Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ x+2009=0
⇔ x=-2009
vậy x=-2009 là nghiệm của pt
a) ( x2 + x )2 + 4( x2 + x ) = 12
<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0
<=> ( x2 + x + 2)2 - 16 = 0
<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0
<=> ( x2 + x + 6 )( x2 + x - 2) = 0
Do : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x
=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1) + 2( x - 1) = 0
<=> ( x - 1)( x + 2 ) = 0
<=> x = 1 hoặc : x = - 2
KL.....
b) Kuroba kaito làm rùi nhé 
\(\Leftrightarrow\dfrac{x}{2005}+1+\dfrac{x-1}{2006}+1=\dfrac{x-2}{2007}+1-1+1\)
\(\Leftrightarrow\dfrac{x+2005}{2005}+\dfrac{x+2005}{2006}=\dfrac{x+2005}{2007}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\right)=0\)
\(\Leftrightarrow x+2005=0\) (vì \(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\ne0\))
\(\Leftrightarrow x=-2005\)
\(\dfrac{x}{2005}+\dfrac{x-1}{2006}=\dfrac{x-2}{2007}-1\)
\(\Leftrightarrow\dfrac{x+2005}{2005}+\dfrac{x+2005}{2006}-\dfrac{x+2005}{2007}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2005}+\dfrac{1}{2006}-\dfrac{1}{2007}\right)=0\)
\(\Leftrightarrow x=-2005\).
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
a: \(x>3:\dfrac{1}{2}=6\)
b: \(x>-2:\left(-\dfrac{1}{3}\right)=6\)
c: \(x>-4:\dfrac{2}{3}=-6\)
d: \(x< -6:\dfrac{3}{5}=-10\)
a: \(x< -9:\dfrac{3}{2}=-9\cdot\dfrac{2}{3}=-6\)
b: 2/3x>-2
hay x>-2:2/3=-3
c: \(2x>\dfrac{9}{5}-\dfrac{4}{5}=1\)
hay x>1/2
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}>6-4=2\)
hay x>2:3/5=2x5/3=10/3
1) \(2\left(3x-1\right)-3x=10\)
<=> \(6x-2-3x=10\)
<=>\(3x-2=10\)
<=> \(3x=12\)
<=> \(x=4\)
Vậy tập nghiệm của pt S={4}
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
ĐKXĐ: x khác 0; x khác 1,-1
<=> \(\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}+\dfrac{x\left(x+1\right)}{x\left(x+1\right)}\)= \(\dfrac{3x^2-x}{x\left(x+1\right)}+\dfrac{1}{x\left(x+1\right)}\)
=> \(\left(x+1\right)^2+x\left(x+1\right)\)= \(3x^2-x+1\)
<=> \(x^2+2x+1+x^2+x=3x^2-x+1\)
<=> \(x^2+x^2+2x+x-3x^2+x\)= \(1-1\)
<=> \(-x^2+4x=0\)
<=>\(4x=x^2\)
<=> \(4=x\) ( TMĐKXĐ)
Vậy tập nghiệm của pt S={4}
c) \(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2}{6}-\dfrac{9x-6}{6}>\dfrac{1}{6}\)
<=> \(\dfrac{4x+2-9x+6}{6}-\dfrac{1}{6}>0\)
<=> \(\dfrac{-5x+7}{6}>0\)
Mà 6>0 . Nên \(-5x+7>0\)
Ta có \(-5x+7>0\)
<=> \(-5x>-7\)
<=> \(x< \dfrac{7}{5}\)
Vậy tập nghiệm của bất phương trình S={x thuộc R| \(x< \dfrac{7}{5}\)}
1)2.(3x-1)-3x=10
6x-2-3x =10
6x-3x =10+2
3x =12
x =4
Vậy S=4
2) \(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
Đkxđ: \(x\ne0\) và \(x\ne-1\)
MTC;x(x+1)
\(\dfrac{x+1}{x}+1=\dfrac{3x-1}{x+1}+\dfrac{1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)\(\dfrac{\left(x+1\right)\left(x+1\right)+x\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x\left(3x-1\right)+1}{x\left(x+1\right)}\)
\(\Leftrightarrow\)(x+1) (x+1)+x(x+1) = x (3x-1)+1
\(\Leftrightarrow\)x2+x+x+1+x2+x =3x2-x+1
\(\Leftrightarrow\)x2+x+x+1+x2+x-3x2+x-1=0
\(\Leftrightarrow\)-x24x=0
\(\Leftrightarrow\)4x-x2=0
\(\Leftrightarrow\)x(4-x)=0
\(\Leftrightarrow\)x=0 hoặc 4-x=0
\(\Leftrightarrow\)x=0 hoặc x =4
3)\(\dfrac{2x+1}{3}-\dfrac{3x-2}{2}>\dfrac{1}{6}\)
\(\Leftrightarrow\)\(\dfrac{2x+1}{3}6-\dfrac{3x-2}{2}6>\dfrac{1}{6}\)6
\(\Leftrightarrow\)2(2x+1)-3(3x-2)>1
\(\Leftrightarrow\)4x+2-9x+6>1
\(\Leftrightarrow\)4x-9x>1-2-6
\(\Leftrightarrow\)-5x>-7
\(\Leftrightarrow\)-5x.\(\dfrac{1}{-5}>-7.\dfrac{1}{-5}\)
\(\Leftrightarrow x>\dfrac{7}{5}\)
sai đề ko bn