giải pt
\(\frac{x+1}{x-1}+\frac{x-2}{x+2}+\frac{x-3}{x+3}+\frac{x+4}{x-4}=4\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NV
0
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 5 2019
\(x\ne\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)
\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)
\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)
\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{-2}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)
A
30 tháng 3 2020
1.\(\frac{x+1}{2013}\)+\(\frac{x+2}{2012}\)=\(\frac{x+3}{2011}\)+\(\frac{x+4}{2010}\)
⇔\(\frac{x+1}{2013}\)+1+\(\frac{x+2}{2012}\)+1=\(\frac{x+3}{2011}\)+1+\(\frac{x+4}{2010}\)+1
⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)=\(\frac{x+2014}{2011}\)+\(\frac{x+2014}{2010}\)
⇔\(\frac{x+2014}{2013}\)+\(\frac{x+2014}{2012}\)-\(\frac{x+2014}{2011}\)-\(\frac{x+2014}{2010}\)=0
⇔(x+2014)(\(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\))=0
Mà \(\frac{1}{2013}\)+\(\frac{1}{2012}\)-\(\frac{1}{2011}\)-\(\frac{1}{2010}\)≠0
⇔x+2014=0
⇔x=-2014
Vậy tập nghiệm của phương trình đã cho là:S={-2014}
A
30 tháng 3 2020
2.\(\frac{3x+2}{4}\)+\(\frac{x+3}{2}\)=\(\frac{x-1}{3}\)-\(\frac{-x-1}{12}\)
⇔\(\frac{3\left(3x+2\right)}{12}\)+\(\frac{6\left(x+3\right)}{12}\)=\(\frac{4\left(x-1\right)}{12}\)+\(\frac{x+1}{12}\)
⇒9x+6+6x+18=4x-4+x+1
⇒15x+24=5x-3
⇒15x-5x=-3-24
⇒10x=-27
⇒ x=-\(\frac{27}{10}\)
Vậy tập nghiệm của phương trình đã cho là S={-\(\frac{27}{10}\)}
1 tháng 6 2020
1) Ta có: x-4=2x+4
\(\Leftrightarrow x-4-2x-4=0\)
\(\Leftrightarrow-x-8=0\)
\(\Leftrightarrow-x=8\)
hay x=-8
Vậy: S={8}
2) Ta có: \(\frac{2x-1}{2}-\frac{x}{3}=x-\frac{x}{6}\)
\(\Leftrightarrow\frac{3\left(2x-1\right)}{6}-\frac{2x}{6}=\frac{6x}{6}-\frac{x}{6}\)
\(\Leftrightarrow3\left(2x-1\right)-2x-6x+x=0\)
\(\Leftrightarrow6x-3-2x-6x+x=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: S={-3}
3) ĐKXĐ: \(x\notin\left\{\frac{-1}{2};3\right\}\)
Ta có: \(\frac{x+3}{2x+1}-\frac{x}{x-3}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-3\right)}{\left(2x+1\right)\left(x-3\right)}-\frac{x\left(2x+1\right)}{\left(x-3\right)\left(2x+1\right)}=\frac{3x^2+x+9}{\left(2x+1\right)\left(x-3\right)}\)
Suy ra: \(x^2-9-\left(2x^2+x\right)-3x^2-x-9=0\)
\(\Leftrightarrow-2x^2-x-18-2x^2-x=0\)
\(\Leftrightarrow-4x^2-2x-18=0\)
\(\Leftrightarrow-4\left(x^2+\frac{1}{2}x+\frac{4}{5}\right)=0\)
\(\Leftrightarrow x^2+\frac{1}{2}x+\frac{4}{5}=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{1}{4}+\frac{1}{16}+\frac{59}{80}=0\)
\(\Leftrightarrow\left(x+\frac{1}{4}\right)^2+\frac{59}{80}=0\)(vô lý)
Vậy: S=\(\varnothing\)
4) Ta có: \(\frac{2x}{3}+\frac{2x-1}{6}=4-\frac{x}{3}\)
\(\Leftrightarrow\frac{4x}{6}+\frac{2x-1}{6}=\frac{24}{6}-\frac{2x}{6}\)
\(\Leftrightarrow4x+2x-1=24-2x\)
\(\Leftrightarrow6x-1-24+2x=0\)
\(\Leftrightarrow8x-25=0\)
\(\Leftrightarrow8x=25\)
hay \(x=\frac{25}{8}\)
Vậy: \(S=\left\{\frac{25}{8}\right\}\)
8 tháng 2 2020
a, Ta có : \(3\left(x-1\right)-2\left(x+3\right)=-15\)
=> \(3x-3-2x-6=-15\)
=> \(3x-3-2x-6+15=0\)
=> \(x=-6\)
Vậy phương trình có nghiệm là x = -6 .
b, Ta có : \(3\left(x-1\right)+2=3x-1\)
=> \(3x-3+2=3x-1\)
=> \(3x-3+2-3x+1=0\)
=> \(0=0\)
Vậy phương trình có vô số nghiệm .
c, Ta có : \(7\left(2-5x\right)-5=4\left(4-6x\right)\)
=> \(14-35x-5=16-24x\)
=> \(14-35x-5-16+24x=0\)
=> \(-35x+24x=7\)
=> \(x=\frac{-7}{11}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{11}\) .
