sai đề rồi không có kết quả tính kiểu gì hả bạn

SAI ĐỀ RỒI BẠN, BẠN XEM LẠI NHÉ!!!

\(\text{Tìm }x\text{ }:\)

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+...+\left(x+2011\right)+\left(x+2012\right)\)

\(=x+1+x+2+x+3+x+4+...+x+2011+x+2012\)

\(\Rightarrow SSH=\left(\left(2012-1\right):1+1\right)=2012\left(SSH\right)\)

\(\Rightarrow\text{Có }2012\cdot x\)

\(=2012\cdot x+\left(2012+1\right)\cdot2012\text{ : }2\)

\(=2012\cdot x+2013\cdot2012\text{ : }2\)

\(=2012\cdot x+4050156\text{ : }2\)

\(=2012\cdot x+2025078\)

\(\text{Mình nghĩ đề bạn bị thiếu rồi bạn à ! Nếu đề bạn đúng thì bạn chỉ cần làm theo cách mình làm ở trên nha!}\)

a) 5x.(x+3/4) = 0

=> x = 0

x+3/4 = 0 => x = -3/4

b) \(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}.\)

\(\Rightarrow\frac{x+7}{2010}+\frac{x+6}{2011}-\frac{x+5}{2012}-\frac{x+4}{2013}=0\)

\(\frac{x+7}{2010}+1+\frac{x+6}{2011}+1-\frac{x+5}{2012}-1-\frac{x+4}{2013}-1=0\)

\(\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)-\left(\frac{x+5}{2012}+1\right)-\left(\frac{x+4}{2013}+1\right)=0\)

\(\frac{x+2017}{2010}+\frac{x+2017}{2011}-\frac{x+2017}{2012}-\frac{x+2017}{2013}=0\)

\(\left(x+2017\right).\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

=> x + 2017 = 0

x = -2017

a) để 2x - 3 > 0

=> 2x > 3

x > 3/2

b) 13-5x < 0

=> 5x < 13

x < 13/5

c) \(\frac{x+3}{2x-1}>0\)

=> x + 3 > 0

x > -3

d) \(\frac{x+7}{x+3}=\frac{x+3+4}{x+3}=1+\frac{4}{x+3}\)

Để x+7/x+3 < 1

=> 1 + 4/x+3 < 1

=> 4/x+3 < 0

=> không tìm được x thỏa mãn điều kiện

a) \(\frac{x+4}{x+3}< 1\)

\(\Leftrightarrow\frac{x+4}{x+3}-1< 0\)

\(\Leftrightarrow\frac{x+4-x-3}{x+3}< 0\)

\(\Leftrightarrow\frac{1}{x+3}< 0\)

\(\Leftrightarrow x+3< 0\)

\(\Leftrightarrow x< -3\)

Vậy \(x< -3\)

b) \(\frac{x+3}{x+4}>1\)

\(\Leftrightarrow\frac{x+3}{x+4}-1>0\)

\(\Leftrightarrow\frac{x+3-x-4}{x+4}>0\)

\(\Leftrightarrow-\frac{1}{x+4}>0\)

\(\Leftrightarrow x+4< 0\)

\(\Leftrightarrow x< -4\)

Vậy \(x< -4\)

c) \(\frac{x+3}{2010}+\frac{x+2}{2011}+\frac{x+1}{2012}+\frac{x+2025}{4}=0\)

\(\Leftrightarrow\left(\frac{x+3}{2010}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2025}{4}-3\right)=0\)

\(\Leftrightarrow\frac{x+2013}{2010}+\frac{x+2013}{2011}+\frac{x+2013}{2012}+\frac{x+2013}{4}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)

\(\Leftrightarrow x+2013=0\) (Vì \(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\ne0\))

\(\Leftrightarrow x=-2013\)

Vậy \(x=-2013\)

Nhớ tick đó ✔✔✔

\(\frac{x+1}{2013}+\frac{x}{2012}+\frac{x-1}{2011}=\frac{x-2}{2010}+\frac{x-3}{2009}+\frac{x-4}{2008}\)

\(\Leftrightarrow\frac{x+1}{2013}-1+\frac{x}{2012}-1+\frac{x-1}{2011}-1=\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}=\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}\)

\(\Leftrightarrow\frac{x-2012}{2013}+\frac{x-2012}{2012}+\frac{x-2012}{2011}-\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2012=0\). Do \(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)

\(\Leftrightarrow x=2012\)

Bai 2: 

a: 2x-3>0

=>2x>3

=>x>3/2

b: =>13-5x<0

=>5x>13

=>x>13/5

c: =>2x-1>0 hoặc x+3<0

=>x>1/2 hoặc x<-3

d: =>(x+7-x-3)/(x+3)<0

=>x+3<0

=>x<-3

sao giống câu hỏi của mình thế chỉ khác số bạn biết làm ko chỉ mình đi