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14 tháng 7 2016

(x + 1)/58 + (x + 2)/57 = (x + 3)/56 + (x + 4)/55

(x + 1)/58 + 1 + (x + 2)/57 + 1 = (x + 3)/56 + 1 + (x + 4)/55 + 1

(x + 59)/58 + (x + 59)/57 = (x + 59)/56 + (x + 59)/55

=> (x + 59)/58 + (x + 59)/57 - (x + 59)/56 - (x + 59)/55 = 0

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

Do 1/56 > 1/58; 1/55 > 1/57 => 1/58 + 1/57 - 1/56 - 1/55 khác 0

=> x + 59 = 0

=> x = -59

14 tháng 7 2016

(x + 1)/58 + (x + 2)/57 = (x + 3)/56 + (x + 4)/55

(x + 1)/58 + 1 + (x + 2)/57 + 1 = (x + 3)/56 + 1 + (x + 4)/55 + 1

(x + 59)/58 + (x + 59)/57 = (x + 59)/56 + (x + 59)/55

=> (x + 59)/58 + (x + 59)/57 - (x + 59)/56 - (x + 59)/55 = 0

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

Do 1/56 > 1/58; 1/55 > 1/57 => 1/58 + 1/57 - 1/56 - 1/55 khác 0

=> x + 59 = 0

=> x = -59

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 .

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

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

\(\Leftrightarrow x-60=0\)

hay 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

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

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

 

 

a) Ta có: \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow-2x+3+x+4=0\)

\(\Leftrightarrow-x+7=0\)

\(\Leftrightarrow-x=-7\)

hay x=7

Vậy: S={7}

b) Ta có: \(\dfrac{2+x}{5}-0.5x=\dfrac{1-2x}{4}+0.25\)

\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{0.5x\cdot20}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{20\cdot0.25}{20}\)

\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)

\(\Leftrightarrow8+4x-10x=5-10x+5\)

\(\Leftrightarrow-6x+8=-10x+10\)

\(\Leftrightarrow-6x+8+10x-10=0\)

\(\Leftrightarrow4x-2=0\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)

d) Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-59}{1}+\dfrac{x-58}{2}+\dfrac{x-57}{3}\)

\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}-1=\dfrac{x-59}{1}-1+\dfrac{x-58}{2}-1+\dfrac{x-57}{3}-1\)

\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{1}+\dfrac{x-60}{2}+\dfrac{x-60}{3}\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}\right)-\left(x-60\right)\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

mà \(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-1-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)

nên x-60=0

hay x=60

Vậy: S={60}