TÌM X
\(b.\frac{2x+1}{2009}+\frac{2x+2}{2008}+\frac{2x+3}{2007}=\frac{2x+10}{2000}+\frac{2x+11}{1999}+\frac{2x+12}{1998}\)
các bạn chịu khó giúp mình nha chứ mình không giỏi toán lắm
\(\)\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
GIÚP MÌNH NHÉ ,MÌNH CẦN GẤP.
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
Tìm x biết
a,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
b, \(\frac{1}{x-1}+\frac{-2}{3}\cdot\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
c,\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
d, \(\left(2x-4\right)\cdot\left(9-3x\right)>0\)
e, \(\left(\frac{3}{2x}-4\right)\cdot\frac{5}{3}>\frac{15}{6}\)
Không cần làm hết đâu , giúp mình mấy câu các bạn biết thôi , trình bày giùm mình nha
cảm ơn , mong mọi người giúp đõ
Mình làm cho bạn 2 câu khó hơn còn mấy câu còn lại dungf phương pháp quy đồng rồi chuyển vế là tính được mà
c, <=> [(x-1)/2009 ]-1 +[ (x-2)/2008] -1 = [(x-3)/2007]-1 +[(x-4)/2006]-1
<=> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/2007 + (x-2010)/2006
<=> (x-2010)*(1/2009+1/2008-1/2007-1/2006)=0
=> x-2010=0 => x=2010
d, TH1 : cả hai cùng âm
=>> 2X-4 <O => X< 2
Và 9-3x<0 =>> x> 3
=>> loại
Th2 cả hai cùng dương
2x-4>O => x>2
Và 9-3x>O => x<3
=>> 2<x<3 (tm)
Tìm x
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
tìm x biết:
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+x+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(\Rightarrow\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
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
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
Giúp mik với !! mik ko hiểu bài này !! giúp mik nhé
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(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
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\right)=0\Leftrightarrow x=-2010\)
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
a,\(\left(x-1\right)\cdot\left(x+2\right)>0\)
b,\(\left(\frac{1}{2}-x\right)\cdot\left(2x-1\right)< 0\)
c,\(\frac{x-1}{2007}+\frac{x-2}{2008}=\frac{x-3}{2009}+\frac{x-4}{2010}\)
d,\(\left(2x-1\right)\cdot\left(2-x\right)>0\)
Tìm x
bạn nào bít thì giúp mình nha!!!
a, \(\left(x-1\right).\left(x+2\right)\)\(>0\Rightarrow\orbr{\begin{cases}x-1< 0;x+2< 0\left(loai\right)\Rightarrow x< 1\\x-1>0;x+2>0\Rightarrow x>1;x>-2\end{cases}}\)
=> -2 < x < 1
Câu b và câu d làm tương tự nha bạn(Câu b thì xét khác dấu)
a) a= 2 và 1
b) = 7
c= 5600 và 7899
d 5 và 6
Giải các phương trình sau :
a) ( 3x - 2 )( 4x + 3 ) = ( 2 - 3x )( x - 1)
b) x2 + ( x + 3 )( 5x - 7 ) = 9
c) 2x2 + 5x + 3 = 0
d) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)
Giúp mik vs !!!
a, (3x - 2)(4x + 3) = (2 - 3x)(x - 1)
\(\Leftrightarrow\) (3x - 2)(4x + 3) - (2 - 3x)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3) + (3x - 2)(x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(4x + 3 + x - 1) = 0
\(\Leftrightarrow\) (3x - 2)(5x + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{-2}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{2}{3}\); \(\frac{-2}{5}\)}
b, x2 + (x + 3)(5x - 7) = 9
\(\Leftrightarrow\) x2 - 9 + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + (x + 3)(5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(x - 3 + 5x - 7) = 0
\(\Leftrightarrow\) (x + 3)(6x - 10) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy S = {-3; \(\frac{5}{3}\)}
c, 2x2 + 5x + 3 = 0
\(\Leftrightarrow\) 2x2 + 2x + 3x + 3 = 0
\(\Leftrightarrow\) 2x(x + 1) + 3(x + 1) = 0
\(\Leftrightarrow\) (x + 1)(2x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy S = {-1; \(\frac{3}{2}\)}
d, \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}=\frac{3-2x}{2009}+\frac{3-2x}{2010}\)
\(\Leftrightarrow\) \(\frac{3-2x}{2006}+\frac{3-2x}{2007}+\frac{3-2x}{2008}-\frac{3-2x}{2009}-\frac{3-2x}{2010}=0\)
\(\Leftrightarrow\) (3 - 2x)\(\left(\frac{1}{2006}+\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)\) = 0
\(\Leftrightarrow\) 3 - 2x = 0
\(\Leftrightarrow\) x = \(\frac{3}{2}\)
Vậy S = {\(\frac{3}{2}\)}
Chúc bn học tốt!!
\(\frac{1}{x^2-2x+2}+\frac{2}{x^2-2x+3}+\frac{3}{x^2-2x+4}\)
các bạn làm giúp mình nha