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31 tháng 3 2023

\(\left(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}\right)-4=0\)

<=>\(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}=4\)

<=>\(\dfrac{x+1}{55}-1+\dfrac{x+2}{56}-1+\dfrac{x+3}{57}+\dfrac{x+4}{58}-1=4-4\)

<=>\(\dfrac{x+1}{55}-\dfrac{55}{55}+\dfrac{x+2}{56}-\dfrac{56}{56}+\dfrac{x+3}{57}-\dfrac{57}{57}+\dfrac{x+4}{58}-\dfrac{58}{58}=0\)

<=>\(\dfrac{x-54}{55}+\dfrac{x-54}{56}+\dfrac{x-54}{57}+\dfrac{x-54}{58}=0\)

<=>\(\left(x-54\right)\left(\dfrac{1}{55}+\dfrac{1}{56}+\dfrac{1}{57}+\dfrac{1}{58}\right)=0\)

<=>x-54=0

<=>x=54

vậy phương trình có tập nghiệm là S={54}

31 tháng 3 2023

ở dòng thứ 6 cậu thêm  \(\dfrac{1}{55}+\dfrac{1}{56}+\dfrac{1}{57}+\dfrac{1}{58}\ne0\) để giải thích nhé .

27 tháng 7 2016

dẽ qua ak nhưng giúp mình làm bài này đi

cho tam giac abc . co canh bc=12cm, duong cao ah=8cm

a> tinh s tam giac abc

b> tren canh bc lay diem e sao cho be=3/4bc. tinh s tam giac abe va s tam giac ace ( bằng nhiều cách

c> lay diem chinh giua cua canh ac va m . tinh s tam giac ame

27 tháng 7 2016

\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)

\(\Rightarrow\left(\frac{x+1}{58}+1\right)+\left(\frac{x+2}{57}+1\right)=\left(\frac{x+3}{56}+1\right)+\left(\frac{x+4}{55}+1\right)\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)

\(\Rightarrow x+59=0\)

\(\Rightarrow x=-59\)

5 tháng 7 2018

\(\dfrac{x+1}{58}+\dfrac{x+2}{57}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)

\(\Leftrightarrow\left(\dfrac{x+1}{58}+1\right)+\left(\dfrac{x+2}{57}+1\right)=\left(\dfrac{x+3}{56}+1\right)+\left(\dfrac{x+4}{55}+1\right)\)

\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)

\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)

\(\Leftrightarrow x+59=0\)

\(\Leftrightarrow x=-59\)

5 tháng 7 2018

\(\dfrac{x+1}{58}+\dfrac{x+2}{59}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)

\(\Leftrightarrow\dfrac{x+1}{58}+1+\dfrac{x+2}{57}+1=\dfrac{x+3}{56}+1+\dfrac{x+4}{55}+1\)

\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}=\dfrac{x+59}{56}+\dfrac{x+59}{55}\)

\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)

\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)

\(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\ne0\)

\(\Rightarrow x+59=0\)

\(\Leftrightarrow x=-59\)

Vậy: \(S=\left\{-59\right\}\)

11 tháng 3 2017

\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}=\dfrac{x-4}{56}-1+\dfrac{x-5}{55}-1+\dfrac{x-6}{54}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{56}+\dfrac{x-60}{55}+\dfrac{x-60}{54}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}-\dfrac{1}{54}\right)=0\)

\(\Leftrightarrow x-60=0\)

\(\Rightarrow x=60\)

vậy \(S=\left\{60\right\}\)

14 tháng 1 2015

(x -1)/59 -1 +(x-2)/58 -1 +(x-3)/57 -1 = (x-3)/56 -1 +(x-4)/55 -1 +(x-5)/54 -1

<=> (x-60)/59 +(x-60)/58 + (X-60)/57 -(x-60)/56 - (X-60)/55 -(X-60)/54 =0

<=> (x-60).(1/59 +1/58 +1/57 -1/56 -1/55 - 1/54)=0

vì 1/59 +1/58 +1/57 -1/56 -1/55 -1/54  <0

nên x-60 =0 <=> x=60

14 tháng 1 2015

đề bài của bạn bi sai vì vế trái không thể bằng vế phải nếu đề đúng thì phải là :

(x-1)/59 +(x-2)/58 +(x-3)/57 =(x-4)/56 +(x-5)/55 +(x-6)/54

khí đó bạn giải cách như trên ,chúc bạn học toán tốt

 

 

12 tháng 7 2016

\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)

\(\Rightarrow\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(\Rightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(\Rightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

Mà \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\)

\(\Rightarrow x+59=0\)

\(\Rightarrow x=-59\)

19 tháng 5 2020

Giải phương trình:

\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)

\(\Leftrightarrow\left(\frac{x+1}{58}+1\right)+\left(\frac{x+2}{57}+1\right)=\left(\frac{x+3}{56}+1\right)+\left(\frac{x+4}{55}+1\right)\)

\(\Leftrightarrow\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(\Leftrightarrow\left(x+59\right)\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)=0\)

\(\Leftrightarrow x+59=0\) \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\ne0\right)\)

\(\Leftrightarrow x=-59\)

Vậy : \(S=\left\{-59\right\}\)

19 tháng 5 2020

\(\frac{x+1}{58}+\frac{x+2}{57}=\frac{x+3}{56}+\frac{x+4}{55}\)

\(\Leftrightarrow\) \(\frac{x+1}{58}+1+\frac{x+2}{57}+1=\frac{x+3}{56}+1+\frac{x+4}{55}+1\)

\(\Leftrightarrow\) \(\frac{x+59}{58}+\frac{x+59}{57}=\frac{x+59}{56}+\frac{x+59}{55}\)

\(\Leftrightarrow\) \(\frac{x+59}{58}+\frac{x+59}{57}-\frac{x+59}{56}-\frac{x+59}{55}=0\)

\(\Leftrightarrow\) (x + 59)(\(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\)) = 0

\(\Leftrightarrow\) x + 59 = 0

\(\Leftrightarrow\) x = -59

Vậy S = {-59}

Chúc bn học tốt!!

