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11 tháng 3 2021

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

<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1

<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)

<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0

<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0

<=> x + 66 = 0

<=> x = -66

11 tháng 3 2021

hình như bn sai đề đúng ko

\(\Leftrightarrow x+66=0\)

hay x=-66

2 tháng 1 2022

chị ơi, chị giải thích rõ được ko ạ?

21 tháng 3 2020

a)\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow2\cdot\left(x^3+1\right)+7x\cdot\left(x+1\right)=0\)

\(\Leftrightarrow2\cdot\left(x+1\right)\cdot\left(x^2+x+1\right)+7x\cdot\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left[2\cdot\left(x^2+x+1\right)+7x\right]=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2-2x+2+7x\right)=0\)

\(\Leftrightarrow\left(x+1\right)\cdot\left(2x^2+5x+2\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-1}{2}\\x=-2\end{matrix}\right.\)

21 tháng 3 2020

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

\(\Leftrightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)

\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)

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

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

\(\Rightarrow x+66=0\)

\(\Rightarrow x=-66\)

6 tháng 8 2017

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

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

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

\(\Leftrightarrow\left(x+66\right)\cdot\left(\dfrac{1}{65}+\dfrac{1}{63}\right)=\left(x+66\right)\cdot\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\)

\(\Rightarrow x=-66\)

Vậy x = -66.

6 tháng 8 2017

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

\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)=0\)\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)\(\Leftrightarrow\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)Nhận xét : Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)

\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)

Vậy để \(\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)\(\Leftrightarrow x+66=0\Leftrightarrow x=-66\)

Vậy....

tik mik nha !!!

28 tháng 4 2015

Cộng cả tử và mẫu với 1=> Kết quả là -66

28 tháng 4 2015

Bạn cộng cả tử và mẫu vs 1 là xong 

29 tháng 3 2020

\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010+18}{6}=0\)

\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010}{6}+3=0\)

\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2010}{6}\right)=0\)

\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+6\right)=0\)

Vì :\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

=> x + 2010 = 0

=> x = -2010

b) \(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)

\(\Rightarrow\frac{x-3}{2011}+\frac{x-2}{2012}-\frac{x-2012}{2}-\frac{x-2011}{3}=0\)

\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-2012}{2}-1\right)-\left(\frac{x-2011}{3}-1\right)=0\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)

\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)

=> x - 2014 = 0

=> x = 2014

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

\(\Rightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)

\(\Rightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)

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

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

Vì :\(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)

=> x + 66 = 0

=> x = -66

a) x - 1 / x + 1 / x + 1 = 2x - 1/x^2 + x

ĐKXĐ:x khác 0;-1

<=>(x - 1)(x + 1) / x(x + 1) + x / x(x+1) = 2x + 1 / x(x + 1)

  =>(x - 1)(x + 1) + x = 2x + 1

<=>x^2 - 1^2 + x = 2x + 1

<=>x^2 + x - 2x - 1 - 1 = 0

<=>x^2 - x - 2 = 0

<=>x^2 + x - 2x - 2 = 0

<=>x(x + 1) - 2(x + 1) = 0

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

<=>x - 2 = 0

      x + 1 = 0

<=>x = 2(t/m)

      x = -1(loại vì ĐKXĐ: x khác 0;-1)

Vậy S = {2}

13 tháng 2 2018

Ta có : 

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

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

\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)

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

Vì \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)

\(\Rightarrow\)\(x+66>0\)

\(\Rightarrow\)\(x>-66\)

Vậy \(x>-66\)

13 tháng 2 2018

mình nhầm câu 2 phải là <=0

18 tháng 2 2020

sê đài

18 tháng 2 2020

Sửa đề:

\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\\\Leftrightarrow \frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\\ \Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\\\Leftrightarrow \left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\\\Leftrightarrow x+66=0\left(Vi\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)\\ \Leftrightarrow x=-66\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-66\right\}\)

Bài 15:

Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)

\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)

\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)

nên x+2010=0

hay x=-2010

Vậy: x=-2010

Bài 17:

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

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

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

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

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

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

\(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)

nên x+66=0

hay x=-66

Vậy: x=-66