Bài 2 :
a, Ta có : \(\frac{x}{30}+\frac{5x-1}{10}=\frac{x-8}{15}-\frac{2x+3}{6}\)
=> \(\frac{x}{30}+\frac{3\left(5x-1\right)}{30}=\frac{2\left(x-8\right)}{30}-\frac{5\left(2x+3\right)}{30}\)
=> \(x+3\left(5x-1\right)=2\left(x-8\right)-5\left(2x+3\right)\)
=> \(x+15x-3=2x-16-10x-15\)
=> \(x+15x-3-2x+16+10x+15=0\)
=> \(24x+28=0\)
=> \(x=\frac{-28}{24}=\frac{-7}{6}\)
Vậy phương trình có nghiệm là \(x=\frac{-7}{6}\) .
b, Ta có : \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)
=> \(\frac{6\left(x+4\right)}{30}-\frac{30x}{30}+\frac{120}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)
=> \(6\left(x+4\right)-30x+120=10x-15\left(x-2\right)\)
=> \(6x+24-30x+120=10x-15x+30\)
=> \(6x+24-30x+120-10x+15x-30=0\)
=> \(-19x+114=0\)
=> \(x=\frac{-114}{-19}=6\)
Vậy phương trình có nghiệm là x = 6 .
N
1 tháng 4 2019
a.\(\Leftrightarrow\left(x+3\right)\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x^2+2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-1\end{matrix}\right.\)
1 tháng 4 2019
(x-2)(x+1)(x+3)=(x+3)(x+1)(2x-58)
\(x^3+2x^2-5x-6\)=\(2x^3+3x^2-14x-15\)
\(-x^3-x^2+9x+9=0\)
\(-x^2\left(x+1\right)+9\left(x+1\right)=0\)
\(\left(x+1\right)\left(9-x^2\right)\)=0
(x+1)(3-x)(3+x)=0
*x+1=0 =>x=-1
*3-x=0=>x=3
*3+x=0=>x=-3
25 tháng 3 2020
Bài 1:
ĐKXĐ: x≠1
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4\left(x-1\right)}{\left(x^2+x-1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4\left(x-1\right)=0\)
\(\Leftrightarrow x^2+x+1+2x^2-5-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
Vì 3≠0
nên \(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 2:
ĐKXĐ: x≠2; x≠3; \(x\ne\frac{1}{2}\)
Ta có: \(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-\left(2x+5\right)}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{x+4}{\left(x-2\right)\left(2x-1\right)}+\frac{x+1-2x-5}{\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow\frac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\frac{\left(-x-4\right)\left(x-2\right)}{\left(x-3\right)\left(2x-1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-12-x^2-2x+8=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(tm)
Vậy: x=-4
Bài 3:
ĐKXĐ: x≠1; x≠-1
Ta có: \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x\left(1-\frac{x-1}{x+1}\right)\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}=3x-\frac{3x\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{x+1}{x-1}-\frac{x-1}{x+1}-3x+\frac{3x\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)-3x\left(x^2-1\right)+3x\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1-3x^3+3x+3x^3-6x^2+3x=0\)
\(\Leftrightarrow-6x^2+10x=0\)
\(\Leftrightarrow2x\left(-3x+5\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\-3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{5}{3}\right\}\)
Bài 4:
ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2x^2+6x+4-\left(2x^2-7x+5\right)=0\)
\(\Leftrightarrow2x^2+6x+4-2x^2+7x-5=0\)
\(\Leftrightarrow13x-1=0\)
\(\Leftrightarrow13x=1\)
hay \(x=\frac{1}{13}\)(tm)
Vậy: \(x=\frac{1}{13}\)
Bài 5:
ĐKXĐ: x≠1; x≠-2
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{x^2+x-2}\)
\(\Leftrightarrow\frac{x+2}{\left(x-1\right)\left(x+2\right)}-\frac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{3}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Leftrightarrow x+2-7\left(x-1\right)-3=0\)
\(\Leftrightarrow x+2-7x+7-3=0\)
\(\Leftrightarrow-6x+6=0\)
\(\Leftrightarrow-6\left(x-1\right)=0\)
Vì -6≠0
nên x-1=0
hay x=1(ktm)
Vậy: x∈∅
Bài 6:
ĐKXĐ: x≠4; x≠2
