TN
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
XO
18 tháng 2 2021
a) ĐKXĐ : \(x\ne-2;x\ne5\)
\(\frac{7}{x+2}=\frac{3}{x-5}\)
<=> 3(x + 2) = 7(x - 5)
<=> 3x + 6 = 7x - 35
<=> 4x = 41
<=>x = 41/4 (tm)
Vậy x = 41/4 là ngiệm phương trình
b) ĐKXĐ \(x\ne\pm3\)
\(\frac{2x-1}{x+3}=\frac{2x}{x-3}\)
<=> \(\frac{\left(2x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
<=> (2x - 1)(x - 3) = 2x(x + 3)
<=> 2x2 - 7x + 3 = 2x2 + 6x
<=> 13x = 3
<=> x = 3/13 (tm)
Vậy x = 3/13 là nghiệm phương trình
c) ĐKXĐ : \(x\ne-7;x\ne1,5\)
Khi đó \(\frac{3x-2}{x+7}=\frac{6x+1}{2x-3}\)
<=> \(\frac{\left(3x-2\right)\left(2x-3\right)}{\left(x+7\right)\left(2x-3\right)}=\frac{\left(6x+1\right)\left(x+7\right)}{\left(x+7\right)\left(2x-3\right)}\)
<=> (3x - 2)(2x - 3) = (6x + 1)(x + 7)
<=> 6x2 - 13x + 6 = 6x2 + 43x + 7
<=> 56x = -1
<=> x = -1/56 (tm)
Vậy x = -1/56 là nghiệm phương trình
d) ĐKXĐ : \(x\ne\pm1\)
Khi đó \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
<=> \(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{5\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
<=> (2x + 1)(x + 1) = 5(x - 1)2
<=> 2x2 + 3x + 1 = 5x2 - 10x + 5
<=> 3x2 - 13x + 4 = 0
<=> 3x2 - 12x - x + 4 = 0
<=> 3x(x - 4) - (x - 4) = 0
<=> (3x - 1)(x - 4) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\left(tm\right)\\x=4\left(tm\right)\end{cases}}\)
Vậy x \(\in\left\{\frac{1}{3};4\right\}\)là nghiệm phương trình
XO
18 tháng 2 2021
e) ĐKXĐ : \(x\ne1\)
Khi đó \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)
<=> \(\frac{3x-5}{x-1}=2\)
<=> 3x - 5 = 2(x - 1)
<=> 3x - 5 = 2x - 2
<=> x = 3 (tm)
Vậy x = 3 là nghiệm phương trình
f) ĐKXĐ : \(x\ne-1\)
\(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> \(\frac{3x+2}{x+1}=3\)
<=> 3x + 2 = 3(x + 1)
<=> 3x + 2 = 3x + 3
<=> 0x = 1
<=> \(x\in\varnothing\)
Vậy tập nghiệm phương trình S = \(\varnothing\)
g) ĐKXĐ : \(x\ne2\)
Khi đó \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
<=>\(\frac{x-2}{x-2}=3\)
<=> (x - 2) = 3(x - 2)
<=> x - 2 = 3x - 6
<=> -2x = -4
<=> x = 2 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
h) ĐKXĐ : \(x\ne7\)
Khi đó \(\frac{1}{7-x}=\frac{x-8}{x-7}-8\)
<=> \(\frac{x-7}{x-7}=8\)
<=> x - 7 = 8(x - 7)
<=> x - 7 = 8x - 56
<=> 7x = 49
<=> x = 7 (loại)
Vậy tập nghiệm phương trình S = \(\varnothing\)
i) ĐKXĐ : \(x\ne0;x\ne6\)
Ta có : \(\frac{x+6}{x}=\frac{1}{2}+\frac{15}{2\left(x-6\right)}\)
<=> \(\frac{x+6}{x}-\frac{15}{2\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2\left(x+6\right)\left(x-6\right)}{2x\left(x-6\right)}-\frac{15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> \(\frac{2x^2-72-15x}{2x\left(x-6\right)}=\frac{1}{2}\)
<=> 4x2 - 144 - 30x = 2x(x - 6)
<=> 2x2 - 18x - 144 = 0
<=> x2 - 9x - 72 = 0
<=> x2 - 9x + 81/4 - 72- 81/4 = 0
<=> \(\left(x-\frac{9}{2}\right)^2-\frac{369}{4}=0\)
<=> \(\left(x-\frac{9}{2}+\sqrt{\frac{369}{4}}\right)\left(x-\frac{9}{2}-\sqrt{\frac{369}{4}}\right)=0\)
<=> \(\orbr{\begin{cases}x=\frac{9}{2}-\sqrt{\frac{369}{4}}\\x=\frac{9}{2}+\sqrt{\frac{369}{4}}\end{cases}}\)(tm)
Vậy x \(\in\left\{\frac{9}{2}-\sqrt{\frac{369}{4}};\frac{9}{2}+\sqrt{\frac{369}{4}}\right\}\)
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!
