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24 tháng 1 2019

\(\frac{2}{x-14}\) -\(\frac{5}{x-13}\)\(\frac{2}{x-9}\)\(\frac{5}{x-11}\)

ĐKXĐ : x\(\ne\)14;13;9;11

(=) \(\frac{2}{x-14}\)\(\frac{5}{x-13}\)\(\frac{2}{x-9}\)-\(\frac{5}{x-11}\)

(=)  \(\frac{2}{x-14}\)-\(\frac{2}{x-9}\)=\(\frac{5}{x-13}\)-\(\frac{5}{x-11}\)

(=) \(\frac{2\left(x-9\right)}{\left(x-14\right)\left(x-9\right)}\)-\(\frac{2\left(x-14\right)}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{5\left(x-11\right)}{\left(x-13\right)\left(x-11\right)}\)-\(\frac{5\left(x-13\right)}{\left(x-13\right)\left(x-11\right)}\)

(=) \(\frac{2x-18-2x+28}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{5x-55-5x+65}{\left(x-13\right)\left(x-11\right)}\)

(=) \(\frac{10}{\left(x-14\right)\left(x-9\right)}\)=\(\frac{10}{\left(x-13\right)\left(x-11\right)}\)

=) ( x - 14 ) ( x - 9 ) = ( x - 13 ) ( x - 11 )

(=) x- 9x - 14x + 126 = x2 - 13x - 11x + 143

(=) x - 17 = 0

 (=)   x = 17

Vậy phương trình có nghiệm là: x = 17

9 tháng 3 2018

\(\dfrac{2}{x-14}-\dfrac{5}{x-13}=\dfrac{2}{x-9}-\dfrac{5}{x-11}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne9;11;13;14\\\left(\dfrac{2}{x-14}-\dfrac{2}{3}\right)-\left(\dfrac{5}{x-13}-\dfrac{5}{4}\right)=\left(\dfrac{2}{x-9}-\dfrac{1}{4}\right)-\left(\dfrac{5}{x-11}-\dfrac{5}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow2\left(\dfrac{x-17}{3\left(x-14\right)}\right)-5\left(\dfrac{x-17}{4\left(x-13\right)}\right)=\left(\dfrac{x-17}{4\left(x-9\right)}\right)-5\left(\dfrac{x-17}{6\left(x-11\right)}\right)\)

\(\left(x-17\right)\left[\dfrac{2}{3\left(x-14\right)}-\dfrac{5}{4\left(x-13\right)}+\dfrac{5}{6\left(x-11\right)}-\dfrac{1}{4\left(x-9\right)}\right]=0\)

[..] vô nghiệm

x=17

AH
Akai Haruma
Giáo viên
10 tháng 3 2018

Lời giải:

Bài của bạn ngonhuminh cơ bản không đúng do không có cơ sở khẳng định biểu thức trong ngoặc vuông vô nghiệm.

ĐKXĐ: \(x\neq \left\{9;11;13;14\right\}\)

\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)

\(\Leftrightarrow 2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)=5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)\)

\(\Leftrightarrow \frac{10}{(x-14)(x-9)}=\frac{10}{(x-13)(x-11)}\)

\(\Rightarrow (x-14)(x-9)=(x-13)(x-11)\)

\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)

\(\Leftrightarrow x-17=0\Leftrightarrow x=17\)

Thử lại thấy thỏa mãn.

Vậy \(x=17\)

17 tháng 1 2023

\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

17 tháng 1 2023

bn ơi ktra lại câu 2 giúp mk đc ko 

5 tháng 3 2023

\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)

\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)

 

 

5 tháng 3 2023

Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)

\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)

\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)

\(\Leftrightarrow\left(x+60\right)\text{=}0\)

\(x\text{=}-60\)

\(Vậy...\)

31 tháng 1 2023

`(x+19)/3 +(x+13)/5 = (x+7)/7 + (x+1)/9`

`<=> x/3 + 19/3 +x/5 +13/5 = x/7 +1 +x/9 +1/9`

`<=> x/3 +x/5 -x/7 -x/9 = 1+1/9 -19/3 -13/5`

`<=> x (1/3 +1/5 -1/7 -1/9) = -118/45`

`<=> x * 88/315 = -352/45`

`<=> x = -28`

Vậy `S={-28}`

31 tháng 1 2023

ủa kì vậy cô tôi dạy kiểu như làm kiếm mẫu số chung gì mà...

Coi lại bài coi đúng không;-;; hoang mang quá đi ạ

12 tháng 8 2017

Mở đầu về phương trình

Mở đầu về phương trình

12 tháng 8 2017

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)

1 tháng 4 2017

(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

1 tháng 4 2017

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

vậy nghiệm của bpt là x<11

f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-x=-5

=>x=5

loading...

11 tháng 3 2023

những bài chứa ẩn ở mẫu thì phải có đkxđ chứ bạn

d: =>\(\dfrac{12x+1}{11x-4}=\dfrac{20x+17-20x+8}{18}=\dfrac{25}{18}\)

=>25(11x-4)=18(12x+1)

=>275x-100=216x+18

=>59x=118

=>x=2

f: =>\(\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

=>28-6(x+2)=-9-5(x-4)

=>28-6x-12=-9-5x+20

=>-6x+16=-5x+11

=>-x=-5

=>x=5

loading...

8 tháng 2 2018

a.

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

\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)

\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)

\(\Leftrightarrow6x+6+3x+9=36-4x-8\)

\(\Leftrightarrow9x+15=28-4x\)

\(\Leftrightarrow9x+4x=28-15\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

8 tháng 2 2018

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

\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)

\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)

\(\Leftrightarrow6x+6+3x+9=36-4x-8\)

\(\Leftrightarrow9x+15=-4x+28\)

\(\Leftrightarrow9x+4x=28-15\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

Vậy ................................