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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!
22 tháng 6 2020
@Nguyễn Lê Phước Thịnh bạn có thể chỉ chỗ mình sai sót được không ạ? Mình mò không ra ._.
23 tháng 4 2020
Bài 1:
a, \(\frac{1}{x+1}+\frac{2}{x-1}=\frac{1+x^2}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)
\(\Leftrightarrow\) \(\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{1+x^2}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow\) x - 1 + 2(x + 1) = 1 + x2
\(\Leftrightarrow\) x - 1 + 2x + 2 - 1 - x2 = 0
\(\Leftrightarrow\) -x2 + 3x = 0
\(\Leftrightarrow\) x(3 - x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=3\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy S = {0; 3}
b, \(\frac{x-2}{x+2}-\frac{x}{x-2}=\frac{8}{x^2-4}\) (ĐKXĐ: x \(\ne\) \(\pm\) 2)
\(\Leftrightarrow\) \(\frac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\frac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{8}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow\) (x - 2)2 - x(x + 2) = 8
\(\Leftrightarrow\) (x - 2)2 - x(x + 2) - 8 = 0
\(\Leftrightarrow\) x2 - 4x + 4 - x2 - 2x - 8 = 0
\(\Leftrightarrow\) -6x - 4 = 0
\(\Leftrightarrow\) x = \(\frac{-2}{3}\) (TMĐKXĐ)
Vậy S = {\(\frac{-2}{3}\)}
c, \(\frac{1}{x}\) + \(\frac{2}{x-3}\) = \(\frac{1-5x}{x^2-3x}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 3)
\(\Leftrightarrow\) \(\frac{x-3}{x\left(x-3\right)}+\frac{2x}{x\left(x-3\right)}=\frac{1-5x}{x\left(x-3\right)}\)
\(\Rightarrow\) x - 3 + 2x = 1 - 5x
\(\Leftrightarrow\) 3x - 3 = 1 - 5x
\(\Leftrightarrow\) 3x + 5x = 1 + 3
\(\Leftrightarrow\) 8x = 4
\(\Leftrightarrow\) x = \(\frac{1}{2}\) (TMĐKXĐ)
Vậy S = {\(\frac{1}{2}\)}
Bài 2:
a, \(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{x^2-4}\)
\(\Leftrightarrow\) \(\frac{1}{x+2}=\frac{-5}{x-2}+\frac{12+x}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\) \(\frac{x-2}{\left(x+2\right)\left(x-2\right)}=\frac{-5\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{12+x}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow\) x - 2 = -5(x + 2) + 12 + x
\(\Leftrightarrow\) x - 2 = -5x - 10 + 12 + x
\(\Leftrightarrow\) x - 2 = -4x + 2
\(\Leftrightarrow\) x + 4x = 2 + 2
\(\Leftrightarrow\) 5x = 4
\(\Leftrightarrow\) x = \(\frac{4}{5}\)
Vậy S = {\(\frac{4}{5}\)}
Chúc bn học tốt!! (Phần b hình như không có gì thì phải)
4 tháng 7 2020
a. (x+4)(\(\frac{1}{4}\)x-1)=0
=>[\(\begin{matrix}x+4=0\\\frac{1}{4}x-1=0\end{matrix}\)
=>[\(\begin{matrix}x=-4\\\frac{1}{4}x=1\end{matrix}\)
=>[\(\begin{matrix}x=-4\left(n\right)\\x=4\left(n\right)\end{matrix}\)
S={-4;4}
b.
⇔\(\frac{x^2+4x+4}{x^2-4}\) -\(\frac{x^2-4x+4}{x^2-4}\) =\(\frac{4}{x^2-4}\)
=>\(x^2+4x+4-x^2+4x-4-4=0\)
⇔ 8x - 4=0
⇔x=\(\frac{1}{2}\) (n)
S=\(\left\{\frac{1}{2}\right\}\)
c.
=>2x-10< 5x+5
=>-3x <15
=> x > 5 (n)
{x/x>5}
KN
5 tháng 3 2020
a) \(2\left(x-1\right)-a\left(x-1\right)=2a+3\)
\(\Leftrightarrow2a-2-ax+a=2a+3\)
\(\Leftrightarrow-2-ax+a=3\)
\(\Leftrightarrow-a\left(x-1\right)=5\)
\(\Leftrightarrow\left(x-1\right)=\frac{-5}{a}\Leftrightarrow x=\frac{a-5}{a}\)
KN
5 tháng 3 2020
b) \(\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=3\)
\(\Leftrightarrow\frac{12x+12+8x+16+6x+18}{24}=3\)
\(\Leftrightarrow12x+12+8x+16+6x+18=72\)
\(\Leftrightarrow26x+46=72\)
\(\Leftrightarrow26x=26\Leftrightarrow x=1\)
MN
12 tháng 2 2020
a) \(\frac{x}{3}-\frac{5x}{6}-\frac{15x}{12}=\frac{x}{4}-5\)
\(\Leftrightarrow\frac{4x-10x-15x}{12}=\frac{3x-60}{12}\)
\(\Leftrightarrow-21x=3x-60\)
\(\Leftrightarrow24x=60\)
\(\Leftrightarrow x=\frac{5}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{5}{2}\right\}\)
b) \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow\frac{\left(8x-3\right)-2\left(3x-2\right)}{4}=\frac{2\left(2x-1\right)+\left(x+3\right)}{4}\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
MN
12 tháng 2 2020
c) \(\frac{x-1}{2}-\frac{x+1}{15}-\frac{2x-13}{6}=0\)
\(\Leftrightarrow\frac{15\left(x-1\right)-2\left(x+1\right)-5\left(2x-13\right)}{30}=0\)
\(\Leftrightarrow15x-15-2x-2-10x+65=0\)
\(\Leftrightarrow3x+48=0\)
\(\Leftrightarrow x=-16\)
Vậy tập nghiệm của phương trình là \(S=\left\{-16\right\}\)
d) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
\(\Leftrightarrow\frac{9\left(3-x\right)+16\left(5-x\right)}{24}=\frac{12\left(1-x\right)-48}{24}\)
\(\Leftrightarrow27-9x+80-16x=12-12x-48\)
\(\Leftrightarrow-25x+107=-12x-36\)
\(\Leftrightarrow-13x+143=0\)
\(\Leftrightarrow x=11\)
Vậy tập nghiệm của phương trình là \(S=\left\{11\right\}\)
AT
28 tháng 3 2020
\(a,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{2\left(x^2+2\right)}{\left(x+2\right)\left(x-2\right)}\)
=> ( x + 1)( x + 2) + ( x - 1)( x - 2) = 2x2 + 4
<=> x2 + 2x + x + 2 + x2 - 2x - x + 2 = 2x2 + 4
<=> x2 + 2x + x + x2 - 2x - x - 2x2 = 4 - 2 - 2
<=> 0x = 0
Vậy phương trình vô số nghiệm
\(\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x^2+1}{\left(x-2\right)\left(x+2\right)}\)
đkxđ : x khác cộng trừ 2
=> x2+2x-5x+10=x2+1
<=> x2+2x-5x-x2=-10+1
<=> -3x=-9
<=> x=3