1 tháng 3 2017

a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Leftrightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)

\(\Leftrightarrow\) \(\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

\(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)

\(\Rightarrow x+95=0\)

\(\Leftrightarrow x=-95\)

Vậy phương trình có một nghiệm x = -95

b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)

\(\Leftrightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)

\(\Leftrightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-60}{55}-\frac{x-60}{54}=0\)

\(\Leftrightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)

\(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)

\(\Rightarrow x-60=0\)

\(\Leftrightarrow x=60\)

Vậy phương trình có một nghiệm x = 60

1 tháng 3 2017

a) \(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)

\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

\(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)

\(\Rightarrow x+95=0\)

\(\Rightarrow x=-95\)

Vậy x = -95

b) \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)

\(\Rightarrow\left(\frac{x-1}{59}-1\right)+\left(\frac{x-2}{58}-1\right)+\left(\frac{x-3}{57}-1\right)=\left(\frac{x-4}{56}-1\right)+\left(\frac{x-5}{55}-1\right)+\left(\frac{x-6}{54}-1\right)\)

\(\Rightarrow\frac{x-60}{59}+\frac{x-60}{58}+\frac{x-60}{57}-\frac{x-60}{56}-\frac{x-5}{55}-\frac{x-6}{54}=0\)

\(\Rightarrow\left(x-60\right)\left(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\right)=0\)

\(\frac{1}{59}+\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}-\frac{1}{54}\ne0\)

\(\Rightarrow x-60=0\)

\(\Rightarrow x=60\)

Vậy x = 60

9 tháng 4 2020

Xin lỗi mình làm hơi tắt nha !!!Còn 1 cách nữa ,nếu bạn muốn thì nói với mình nha !!

Ta có : \(\frac{x-1}{59}+\frac{x-2}{58}+\frac{x-3}{57}=\frac{x-4}{56}+\frac{x-5}{55}+\frac{x-6}{54}\)

\(\Leftrightarrow\frac{x}{59}+\frac{x}{58}+\frac{x}{57}-\frac{x}{56}-\frac{x}{55}-\frac{x}{54}=\frac{1}{59}+\frac{2}{58}+\frac{3}{57}-\frac{4}{56}-\frac{5}{55}-\frac{6}{54}\)

<=> x = 60 

Vậy x = 60

9 tháng 4 2020

Bạn kiểm tra lại đề nhé. Chỗ

\(.....=\frac{x-4}{56}+\frac{x-5}{56}+\frac{x-6}{54}\)

11 tháng 1 2017

Bài 1:
\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)

\(\Rightarrow\left(\frac{x+1}{94}+1\right)+\left(\frac{x+2}{93}+1\right)+\left(\frac{x+3}{92}+1\right)=\left(\frac{x+4}{91}+1\right)+\left(\frac{x+5}{90}+1\right)+\left(\frac{x+6}{89}+1\right)\)

\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)

\(\Rightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)

\(\Rightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)

\(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)

\(\Rightarrow x+95=0\)

\(\Rightarrow x=-95\)

Vậy x = -95

Bài 2: tương tự

11 tháng 1 2017

\(\frac{x+1}{94}\)+\(\frac{x+2}{93}\)+\(\frac{x +3}{92}\)= \(\frac{x+4}{91}\)+ \(\frac{x+5}{90}\)+ \(\frac{x+6}{89}\)

<=> [(\(\frac{x+1}{94}\)+\(\frac{x+2}{93}\)+\(\frac{x+3}{92}\)+3)]= [(\(\frac{x+4}{91}\)+\(\frac{x+5}{90}\)+\(\frac{x+6}{89}\)+3)]

<=> [(\(\frac{x+1}{94}\)+1)+(\(\frac{x+2}{93}\)+1)+(\(\frac{x+3}{92}\)+1)]- [(\(\frac{x+4}{91}\)+1)+(\(\frac{x+5}{90}\)+1)+(\(\frac{x+6}{89}\)+1)] =0

<=> [(\(\frac{x+1}{94}\)+ \(\frac{94}{94}\))+(\(\frac{x+2}{93}\)+\(\frac{93}{93}\))+(\(\frac{x+3}{92}\)+\(\frac{92}{92}\))] -[(\(\frac{x+4}{91}\)+\(\frac{91}{91}\))+(\(\frac{x+5}{90}\)+\(\frac{90}{90}\))+(\(\frac{x+6}{89}\)+\(\frac{89}{89}\))] =0

<=> (\(\frac{x+95}{94}\)+\(\frac{x+95}{93}\)+\(\frac{x+95}{92}\)) -(\(\frac{x+95}{91}\)+\(\frac{x+95}{90}\)+\(\frac{x+95}{89}\)) =0

<=> (x+95)( \(\frac{1}{94}\)+\(\frac{1}{93}\)+\(\frac{1}{92}\)-\(\frac{1}{91}\)-\(\frac{1}{90}\)-\(\frac{1}{89}\)) =0

Vì (\(\frac{1}{94}\)+\(\frac{1}{93}\)+\(\frac{1}{92}\)-\(\frac{1}{91}\)-\(\frac{1}{90}\)-\(\frac{1}{89}\)) \(\ne\) 0

=> x+95=0

<=> x= -95

Vậy S={-95}