(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

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

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

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

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

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

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

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

\(\Rightarrow x+59=0\)

\(\Rightarrow x=-59\)

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

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

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

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

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

\(\Rightarrow x+95=0\)

\(\Leftrightarrow x=-95\)

Vậy phương trình có một nghiệm x = -95

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

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

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

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

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

\(\Rightarrow x-60=0\)

\(\Leftrightarrow x=60\)

Vậy phương trình có một nghiệm x = 60

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

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

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

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

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

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

\(\Rightarrow x+95=0\)

\(\Rightarrow x=-95\)

Vậy x = -95

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

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

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

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

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

\(\Rightarrow x-60=0\)

\(\Rightarrow x=60\)

Vậy x = 60

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

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

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

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

\(\Leftrightarrow\) (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!!

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 : \(S=\left\{-59\right\}\)

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

Bạn kiểm tra lại đề nhé. Chỗ

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

tae tae ơi khó quá hổng hiểu j hết trơn

mình làm câu cuối thôi nhé , những câu còn lại bạn tự làm đi , dễ mà :)))) chỉ cần quy đồng mẫu lên là được 

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

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

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

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

Vì \(\left(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\right)\) luôn khác 0 

<=> x + 59 = 0 

<=> x=-59 

Bài 1

a) 7x - 5 = 13 - 5x

7x + 5x = 13 + 5

12x = 18

x = 18 : 12

x = 1,5

b) 13 - 7x = 4x - 20

-7x - 4x = -20 - 13

-11x = -33

x = (-33) : (-11)

x = 3

c) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)

<=> \(\frac{7\left(2x-1\right)}{7.3}-\frac{3\left(5x+2\right)}{3.7}=\frac{21.x}{21.1}+\frac{21.13}{21.1}\)

<=> 14x - 7 - 15x - 6 = 21x + 273

<=> 14x - 15x - 21x = 273 + 7 + 6

<=> -22x = 286

<=> x = 286 : (-22)

<=> x = -13

d) \(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)

<=> \(\frac{10\left(2x-3\right)}{10.3}-\frac{5\left(x-3\right)}{5.6}=\frac{6\left(4x+3\right)}{6.5}-17\)

<=> 20x - 30 - 5x + 15 = 24x + 18 - 17

<=> 20x - 5x - 24x = 18 - 17 + 30 - 15

<=> -9x = 16

<=> x = \(\frac{-16}{9}\)

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

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

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

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

Rõ ràng, \(\frac{1}{58}+\frac{1}{57}-\frac{1}{56}-\frac{1}{55}\)\(\) \(\)khác 0, do đó x + 59 = 0, suy ra x = -59

Bài 2

Ta có 3x + 3 = 0

<=> 3x = -3

<=> x = -1

Thay x = -1 vào phương trình (2) ta được

5 - k . (-1) = 7

<=> k . (-1) = 5 - 7

<=> k . (-1) = -2

<=> k = (-2) : (-1)

<=> k = 2

Vậy k = 2 thì nghiệm của phương trình (1) là nghiệm của phương trình (2)

Ai tick cho mình tròn 40 với

các bạn không giải thì làm ơn đừng trả lời 

\(h.\)  \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)

\(\Leftrightarrow\)  \(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}+4=0\)

\(\Leftrightarrow\)  \(\left(\frac{99-x}{101}+1\right)+\left(\frac{97-x}{103}+1\right)+\left(\frac{95-x}{105}+1\right)+\left(\frac{93-x}{107}\right)=0\)

\(\Leftrightarrow\)  \(\frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)

\(\Leftrightarrow\)  \(\left(200-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)

\(\Leftrightarrow\)   \(200-x=0\)  \(\Leftrightarrow\)  \(x=200\)