Question

giải bất pt;

a)\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)

b) (x-5)(x-9)>0

c)\(\dfrac{x-5}{x-8}>2\)

Answer 1

(x-5)(x-9)>0\(\Leftrightarrow\left\{{}\begin{matrix}x-5>0\Leftrightarrow x>5\\x-9>0\Leftrightarrow x>9\end{matrix}\right.\)

Vậy x>9 thì (x-5)(x-9)>0

Answer 2

có

\(\dfrac{x-5}{x-8}>2\\ < =>x-5>2\left(x-8\right)\\ < =>x-5>2x-16\\ < =>-x>-11\\ < =>x< 11\)

vậy nghiệm của bpt là x<11

Answer 3

a/

\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)

\(\Leftrightarrow\dfrac{x+2014-2011}{2011}+\dfrac{x+2014-2013}{2013}\ge\dfrac{x+2014-2004}{2004}+\dfrac{x+2014-2001}{2001}\)

\(\Leftrightarrow-1+\dfrac{x+2014}{2011}-1+\dfrac{x+2014}{2013}\ge-1+\dfrac{x+2014}{2004}-1+\dfrac{x+2014}{2001}\)

\(\Leftrightarrow\dfrac{x+2014}{2011}+\dfrac{x+2014}{2013}-2\ge\dfrac{x+2014}{2004}+\dfrac{x+2014}{2001}-2\)

\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)

\(\Leftrightarrow\dfrac{1}{2011}+\dfrac{1}{2013}>\dfrac{1}{2004}+\dfrac{1}{2001}\) hoặc \(\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}\right)\ge\left(x+2014\right)\left(\dfrac{1}{2004}+\dfrac{1}{2001}\right)\)

(với mọi x>0) \(\Leftrightarrow x=2014\)

Answer 4

a,(\(\dfrac{x+3}{2011}\)+1)+(\(\dfrac{x+1}{2013}\)+1)>=(\(\dfrac{x+10}{2004}\)+1)+(\(\dfrac{x+13}{2001}\)+1)

<=>\(\dfrac{x+2014}{2011}\)+\(\dfrac{x+2014}{2013}\)>=\(\dfrac{x+2014}{2004}\)+\(\dfrac{x+2014}{2001}\)

<=>(x+2014)(1/2011+1/2013)>=(x+2014)(1/2004+1/2001)

vì 1/2011+1/2013<x+2014+1/2001

=>x+2014< hoặc =0<=>x < hoặc =-2014

Answer 5

b,để (x-5)(x-9)>0=>x-5 và x-9 phải có gt cùng dấu

th1:x-5 và x-9>0 =>x>9

th2:x-5 và x-9<0 =>x<5

Answer 6

c,đkxđ: x-8 khác 0 =>x khác 8

x-5>2(x-8) <=> x-5>2x-16

<=>-x>-11

<=>x<11

Answer 7

a)\(\dfrac{x+3}{2011}+\dfrac{x+1}{2013}\ge\dfrac{x+10}{2004}+\dfrac{x+13}{2001}\)

\(\Leftrightarrow\dfrac{x+2014}{2011}+\dfrac{x+2014}{2013}-2\ge\dfrac{x+2014}{2004}+\dfrac{x+2014}{2001}\)

\(\Leftrightarrow\left(x+2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\right)\ge0\)

ta chia hai vế với \(\dfrac{1}{2011}+\dfrac{1}{2013}-\dfrac{1}{2004}-\dfrac{1}{2001}\). Khi đó:

\(x+2014\ge0\\ \Leftrightarrow x\ge-2014\)

vậy phương trình có tập nghiệm là \(S=\left\{x|x\ge-2014\right\}\)

Answer 8

c) \(\dfrac{x-5}{x-8}>2\left(ĐKXĐ:x\ne8\right)\)

\(\Leftrightarrow x-5>2x-16\\ \Leftrightarrow-x>-11\\ \Leftrightarrow x>11\left(TMĐK\right)\)

vậy bất phương trình có tập nghiệm là \(S=\left\{x|x>11\right\}\)

Answer 9

b) (x-5)(x-9)>0

\(\Leftrightarrow\) \(\left[\begin{array}{} \begin{cases} x-5>0\\ x-9>0 \end{cases}\\ \begin{cases} x-5<0\\ x-9<0 \end{cases} \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} \begin{cases} x>5\\ x>9 \end{cases}\\ \begin{cases} x<5\\ x<9 \end{cases} \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x>9\\ x<5 \end{array} \right.\)

Từ đó suy ra đpcm.