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2: \(\Leftrightarrow\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x-1\right)\left(x+1\right)\)
=>x^2-3x-4+x^2+3x-4=2x^2-2
=>2x^2-8=2x^2-2(loại)
3: \(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)+x^2\left(x+3\right)=-7x^2+3x\)
=>x^3-3x^2-x^2+3x+x^3+3x^2+7x^2-3x=0
=>2x^3+6x^2=0
=>2x^2(x+3)=0
=>x=0(nhận) hoặc x=-3(loại)
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ 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 
1, \(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\\ \\ < =>\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\\ \\ < =>\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\\ \\ < =>\left(x-2014\right).\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\\ \\ < =>x-2014=0< =>x=2014\)
2, \(x^2+1=x\\ \\ < =>x^2-x+1=0\\ \\ < =>x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=0\\ \\ < =>\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\)
có vế trái luôn dương, vế phải = 0 => vô nghiệm
Ta có: 8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)^2\)\(\left(x+\dfrac{1}{x}\right)^2\)=(x+4)2
ĐKXĐ: x khác 0
<=>8\(\left(x+\dfrac{1}{x}\right)^2\)+4\(\left(x^2+\dfrac{1}{x^2}\right)\)\(\left(x^2+\dfrac{1}{x^2}-x^2-2-\dfrac{1}{x^2}\right)\)=(x+4)2
<=>8\(\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
<=>8\(\left(x^2+2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}\right)\)=(x+4)2
=>(x+4)2=16
Vậy có 2 TH:
+) x+4=4 => x=0(KTMĐKXĐ)
+)x+4=-4 => x=-8(TMĐKXĐ)
Vậy tập nghiệm của phương trình S={-8}
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
a)\(\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\left(đkxđ:x\ne-1;-4;-6;3\right)\)
\(\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x+6\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{4}{3}+\dfrac{1}{x-3}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)
\(\Leftrightarrow\dfrac{-4}{\left(x+1\right)\left(x-3\right)}=\dfrac{4}{3}\)
\(\Leftrightarrow\left(x+1\right)\left(3-x\right)=3\)
\(\Leftrightarrow2x-x^2+3=3\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\left(tm\right)\)
b)\(x^2-y^2+2x-4y-10=0\)
\(\Leftrightarrow x^2+2x+1-y^2-4y-4-7=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\)
\(\Leftrightarrow\left(x-y-1\right)\left(x+y+3\right)=7\)
Mà x,yEN*=>x-y-1<x+y+3
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y-1=1\\x+y+3=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-y-1=-7\\x+y+3=-1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy ...
Để mình giúp nha
\(x^2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+7=0\)
ĐKXD: x\(\ne0\)
\(\Leftrightarrow x^2+2+\dfrac{1}{x^2}-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-\dfrac{9}{2}\left(x+\dfrac{1}{x}\right)+5=0\)
Đặt \(a=x+\dfrac{1}{x}\) khi đó phương trình trở thành
\(a^2-\dfrac{9}{2}a+5=0\)
\(\Leftrightarrow\left(a\right)^2-2.a.\dfrac{9}{4}+\left(\dfrac{9}{4}\right)^2-\dfrac{81}{16}+5=0\)
\(\Leftrightarrow\left(a+\dfrac{9}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}a-\dfrac{9}{4}=\dfrac{1}{4}\\a-\dfrac{9}{4}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{5}{2}\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5}{2}\\x+\dfrac{1}{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x^2+1}{x}=\dfrac{5}{2}\\\dfrac{x^2+1}{x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-\dfrac{5}{2}x+1=0\\x^2-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x\right)^2-2.x.\dfrac{5}{4}+\left(\dfrac{5}{4}\right)^2-\dfrac{25}{16}+1=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{5}{4}\right)^2-\dfrac{9}{16}=0\\x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=\dfrac{1}{2}\left(n\right)\\x=1\left(n\right)\end{matrix}\right.\)
Vậy S=\(\left\{1;2;\dfrac{1}{2}\right\}\)
1a)\(\left(x+2\right)^2-6\left(x+2\right)\le x^2-4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\ge\left(x+2\right)\left(x+2-6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x+2\right)\left(x-4\right)\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+4\right)\ge0\)
\(\Leftrightarrow\left(x+2\right)\cdot2\ge0\)
\(\Leftrightarrow x+2\ge0\)
\(\Leftrightarrow x\ge-2\)
b)\(-\dfrac{2}{x-1}>0\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{2}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)
\(\Leftrightarrow x< 1\)
1)
a) (x+2)2 - 6(x+2) \(\le\) x2 - 4
<=> (x+2).(x+2 - 6) \(\le\) (x+2)(x-2)
<=> x2 + 2x - 6x + 2x + 4 - 12 \(\le\) x2 - 2x +2x - 4
<=> x2 - x2 + 2x + 2x - 6x + 2x - 2x \(\le\) 12 - 4 - 4
<=> -2x \(\le\) 4
<=> x \(\ge\) -2
Vậy bpt có nghiệm x \(\ge\) -2
b) Để \(\dfrac{-2}{x-1}\) nhận giá trị không âm
=>\(\dfrac{-2}{x-1}\) \(\ge\) 0
<=> -2 . (x-1) \(\ge\) 0
<=> -2x + 2 \(\ge\) 0
<=> -2x \(\ge\) -2
<=> x \(\le\) 1
Vậy với x \(\le\) 1 thì biểu thức \(\dfrac{-2}{x-1}\) nhận giá trị không âm.
2)
a) \(|x+1|\) = 2x - 1
+) Nếu x+1 \(\ge\) 0 => x \(\ge\) -1 thì phương trình :
x + 1 = 2x -1
<=> -x = -2
<=> x = 2 (thỏa mãn)
+) Nếu x+1 < 0 => x < -1 thì phương trình :
-(x + 1) = 2x - 1
<=> -x -1 = 2x -1
<=> -3x = 0
<=> x = 0 (không thỏa mãn)
Vậy phương trình có nghiệm x = 2.
c) \(\dfrac{x+1}{x-2}\) - \(\dfrac{5}{x+2}\) = \(\dfrac{12}{x^2-4}\)+ 1 (ĐKXĐ: x \(\ne\) \(\pm\) 2)
<=>\(\dfrac{\left(x+1\right).\left(x+2\right)-5.\left(x-2\right)}{x^2-4}\) = \(\dfrac{12+x^2-4}{x^2-4}\)
<=> \(\dfrac{x^2-2x+12}{x^2-4}\) = \(\dfrac{x^2+8}{x^2-4}\)
<=> x2 - 2x +12 = x2 +8
<=> -2x = 8-12
<=> x = 2 ( không thỏa mãn ĐKXĐ)
Vậy phương trình vô nghiệm.
An Lê Khánh haizzzzz \(x+\dfrac{1}{x}=x\left(x+\dfrac{1}{x}\right)\)
\(\Leftrightarrow1=x\)
\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\\ \Rightarrow\dfrac{x^2+1}{x}=\dfrac{x^4+1}{x^2}\\ \Rightarrow\dfrac{x^3+x}{x^2}=\dfrac{x^4+1}{x^2}\\ \Rightarrow x^3+x=x^4+1\\ \Rightarrow x^4+1-x^3-x=0\\ \Rightarrow\left(x^4-x^3\right)-\left(x-1\right)=0\\ \Rightarrow x^3\left(x-1\right)-\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x^3-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x-1\right)\left(x^2+x+1\right)=0\\ \Rightarrow\left(x-1\right)^2\left(x^2+x+1\right)=0\\ \Rightarrow\left(x-1\right)^2=0\\ \Rightarrow x=1\)