x+3)>0

TH1: \(\begin{cases}x-2>0\\ x+3>0\end{cases}\Rightarrow\begin{cases}x>2\\ x>-3\end{cases}\Rightarrow x>2\)

TH2: \(\begin{cases}x-2<0\\ x+3<0\end{cases}\Rightarrow\begin{cases}x<2\\ x<-3\end{cases}\)

=>x<-3

b: (2x-1)(-x+1)>0

=>(2x-1)(x-1)<0

TH1: \(\begin{cases}2x-1>0\\ x-1<0\end{cases}\Longrightarrow\begin{cases}x>\frac12\\ x<1\end{cases}\)

=>\(\frac12

TH2: \(\begin{cases}2x-1<0\\ x-1>0\end{cases}\Rightarrow\begin{cases}x<\frac12\\ x>1\end{cases}\)

=>x∈∅

c: (x+1)(3x-6)<0

=>3(x+1)(x-2)<0

=>(x+1)(x-2)<0

TH1: \(\begin{cases}x+1>0\\ x-2<0\end{cases}\Rightarrow\begin{cases}x>-1\\ x<2\end{cases}\Rightarrow-1

TH2: \(\begin{cases}x+1<0\\ x-2>0\end{cases}\Rightarrow\begin{cases}x<-1\\ x>2\end{cases}\)

=>x∈∅

(x-7/2012 +1)+(x-6/2013 +1) =(x-5/2014 +1) +(x-4/2015 +1)

x-2019 /2012 + x-2019 /2013 = x-2019 / 2014 + x-2019 /2015

x-2019 /2012+ x-2019 /2013 - x-2019 /2014 - x-2019 /2015 = 0

(x-2019) * (1/2012 + 1/2013 - 1/2014 -1/2015) = 0

Vì x khác 0 =>1/2012 + 1/2013 -1/2014 -1/2015 khác 0

=> x-2019=0

    x        = 0+2019

    x        = 2019

Vậy x=2019

tick mình nha

Thay x = 2016 vào biểu thức B, ta có:

B = 2016²⁰¹⁶ - 2015.2016²⁰¹⁵ - 2015.2016²⁰¹⁴ - ... - 2015.2016² - 2015.2016 + 1

B = 2016²⁰¹⁶ - (2016 - 1).2016²⁰¹⁵ - (2016 - 1).2016²⁰¹⁴ - ... - (2016 - 1).2016² - (2016 - 1).2016 + 1

B = 2016²⁰¹⁶ - 2016²⁰¹⁶ + 2016²⁰¹⁵ - 2016²⁰¹⁵ + 2016²⁰¹⁴ - ... - 2016³ + 2016² - 2016² + 2016 + 1

B = (2016²⁰¹⁶ - 2016²⁰¹⁶) + (2016²⁰¹⁵ - 2016²⁰¹⁵) + ... + (2016² - 2016²) + (2016 + 1)

B = 2016 + 1 = 2017

Vậy ...

=> (x-2015).2014 = (y-2014).2015

=> 2014x-2015.2014 = 2015y - 2014.2015

=> 2014x = 2015y

=> x/y=2015/2014

=> (x-2015).2014 = (y-2014).2015

=> 2014x-2015.2014 = 2015y - 2014.2015

=> 2014x = 2015y

=> x/y=2015/2014

\(A = | x + 2014 | + | x + 2015| + 2015\)

\(A = | x + 2014 | + | x + 2015 | + 2015 \)\(\ge\)

\(2015\)

\(Dấu " = " xảy \)  \(ra\) \(\Leftrightarrow\)\(x + 2014 = 0 hoặc x + 2015= 0\)

\(\Leftrightarrow\)\(x = - 2014 hoặc x = - 2015\)

\(Min A = 2015\) \(\Leftrightarrow\)\(x = - 2014 hoặc x = - 2015\)

\(A=\left|x+2014\right|+\left|x+2015\right|+2015\)

\(=\left|x+2014\right|+\left|-x-2015\right|+2015\)

Ta có: \(\left|x+2014\right|+\left|-x-2015\right|\ge\left|x+2014-x-2015\right|=1\)

\(\Rightarrow\left|x+2014\right|+\left|-x-2015\right|+2015\ge2016\)

Dấu"="xảy ra \(\Leftrightarrow\left(x+2014\right)\left(-x-2015\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x+2014\ge0\\-x-2015\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+2014< 0\\-x-2015< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge-2014\\x\le-2015\end{cases}}\)hoặc \(\hept{\begin{cases}x< -2014\\x>-2015\end{cases}\left(loai\right)}\)

\(\Leftrightarrow-2014\le x\le-2015\)

Vậy \(A_{min}=2016\)\(\Leftrightarrow-2014\le x\le-2015\)

Ta có : \(\frac{x-2015}{2015}=\frac{y-2014}{2014}\)

=> \(\frac{x}{2015}-\frac{2015}{2015}=\frac{y}{2014}-\frac{2014}{2014}\)

=> \(\frac{x}{2015}-1=\frac{y}{2014}-1\)

=> \(\frac{x}{2015}=\frac{y}{2014}\)

=> \(\frac{x}{y}=\frac{2015}{2014}\)

cx ko bt cách tl ntn, thôg cảm