Bài : Giải phương trình sau :
\(\dfrac{x+1}{x-1}\)- \(\dfrac{x-2}{x+3}\) = 3
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Những câu hỏi liên quan
H
26 tháng 3 2021
Bài 4 :
24 phút = \(\dfrac{24}{60} = \dfrac{2}{5}\) giờ
Gọi thời gian dự định đi từ A đến B là x(giờ) ; x > 0
Suy ra quãng đường AB là 36x(km)
Khi vận tốc sau khi giảm là 36 -6 = 30(km/h)
Vì giảm vận tốc nên thời gian đi hết AB là x + \(\dfrac{2}{5}\)(giờ)
Ta có phương trình:
\(36x = 30(x + \dfrac{2}{5})\\ \Leftrightarrow x = 2\)
Vậy quãng đường AB dài 36.2 = 72(km)
7 tháng 11 2023
\(a,\sqrt{\left(x-1\right)^2-\left(x^2-3\right)}=3\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x^2-3\right)=9\)
\(\Leftrightarrow x^2-2x+1-x^2+3=9\)
\(\Leftrightarrow4-2x=9\)
\(\Leftrightarrow x=\dfrac{-5}{2}\)
\(b,\dfrac{x+3}{x}+\dfrac{x-3}{x-2}=2\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(2x-2\right)}{x\left(x-2\right)}=2\)
\(\Leftrightarrow\left(x-3\right)\left(2x-2\right)=2x\left(x-2\right)\)
\(\Leftrightarrow2x^2-8x+6=2x^2-4x\)
\(\Leftrightarrow-4x=-6\)
\(\Leftrightarrow x=1,5\)
5 tháng 2 2022
a. ĐKXĐ: \(x\ne2\).
\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
⇔\(\dfrac{1}{x-2}+\dfrac{3x-6}{x-2}=\dfrac{3-x}{x-2}\)
⇔\(1+3x-6=3-x\)
⇔\(4x-8=0\)
⇔\(x=2\) (không thỏa mãn)
-Vậy S=∅.
b. ĐKXĐ: \(x\ne-1\)
\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+1=-\dfrac{6}{x+1}\)
⇔\(\dfrac{5x}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}=-\dfrac{12}{2\left(x+1\right)}\)
⇔\(5x+2\left(x+1\right)=-12\)
⇔\(5x+2x+2+12=0\)
⇔\(7x+14=0\)
⇔\(x=-2\) (thỏa mãn).
-Vậy \(S=\left\{-2\right\}\)
L
5 tháng 2 2022
a, \(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3.\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\\ \Leftrightarrow1+3x-6=3-x\)
\(\Leftrightarrow3x+x=3-1+6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=\dfrac{8}{4}=2\\ Vậy.S=\left\{2\right\}\)
b, \(\Leftrightarrow\)\(\dfrac{5x}{2x+2}+\dfrac{2x+2}{2x+2}=\dfrac{-6.2}{2.\left(x+1\right)}\)
\(\Leftrightarrow5x+2x+2=-12\\ \Leftrightarrow7x=-12-2\\ \Leftrightarrow7x=-14\\ \Leftrightarrow x=-\dfrac{14}{7}=-2\\ Vậy.S=\left\{-2\right\}\)
28 tháng 1 2022
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)
\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)
Kết hợp ĐKXĐ.
\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)
\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)
Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)
Ta có bảng sau:
| \(x\) | \(-\infty\) \(-\dfrac{3}{2}\) \(-\dfrac{1}{2}\) \(1\) \(2\) \(+\infty\) |
| \(1+2x\) | - | - 0 + | + | + |
| \(x-2\) | - | - | - | - 0 + |
| \(2x+3\) | - 0 + | + | + | + |
| \(1-x\) | + | + | + 0 - | - |
| \(f\left(x\right)\) | - || + 0 - || + 0 - |
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].
6 tháng 4 2021
Bài 2:
a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)
\(=5m^2-2m+9>0\forall m\)
Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m
H
6 tháng 4 2021
Bài 1:
ĐKXĐ \(2x\ne y\)
Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)
HPT trở thành
\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)
17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)
\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)
\(\Leftrightarrow12x-9=29x-145\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x+136=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\left(tm\right)\)
Vậy \(S=\left\{8\right\}\)
\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)
\(\Rightarrow2x-1=2\left(5-3x\right)\)
\(\Leftrightarrow2x-1=10-6x\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x-11=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{11}{8}\right\}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{3x-2}{x-1}\)
\(\Rightarrow4x-5=3x-2\)
\(\Leftrightarrow4x-5-3x+2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy \(S=\left\{3\right\}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)
\(\Rightarrow15x+25=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)
Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)
Y
17 tháng 1 2023
\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)
\(\Leftrightarrow12x-9-29x+145=0\)
\(\Leftrightarrow-17x=-136\)
\(\Leftrightarrow x=8\)
\(2,\dfrac{2x-1}{5-3x}=2\)
\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)
\(\Leftrightarrow2x-1-10+6x=0\)
\(\Leftrightarrow8x=11\)
\(\Leftrightarrow x=\dfrac{11}{8}\)
\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)
\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)
\(\Leftrightarrow4x-5-2x+2+2x=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\dfrac{3}{4}\)
\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)
\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)
\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)
\(\Leftrightarrow15x=-25\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
3 tháng 2 2021
Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4
1 tháng 3 2022
a: =>3x-9+5+10x=90
=>13x-4=90
=>13x=94
hay x=94/13
b: \(\Leftrightarrow2x-4-x-1=3x-11\)
=>3x-11=x-5
=>2x=6
hay x=3(nhận)
\(\dfrac{x+1}{x-1}-\dfrac{x-2}{x-3}=3\) (ĐK: \(x\ne1;x\ne-3\))
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=3\)
\(\Leftrightarrow\dfrac{\left(x^2+x+3x+3\right)-\left(x^2-x-2x+2\right)}{\left(x+3\right)\left(x-1\right)}=3\)
\(\Leftrightarrow\dfrac{x^2+x+3x+3-x^2+x+2x-2}{\left(x+3\right)\left(x-1\right)}=3\)
\(\Leftrightarrow7x+1=3\left(x^2-x+3x-3\right)\)
\(\Leftrightarrow3x^2+6x-9-7x-1=0\)
\(\Leftrightarrow3x^2-x-10=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{5}{3}\end{matrix}\right.\)