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NN
2
22 tháng 1 2018
ĐKXĐ x≠3 ; x≠-3
\(\dfrac{2x-1}{x+3}=\dfrac{2x+1}{x-3}\)
=> (2x-1)(x-3)=(2x+1)(x+3)
⇔2x2-6x-x+3=2x2+6x+x+3
⇔2x2-2x2-7x-6x=3-3
⇔ -13x=0
⇔x=0 (tm)
vậy phương trình trên có tập no S={0}
NH
1
20 tháng 3 2017
Ta có : 1+\(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}\) (đkxđ x\(\ne\pm2\) )
\(\Leftrightarrow\) \(\dfrac{1}{x+2}\) = \(\dfrac{12}{8-x^3}-1\)
\(\Leftrightarrow\)\(\dfrac{1}{x+2}=\dfrac{12-\left(8-x^3\right)}{8-x^3}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{x^3+4}{8-x^3}\)
\(\Leftrightarrow8-x^3=\left(x+2\right)\left(x^3+4\right)\)
\(\Leftrightarrow8-x^3=x^4+4x+2x^3+8\)
\(\Leftrightarrow-x^3-x^4-4x-2x^3=8-8\)
\(\Leftrightarrow-x^4-3x^3-4x=0\)
\(\Leftrightarrow-x\left(x^3+3x^2+4\right)=0\)
\(\Rightarrow-x=0\)\(\Rightarrow x=0\) (TM x\(\ne\pm2\))
MT
2
DM
23 tháng 4 2017
Pt trên có MSC là \(\left(x-1\right)\left(x^2+x+1\right)\)
Quy đồng mẫu số :
\(\dfrac{1}{x-1}+\dfrac{7x-10}{x^3-1}-\dfrac{3}{x^2+x+1}=0\)
( ĐKXĐ \(x\ne1\))
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{x^3-1}-\dfrac{3x-3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+x+1+7x-10-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\) \(\dfrac{x^2+5x-6}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=-6\left(TMĐK\right)\end{matrix}\right.\)
Vậy \(S=\left\{-6\right\}\)
TS
23 tháng 4 2017
ĐKXĐ: \(x\ne1\); \(x\ne-1\)
\(\dfrac{1}{x-1}+\dfrac{7x-10}{x^3-1}-\dfrac{3}{x^2+x+1}=0\)
\(\Leftrightarrow\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{7x-10}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Rightarrow x^2+x+1+7x-10-3x+3=0\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow x-1=0\) ; \(x+6=0\)
+) \(x-1=0\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
+) \(x+6=0\)
\(\Leftrightarrow x=-6\) (Thỏa mãn ĐKXĐ)
Tập nghiệm: \(S=\left\{-6\right\}\)
LL
1 tháng 5 2018
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
WJ
11 tháng 5 2018
5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6
\(\Leftrightarrow\) 5x-2x>6+2
\(\Leftrightarrow\)3x>8
\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)
0 8/3
Chúc bn học tốt❤
ND
8 tháng 2 2018
a.
\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)
\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=28-4x\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
GH
8 tháng 2 2018
a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)
\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)
\(\Leftrightarrow6x+6+3x+9=36-4x-8\)
\(\Leftrightarrow9x+15=-4x+28\)
\(\Leftrightarrow9x+4x=28-15\)
\(\Leftrightarrow13x=13\)
\(\Leftrightarrow x=1\)
Vậy ................................
VT
5
13 tháng 3 2017
với x>0 thì pt luôn xác định.
\(\Rightarrow\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}=\dfrac{12}{x^3+8}\)
\(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(x\left(x^2+x-2\right)=0\Rightarrow x=0\) hoặc \(x^2+x-2=0\)
x=0 hoac (x\(^2\)-1) +(x-1) =0
x=0 hoặc (x-1)(x+2)=0
x=0 hoax x=1 hoặc x=2 vỉ x>0 nên pt có 2 nghiệm là x=1 , x=2.
LT
1
KT
14 tháng 3 2017
\(1+\dfrac{1}{x+2}=\dfrac{12}{x^3+8}\Leftrightarrow\dfrac{\left(x^3+8\right)\left(x+2\right)}{\left(x^3+8\right)\left(x+2\right)}+\dfrac{\left(x^3+8\right)}{\left(x^3+8\right)\left(x+2\right)}=\dfrac{12\left(x+2\right)}{\left(x^3+8\right)\left(x+2\right)}\)
\(\Rightarrow x^4+2x^3+8x+16+x^3+8=12x+24\)
\(\Leftrightarrow x^4+3x^3-4x=0\\ \Leftrightarrow x\left(x^3+3x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x^3+3x^2-4=0\end{matrix}\right.\)
\(x^3+3x^2-4=0\Leftrightarrow\left(x^3+4x^2+4x\right)-\left(x^2+4x+4 \right)=0\)
\(\left(x-1\right)\left(x^2+4x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x^2+4x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left(x+2\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\left(loại\right)\end{matrix}\right.\)
vậy phương trình có tập nghiệm là S={1}
\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\) ( ĐK : \(x\ne-2\) )
\(\Leftrightarrow\dfrac{12}{x^3+2^3}=1+\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{12}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}+\dfrac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow12=\left(x+2\right)\left(x^2-2x+4\right)+x^2-2x+4\)
\(\Leftrightarrow x^3+8+x^2-2x+4=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(N\right)\\x=1\left(N\right)\\x=-2\left(L\right)\end{matrix}\right.\)
Vậy \(S=\left\{0;1\right\}\)
Thank you ! <3 !! :))