\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

x=-2010 nhé

chúc bạn hc tốt

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+x+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)

\(\Rightarrow\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

Vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)

Nên x + 2010 = 0 => x = -2010

x+1/2009+1+x+2/2008+1+x+3/2007+1=x+10/2000+1+x+11/1999+1+x+12/1998

x+2010/2009+x+2010/2008+x+2010/2007=x+2010/2000+x+2010/1999+x+2010/1998

x+2010*(1/2009+1/2008+1/2007-1/200-1/1999-1/1998)=0

x+2010=0

x=-2010

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)

\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)

\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)

\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)

\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

x+2010=0

x=-2010

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)

\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)

\(=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\right)=0\Leftrightarrow x=-2010\)

P/s: Công vào 6 phân thức trên, mỗi phân thức công thêm 1 rồi quy đồng lên ta được:

\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

Ta xét: \(\hept{\begin{cases}\frac{1}{2009}< \frac{1}{2000}\\\frac{1}{2008}< \frac{1}{1999}\\\frac{1}{2007}< \frac{1}{1998}\end{cases}}\Rightarrow\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)

=> \(x+2010=0\Rightarrow x=-2010\)

Vậy x = -2010

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)

\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

           Bài làm :

\(...\)

\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}\right)\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

\(\text{Vì : }\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\)

\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\)

Vậy x=-2010

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

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

<=>  \(x+1=0\)  (do  1/2 + 1/3 + 1/4 - 1/5 - 1/6 khác 0)

<=>  \(x=-1\)

Vậy...

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

<=>  \(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)

<=>  \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

<=>  \(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

<=>  \(x+2010=0\)  (do  1/2009 + 1/2008 + 1/2007 - 1/2000 - 1/1999 - 1/1998 khác 0)

<=>  \(x=-2010\)

Vậy....

d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)

<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)

<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)

<=> x = -2010

b./ \(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)

\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)(b)

Mà \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}< 0\)

(b) \(\Leftrightarrow x+2010=0\Leftrightarrow x=-2010\)

a./

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

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

Mà \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)

(a) \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

c./ \(\Leftrightarrow2\cdot2^2\cdot2^3\cdot...\cdot2^x=2^{10}\Leftrightarrow2^{1+2+3+...+x}=2^{10}\Leftrightarrow1+2+3+...+x=10\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2}=10\Leftrightarrow x\left(x+1\right)=20=4\cdot5\Rightarrow x=4\)

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

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

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

Vì\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}>0\)nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

1/�+12+�+13+�+14=�+15+�+162x+1​+3x+1​+4x+1​=5x+1​+6x+1​

⇔�+12+�+13+�+14−�+15−�+16=0⇔2x+1​+3x+1​+4x+1​−5x+1​−6x+1​=0

⇔(�+1)(12+13+14−15−16)=0⇔(x+1)(21​+31​+41​−51​−61​)=0

Vì12+13+14−15−16>021​+31​+41​−51​−61​>0nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10

1/�+12+�+13+�+14=�+15+�+162x+1​+3x+1​+4x+1​=5x+1​+6x+1​

⇔�+12+�+13+�+14−�+15−�+16=0⇔2x+1​+3x+1​+4x+1​−5x+1​−6x+1​=0

⇔(�+1)(12+13+14−15−16)=0⇔(x+1)(21​+31​+41​−51​−61​)=0

Vì12+13+14−15−16>021​+31​+41​−51​−61​>0nên để biểu thức có giá trị là 0 thì x+1=0 <=> x=-1

2/Tương tự bài 2 bạn cộng mỗi vế cho 3 mỗi biểu thức cộng cho 1 thỳ bn sẽ tìm đc kq là -2010

3/ trừ 2 cho mỗi vế, mỗi biểu thức trừ cho 1, lập luận ta có x=100

4/ bài này chuyển -3 qua vế trái thành 3 rồi tách, nhóm mỗi biểu thức với 1 ta có x=-10