Ta có: \(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{6x-8-x^2}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}-\frac{2}{-\left(x^2-6x+8\right)}=0\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}+\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-4\right)\left(x-2\right)}=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
Vì 2≠0
nên \(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Vậy: x=0
Bài 7:
ĐKXĐ: x≠1; x≠-2; x≠-1
Ta có: \(\frac{1}{x-1}-\frac{7}{x+2}=\frac{3}{1-x^2}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{7}{x+2}+\frac{3}{x^2-1}=0\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}-\frac{7\left(x-1\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)\left(x+1\right)}+\frac{3\left(x+2\right)}{\left(x-1\right)\left(x+1\right)\left(x+2\right)}=0\)
\(\Leftrightarrow x^2+3x+2-7\left(x^2-1\right)+3x+6=0\)
\(\Leftrightarrow x^2+3x+2-7x^2+7x+3x+6=0\)
\(\Leftrightarrow-6x^2+13x+8=0\)
\(\Leftrightarrow-6x^2+16x-3x+8=0\)
\(\Leftrightarrow2x\left(-3x+8\right)+\left(-3x+8\right)=0\)
\(\Leftrightarrow\left(-3x+8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+8=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-8\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{8}{3}\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{8}{3};\frac{-1}{2}\right\}\)
25 tháng 3 2020
\( 1)\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\ DK:x \ne 1\\ \Leftrightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^3} - 1}}\\ \Leftrightarrow {x^2} + x + 1 + 2{x^2} - 5 = 4x - 4\\ \Leftrightarrow 3{x^2} - 3x = 0\\ \Leftrightarrow 3x\left( {x - 1} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\left( {tm} \right)\\ x = 1\left( {ktm} \right) \end{array} \right.\\ 2)\dfrac{{x + 4}}{{2{x^2} - 5x + 2}} + \dfrac{{x + 1}}{{2{x^2} - 7x + 3}} = \dfrac{{2x + 5}}{{2{x^2} - 7x + 3}}\\ + DK:x \ne \dfrac{1}{2};x \ne 2;x \ne 3\\ \Leftrightarrow \dfrac{{x + 4}}{{\left( {2x - 1} \right)\left( {x - 2} \right)}} + \dfrac{{x + 1}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}} = \dfrac{{2x + 5}}{{\left( {x - 3} \right)\left( {2x - 1} \right)}}\\ \Leftrightarrow \left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 1} \right)\left( {x - 2} \right) = \left( {2x + 5} \right)\left( {x - 2} \right)\\ \Leftrightarrow {x^2} + x - 12 + {x^2} - x - 2 = 2{x^2} + x - 10\\ \Leftrightarrow x = - 4\left( {tm} \right)\\ 3)\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}} = 3x\left( {1 - \dfrac{{x - 1}}{{x + 1}}} \right)\\ DK:x \ne \pm 1\\ \Leftrightarrow {\left( {x + 1} \right)^2} - {\left( {x - 1} \right)^2} = 3x\left( {x - 1} \right)\left( {x + 1 - x + 1} \right)\\ \Leftrightarrow {x^2} + 2x + 1 - {x^2} + 2x - 1 = 6x\left( {x - 1} \right)\\ \Leftrightarrow 4x = 6{x^2} - 6x\\ \Leftrightarrow 2x\left( {3x - 5} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \dfrac{5}{3} \end{array} \right.\left( {tm} \right) \)
Còn lại tương tự mà làm nhé!
Sửa lại đề \(\frac{x+1}{x-1}+\frac{x-2}{x+2}+\frac{x-3}{x+3}+\frac{x+4}{x-4}=-4\)
ĐK \(x\ne\left\{1;-2;-3;4\right\}\)
\(\Leftrightarrow\left(\frac{x+1}{x-1}+1\right)+\left(\frac{x-2}{x+2}+1\right)+\left(\frac{x-3}{x+3}+1\right)+\left(\frac{x+4}{x-4}+1\right)=0\)
\(\Leftrightarrow\frac{2x}{x-1}+\frac{2x}{x+2}+\frac{2x}{x+3}+\frac{2x}{x-4}=0\)
\(\Leftrightarrow2x\left(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\right)=0\Leftrightarrow x=0\)vì \(\frac{1}{x-1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x-4}\ne0\)
Vậy pt có nghiệm \(x=0\)
cô giáo in đề cho mk là =4 mà
nếu k thì mk xong lâu r