x+2/2008 + x+3/2007 + x+4/2006 + x+2028/6 =0
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)0
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
\(\Rightarrow x+2010=0\Rightarrow x=-2010\)
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy x = -2010
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\dfrac{x+2}{2008}+1+\dfrac{x+3}{2007}+1+\dfrac{x+4}{2006}+1+\dfrac{x+2028}{6}-3=0\)
\(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+2010=0\). Do \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x=-2010\)
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0+1+1+1-3\) \(\Leftrightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
Vậy S={-2010}
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\frac{x+2}{2008}\)+ 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
\(\Leftrightarrow\)(x +2010).\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)\)=0
\(\Leftrightarrow\)x + 2010 = 0 \(\left(vì\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\right)\)
\(\Leftrightarrow\) x = -2010
Vậy S = \(\left\{-2010\right\}\)
⇔ \(\dfrac{x+2}{2008}\) +1 +\(\dfrac{x+3}{2007}\) +1+\(\dfrac{x+4}{2006}\)+1 +\(\dfrac{2028}{6}\)-3 =0
⇔\(\dfrac{x+2}{2008}+\dfrac{2008}{2008}+\dfrac{x+3}{2007}+\dfrac{2007}{2007}+\dfrac{x+4}{2006}+\dfrac{2006}{2006}+\dfrac{x+2028}{6}-\dfrac{18}{6}=0\)
⇔\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
⇔(x+2010)\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)=0\)
Mà \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right)\)≠0
⇒x+2010=0
⇔x=-2010
Vậy phương trình có nghiệm x=-2010
Bạn tham khảo link này,định copy ra nhưng cơ mà thôi,tự xem nhé.
Giải PT: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\\ \Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\\ \Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\\ \Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\\ \Leftrightarrow x+2010=0\\ \Leftrightarrow x=-2010\)
Vậy pt có tập nghiệm \(S=\left\{-2010\right\}\)
Giải phương trình sau:
x+2/2008 + x+3/2007 + x+4/2006 + x+2028/6 =0
Ta có :
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\)\(\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\) ( quy đồng )
\(\Leftrightarrow\)\(\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
\(\Rightarrow\)\(x+2010=0\)
\(\Rightarrow\)\(x=-2010\)
Vậy \(x=-2010\)
Chúc bạn học tốt ~
cho mình hỏi tại sao x+2028/6 lại trừ 3 và sao nó lại ra x+2010/6
Giải pt:
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)
\(\Leftrightarrow x+2010=0\) vì \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)
=> x=-2010
vậy.....
giải phương trình
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
(x+2/2008+1)+(x+3/2007+1)+(x+4/2006+1)+(x+2028/6-3)=0
=x+2010/2008+ x+2010/2007+ x+2010/2006+ x+2010/6=0
=(x+2010)(1/2008+1/2007+1/2006+1/6)=
VÌ 1/2008 +1/2007 +1/2006+1/6 khác 0
=>x+2010=0=>x=-2010
giải phương trình hộ nhé
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
3 hạng tử đầu , mỗi hạng tử cùng cộng 1
Hạng tử cuối trừ 3
Nhân tử chung : x + 2010
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
\(\Rightarrow x+2010=0\Leftrightarrow x=-2010\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
Vậy....
giải phương trình sau
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\) O
Để em trình bày dễ hiểu có chú thíck lun cho chụy :)
Ta có :
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2028}{6}-3\right)=0\) ( cộng 3 phân số đầu cho 3, trừ phân số cuối cho 3 )
\(\Leftrightarrow\)\(\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\) ( quy đồng )
\(\Leftrightarrow\)\(\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\) ( vì tổng lớn hơn 0 nên khác 0 )
Nên \(x+2010=0\)
\(\Rightarrow\)\(x=-2010\) ( chuyển vế )
Vậy \(x=-2010\)
Chúc chụy học tốt ~
x + 2 x + 3 x + 4 x + 2028
▬▬▬ + ▬▬▬ + ▬▬▬▬ + ▬▬▬▬▬ = 0
2008 2007 2006 6
<=> 2007.2006.6.x + 2.2007.2006.6 + 2008.2006.6x + 3.2008.2006.6 + 2008.2007.6x + 4.2008.2007.6 + 2008.2007.2006x + 2028.2008.2007.2006 = 0
<=> ( 2007.2006.6 + 2008.2006.6 + 2008.2007.6 + 2008.2007.2006 )x = -( 2.2007.2006.6 + 3.2008.2006.6 + 4.2008.2007.6 + 2028.2008.2007.2006 )
<=> x = -( 2.2007.2006.6 + 3.2008.2006.6 + 4.2008.2007.6 + 2028.2008.2007.2006 ) / ( 2007.2006.6 + 2008.2006.6 + 2008.2007.6 + 2008.2007.2006 ) = -2010
cộng 1 vào 3 phân thức đầu và trừ 3 ở phân thức thứ tư thì ta sẽ được các tử bằng nhau rồi dùng phân phối sẽ ra.