có 1/x(x+1)= 1/x + 1/2011

 => 1/x - 1/x+1=1/x+ 1/2011

=> -1/x+1=1/x-1/x+1/2011

=>-1/x+1=1/2011

=> -(x+1)=2011

=> x+1 = -2011

=>x=-2011-1=-2012

vậy x= -2012

2010

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\frac{1}{x\left(x+1\right)}-\frac{1}{x}=2011\)

\(\frac{1-x+1}{x\left(x+1\right)}=\frac{1}{2011}\)

\(\frac{x}{x\left(x+1\right)}=\frac{1}{2011}\)

x(x+1)=2011x

=>x+1=2011

=>x=2010

a)

\(2^x\left(1+2+2^2+2^3\right)=480\)

\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)

Chính Xác 100% là X=5 

k cho mink nhé các pạn

Cái này lớp 6 : 

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

<=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{4026}=\frac{1}{2013}\)

\(\Leftrightarrow x+1=2013\)

=> x = 2012

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}\)

\(\Rightarrow1-\frac{2}{x+1}=\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{x+1}=1-\frac{2011}{2013}\)

\(\Rightarrow\frac{2}{x+1}=\frac{2}{2013}\)

\(\Rightarrow x+1=2013\)

\(\Rightarrow x=2013-1\)

\(\Rightarrow x=2012\)

Vậy \(x=2012\)

~ Ủng hộ nhé 

(x khác 0,-1)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\Leftrightarrow-\frac{1}{x+1}=\frac{1}{2011}\Leftrightarrow x+1=-2011\Leftrightarrow x=-2012.\)

\(\Rightarrow\frac{1}{x\left(x+1\right)}-\frac{1}{x}=\frac{1}{2011}\)

\(\Rightarrow\frac{1}{x\left(x+1\right)}-\frac{x+1}{x\left(x+1\right)}=\frac{1}{2011}\)

\(\Rightarrow\frac{1-x-1}{x\left(x+1\right)}=\frac{1}{2011}\)

\(\Rightarrow-\frac{x}{x\left(x+1\right)}=\frac{1}{2011}\)

\(\Rightarrow\frac{-1}{x+1}=\frac{1}{2011}\)

\(\Rightarrow2011.\left(-1\right)=\left(x+1\right).1\)

\(\Rightarrow-2011=x+1\)

\(\Rightarrow x=-2011-1\)

\(\Rightarrow x=-2012\)

\(\frac{1}{x\left(x+1\right)}=\frac{1}{x}+\frac{1}{2011}\)

\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Rightarrow\frac{1}{x}+\frac{-1}{x+1}=\frac{1}{x}+\frac{1}{2011}\)

\(\Rightarrow\frac{-1}{x+1}=\frac{-1}{-2011}\)

\(\Rightarrow x+1=-2011\)

\(\Rightarrow x=-2011-1=-2012\)

mình chưa tìm ra câu trả lời xin lỗi

Anh chỉ giải câu a thôi, câu b anh thấy nó bình thường mà.

Cộng vào mỗi phân số thêm 1 đơn vị được:

\(\frac{x+2013}{2009}+\frac{x+2013}{2010}=\frac{x+2013}{2011}+\frac{x+2013}{2012}\).

Tới đây tự làm tiếp nhá.

\(\frac{1}{2011}.x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2010}\right).\left(1-\frac{1}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2009}{2010}.\frac{2010}{2011}\)

\(\frac{1}{2011}.x=\frac{1.2.3...2009.2010}{2.3.4...2010.2011}\)\(=\frac{1}{2011}\)

\(x=\frac{1}{2011}:\frac{1}{2011}=1\)

Vậy x=1

\(\frac{1}{2011}.x=\frac{1}{2}.\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)......\left(\frac{2010}{2011}\right)\)

\(\frac{1}{2011}.x=\frac{2}{4}.\left(\frac{4}{6}\right).\left(\frac{6}{8}\right).......\left(\frac{4018}{4020}\right).\left(\frac{4020}{4022}\right)\)

\(\frac{1}{2011}.x=\frac{2.4.6.8.....4018.4020}{4.6.8.10.....4020.4022}\)

\(\frac{1}{2011}.x=\frac{2}{4022}\)

\(\Rightarrow\)\(x=\frac{2}{4022}:\frac{1}{2011}=1\)

Ai thấy đún thì ủng hộ mink nha !!!

Thanks you very much !!

Chúc các bạn luôn học giỏi !!!

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<