TN
13 tháng 3 2020
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
10 tháng 4 2020
Bài làm
j) \(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\) ĐKXĐ: \(x\ne\pm5\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}=\frac{20}{x^2-25}\)
\(\Rightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
k) \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{x^2-16}+\frac{5x-2}{x^2-16}=\frac{4\left(x-4\right)}{x^2-16}\)
\(\Rightarrow3x+12+5x-2=4x-16\)
\(\Leftrightarrow4x=-26\)
<=> \(x=-\frac{13}{2}\)
Vậy x = -13/2 là nghiệm phương trình.
l) \(\frac{2x-1}{3}-\frac{5x+2}{4}=2x\)
\(\Leftrightarrow4x-4-15x-6=24x\)
\(\Leftrightarrow-35x=10\)
\(\Leftrightarrow x=-\frac{2}{7}\)
Vậy x = -2/7 là nghiệm phương trình.
10 tháng 4 2020
Bài làm
2 - x = 3x + 1
<=> - x - 3x = -2 + 1
<=> -4x = -1
<=> x = 1/4
Vậy x = 1/4 là nghiệm phương trình.
4x + 7( x - 2 ) = -9x + 5
<=> 4x + 7x - 14 = -9x + 5
<=> 4x + 7x + 9x = 14 + 5
<=> 20x = 19
<=> x = 19/20
Vậy x = 19/20 là nghiệm phương trình.
5x - 2( 3x - 5 ) = 7x + 11
<=> 5x - 6x + 10 = 7x + 11
<=> 5x - 6x - 7x = 11 - 10
<=> -8x = -21
<=> x = 21/8
Vậy x = 21/8 là nghiệm phương trình.
( 5x + 2 )( x - 7 ) = 0
<=> \(\left[{}\begin{matrix}5x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{2}{5}\\x=7\end{matrix}\right.\)
Vậy tập nghiệm phương trình S = { -2/5; 7 }
2x( x - 5 ) + 3( x - 5 ) = 0
<=> ( 2x + 3 )( x - 5 ) = 0
<=> \(\left[{}\begin{matrix}2x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=5\end{matrix}\right.\)
Vậy tập nghiệm phương trìh S = { -3/2; 5 }
\(\frac{5x-3}{6}=\frac{-2x+5}{9}\)
\(\Rightarrow6\left(-2x+5\right)=9\left(5x-3\right)\)
\(\Leftrightarrow-12x+30=45x-27\)
\(\Leftrightarrow-57x=-57\)
\(\Leftrightarrow x=1\)
Vậy x = 1 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{5x}{6}\)
\(\Leftrightarrow2x-3\left(2x+1\right)=5x\)
\(\Leftrightarrow2x-6x-3=5x\)
\(\Leftrightarrow-9x=3\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy x = -1/3 là nghiệm phương trình.
\(\frac{x}{3}-\frac{2x+1}{2}=\frac{x}{6}-x\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow2x-6x-3=x-6x\)
\(\Leftrightarrow2x=3\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy x = 3/2 là nghiệm phương trình.
\(\frac{3}{x+1}=\frac{5}{2x+2}\) ĐKXĐ: x khác 1
<=> \(\frac{6}{2x+2}=\frac{5}{2x+2}\)( vô lí )
Vậy phương trình trên vô nghiệm.
# Học tốt #