K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

30 tháng 3 2017

\(\dfrac{2-x}{2007}\) - 1 = \(\dfrac{1-x}{2008}\) - \(\dfrac{x}{2009}\)

<=> \(\dfrac{2-x}{2009}\) +1 -1 +1 = \(\dfrac{1-x}{2008}\) +1 - \(\dfrac{x}{2009}\) +1

<=> \(\dfrac{2-x+2007}{2007}\) = \(\dfrac{1-x+2008}{2008}\) + \(\dfrac{-x+2009}{2009}\)

<=> \(\dfrac{2009-x}{2007}\) = \(\dfrac{2009-x}{2008}\) + \(\dfrac{2009-x}{2009}\)

<=> (2009-x)(\(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) - \(\dfrac{1}{2009}\) ) = 0

<=> 2009 -x = 0

hoặc: \(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) -\(\dfrac{1}{2009}\) = 0

\(\dfrac{1}{2007}\) \(\ne\) \(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\)

=> \(\dfrac{1}{2007}\) - (\(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\) ) \(\ne\) 0

=> 2009 -x =0

<=> x =2009

30 tháng 3 2017

\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2009-x}{2007}-2=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}-2\)

\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)

\(\Rightarrow2009-x=0\Leftrightarrow x=2009\)

AH
Akai Haruma
Giáo viên
28 tháng 1 2021

Lời giải:

a) 

PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)

\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)

\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)

\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)

\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)

\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)

b) Bạn kiểm tra lại xem có sai đề không?

6 tháng 3 2017

đề ko có vấn đề nhỉ?

7 tháng 3 2017

Không chẳng có vấn đề gì cả. có thể sai so với cái đề nào đó "nội hàm nó đúng"

\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{-x+4}{2006}+\dfrac{-x-2008}{6}\)

\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).x=\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)\)\(x=\dfrac{\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)}{\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).}\)

Thích thì rút gọn chẳng thích thì kệ nó

15 tháng 5 2018

Giải:

\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)

\(\Leftrightarrow\dfrac{2-x}{2007}-1+2=\dfrac{1-x}{2008}-\dfrac{x}{2009}+2\)

\(\Leftrightarrow\dfrac{2-x}{2007}+1=\dfrac{1-x}{2008}+1-\dfrac{x}{2009}+1\)

\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)

\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}\)

\(\Leftrightarrow\dfrac{2009-x}{2007}-\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)

\(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\ne0\)

\(\Leftrightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy ...

15 tháng 5 2018

\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)

\(\Leftrightarrow\left(\dfrac{2-x}{2007}+1\right)-\left(1+1\right)=\left(\dfrac{1-x}{2008}+1\right)-\left(\dfrac{x}{2009}+1\right)\)

\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)

\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}+\dfrac{-x+2009}{2009}\)

\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\)

\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\)

\(\Leftrightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

22 tháng 4 2018

ta có (x+1/2009 +1) + ( x+3/2007 + 1)- (x+5/2005 +1) - (x+7/1993 + 1) = 0

=>(x +100/ 2009) + (x+100/2007) - (x+100/2005)-(x+100/1993)

=> (x +100) * (1/2009 + 1/2007+ 1/2005 + 1/1993) = 0

=> x = -100

Bạn cứ tinh ý để ý đến phần tử và mẫu cộng lại bằng 100. Khi bạn bỏ phần x + 100 ra thì còn lại như trên. Sau đó lược bỏ còn lại x = -100

22 tháng 4 2018

Mạn phép mk không chép đề , mk làm luôn nhé

\(\dfrac{x+1}{2009}+1+\dfrac{x+3}{2007}+1=\dfrac{x+5}{2005}+1+\dfrac{x+7}{1993}+1\)

\(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2005}-\dfrac{x+2010}{1993}=0\)

⇔( x + 2010 )\(\left(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}\right)=0\)

Ta thấy : \(\dfrac{1}{2009}< \dfrac{1}{2007}< \dfrac{1}{2005}< \dfrac{1}{1993}\)

\(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}< 0\)

⇒ x + 2010 = 0

⇒ x = -2010

KL....

13 tháng 2 2018

\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)

\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)

\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)

\(\Rightarrow x+2010=0\)

\(\Rightarrow x=-2010\)

Vậy pt có nghiệm duy nhất \(x=-2010\)

14 tháng 3 2018

giải phương trình

x^2+x-2=0

vậy kết quả bằng mấy vậy

31 tháng 7 2018

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

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

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

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

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

=> x=-2010

vậy.....

12 tháng 8 2017

Mở đầu về phương trình

Mở đầu về phương trình

12 tháng 8 2017

2.

\(\dfrac{x+5}{2006}+\dfrac{x+4}{2007}+\dfrac{x+3}{2008}< \dfrac{x+9}{2002}+\dfrac{x+10}{2001}+\dfrac{x+11}{2000}\\ \Leftrightarrow\dfrac{x+5}{2006}+1+\dfrac{x+4}{2007}+1+\dfrac{x+3}{2008}+1< \dfrac{x+9}{2002}+1+\dfrac{x+10}{2001}+1+\dfrac{x+11}{2000}+1\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}< \dfrac{x+2011}{2002}+\dfrac{x+2011}{2001}+\dfrac{x+2011}{2000}\\ \Leftrightarrow\dfrac{x+2011}{2006}+\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}-\dfrac{x+2011}{2002}-\dfrac{x+2011}{2001}-\dfrac{x+2011}{2000}< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2002}-\dfrac{1}{2001}-\dfrac{1}{2000}\right)< 0\\ \Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\left\{{}\begin{matrix}\dfrac{1}{2006}< \dfrac{1}{2002}\\\dfrac{1}{2007}< \dfrac{1}{2001}\\\dfrac{1}{2008}< \dfrac{1}{2000}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2006}-\dfrac{1}{2002}< 0\\\dfrac{1}{2007}-\dfrac{1}{2001}< 0\\\dfrac{1}{2008}-\dfrac{1}{2000}< 0\end{matrix}\right.\Rightarrow\left(\dfrac{1}{2006}-\dfrac{1}{2002}+\dfrac{1}{2007}-\dfrac{1}{2001}+\dfrac{1}{2008}-\dfrac{1}{2000}\right)< 0\)

\(\Rightarrow x>0\)

Vậy \(x>0\)