K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

16 tháng 7 2016

ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)

\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)

\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)

\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)

\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)

mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)

=> 2x + 7 = 0 => x = -7/2 

                                                                              Vậy x = -7/2

19 tháng 7 2017

các bạn giúp mình với. cảm ơn 

19 tháng 7 2017

giúp mình với

23 tháng 3 2019

a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)

\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)

\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)

\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0

\(x-1=0\)

\(x=1\)

30 tháng 3 2020

\(\text{GIẢI :}\)

ĐKXĐ : \(x\ne\pm1\)

\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)

\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}=\frac{x+1}{x^2-1}\)

\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}-\frac{x+1}{x^2-1}=0\)

\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow\text{ }2\left(x-1\right)+x\left(x+1\right)-(x+1)=0\)

\(\Leftrightarrow\text{ }2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2+x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1\text{ (loại)}\\x=-3\text{ (Chọn)}\end{cases}}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-3\right\}\).

24 tháng 5 2020

\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}.\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)\(đk:x\ne\pm1\)

\(< =>\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left[\frac{7}{6}.\frac{6}{7}+\left(1\right)\right]x+1}{x^2-1}\)

\(< =>\frac{2x-2+x^2+x}{x^2+x-x-1}=\frac{2x+1}{x^2-1}\)\(< =>\frac{x^2+3x-2}{x^2-1}=\frac{2x-1}{x^2-1}\)

\(< =>x^2+2x-2=2x-1\)\(< =>x^2+2x-2x-2+1=0\)

\(< =>x^2-1=0< =>x^2=1\)\(< =>x=\pm1\)\(\left(ktmđk\right)\)

Vậy phương trình trên vô nghiệm

3 tháng 1 2021

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x2 + 5x - 2) - (x3 - 1) = 0

=> (x - 1)(x2 + 5x - 2) - (x - 1)(x2 + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50

3 tháng 1 2021

b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)

\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)

f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)

\(\Leftrightarrow x=-66\)