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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!
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={-\(\frac{27}{10}\)}
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
a)
\(\begin{array}{l}\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{2}{{3{\rm{x}}}} - \frac{x}{{1 - x}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4\left( {1 - x} \right) - 6{{\rm{x}}^2} + 3\left( {6{{\rm{x}}^2} - 4} \right)}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4 - 4{\rm{x}} - 6{{\rm{x}}^2} + 18{{\rm{x}}^2} - 12}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{12{{\rm{x}}^2} - 4{\rm{x}} - 8}}{{6{\rm{x}}\left( {1 - x} \right)}}\end{array}\)
b)
\(\begin{array}{l}\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{x}{{{x^3} - 1}}\end{array}\)
c)
\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{ - 4{\rm{x - 2}}}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{\left( { - 4{\rm{x}} - 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 8{\rm{x}} - 2{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 6{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {4{{\rm{x}}^2} - 1} \right)}}\end{array}\)
d)
\(\begin{array}{l}1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}\\ = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}\\ = 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}\\ = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\\ = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\\ = \frac{1}{{{x^2} + 1}}\end{array}\)
HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
\(\begin{array}{l}a)\frac{{{x^2} + 4{\rm{x}} + 4}}{{{x^2} - 4}} + \frac{x}{{2 - x}} + \frac{{4 - x}}{{5{\rm{x}} - 10}}\\ = \frac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} - \frac{x}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}\\ = \frac{{x + 2}}{{x - 2}} - \frac{x}{{x - 2}} + \frac{{4 - x}}{{5\left( {x - 2} \right)}}\\ = \frac{{5\left( {x + 2} \right) - 5x + 4 - x}}{{5\left( {x - 2} \right)}} = \frac{{ - x + 14}}{{5\left( {x - 2} \right)}}\end{array}\)
\(\begin{array}{l}b)\frac{x}{{{x^2} + 1}} - \left( {\frac{3}{{x + 6}} + \frac{{x - 2}}{{x + 4}}} \right) + \left[ {\frac{3}{{x + 6}} - \left( {\frac{1}{{{x^2} + 1}} - \frac{{x - 2}}{{x + 4}}} \right)} \right]\\ = \frac{x}{{{x^2} + 1}} - \frac{3}{{x + 6}} - \frac{{x - 2}}{{x + 4}} + \frac{3}{{x + 6}} - \frac{1}{{{x^2} + 1}} + \frac{{x - 2}}{{x + 4}}\\ = \frac{x}{{{x^2} + 1}} - \frac{1}{{{x^2} + 1}} = \frac{{x - 1}}{{{x^2} + 1}}\end{array}\)
24 tháng 1 2022
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
A
15 tháng 4 2020
1, Đk x≠2;-2
\(\frac{x+2}{2x-4}-\frac{4x}{x^2-4}=0\\ =>\frac{x+2}{2\left(x-2\right)}-\frac{4x}{\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{\left(x+2\right)^2}{2\left(x^2-4\right)}-\frac{8x}{2\left(x-2\right).\left(x+2\right)}=0\\ =>\frac{x^2+4x+4-8x}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}=0\\ =>\frac{x-2}{2\left(x+2\right)}=0\\ =>x-2=0\\ =>x=2\left(loại\right)\)
16 tháng 3 2023
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
\(\frac{x-1}{4}+\frac{x-2}{3}+\frac{x-3}{2}+\frac{x-4}{1}=4\)\(\Leftrightarrow\left(\frac{x-1}{4}-1\right)+\left(\frac{x-2}{3}-1\right)+\left(\frac{x-3}{2}-1\right)+\left(\frac{x-4}{1}-1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{4}+\frac{1}{3}+\frac{1}{2}+1\right)=0\Leftrightarrow x=5\)
Đáp án : 2\(\frac{29}{48}\)( 2,6041 )