Giải phương trình:
\(\dfrac{3}{x^2-1}-2\left(x+3\right)=\dfrac{5}{x+1}\)
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28 tháng 1 2022
\(1,\) thiếu đề
\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)
\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)
\(\Leftrightarrow25x+10-80x+10=24x+12-150\)
\(\Leftrightarrow-55x+20=24x-138\)
\(\Leftrightarrow24x-138+55x-20=0\)
\(\Leftrightarrow79x-158=0\)
\(\Leftrightarrow x=2\)
\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x-1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x-1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4\left(x-1\right)}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)
\(\Leftrightarrow x.\dfrac{x^2-1+x^2-2x-3-4x+4}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x^2-6x}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{2x\left(x-3\right)}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x=0\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1a.
ĐKXĐ: \(x\ne\left\{1;3\right\}\)
\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)
\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)
\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)
b.
ĐKXĐ: \(x\ne\left\{-1;2\right\}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)
\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 3 2021
1c.
ĐKXĐ: \(x\ne\left\{2;5\right\}\)
\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)
\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)
\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)
2a.
\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)
\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
2b.
\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)
12 tháng 11 2023
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
5 tháng 1
a: ĐKXĐ: \(x\notin\left\{2;5\right\}\)
\(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(\dfrac{6x+1}{\left(x-2\right)\left(x-5\right)}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
=>\(6x+1+5\left(x-5\right)=3\left(x-2\right)\)
=>6x+1+5x-25-3x+6=0
=>8x-18=0
=>8x=18
=>\(x=\dfrac{9}{4}\left(nhận\right)\)
b: Đề thiếu vế phải rồi bạn
c: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)
\(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{1}{x+1}-\dfrac{x}{x-3}=\dfrac{-\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\)
=>\(\dfrac{x+1}{x-3}+\dfrac{1}{x+1}=\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x+1\right)}\)
=>\(\left(x+1\right)^2+x-3=\left(x-1\right)^2\)
=>\(x^2+2x+1+x-3=x^2-2x+1\)
=>\(3x-2=-2x+1\)
=>5x=3
=>\(x=\dfrac{3}{5}\left(nhận\right)\)
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)
15 tháng 5 2023
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
9 tháng 2 2021
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
29 tháng 8 2021
a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)
\(\Leftrightarrow4x-2+2x=5x-20\)
\(\Leftrightarrow x=-18\)
b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-3x=-1\)
hay \(x=\dfrac{1}{3}\)
c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
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].
HP
16 tháng 3 2021
1.
ĐK: \(x\ne7;x\ne-1;x\ne3\)
\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)
TH1: \(x< -1\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\ge0\)
\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)
TH2: \(-1< x< 3\)
\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)
\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)
\(\Leftrightarrow-x^2+5x-22\ge0\)
\(\Rightarrow\) vô nghiệm
TH3: \(3< x< 7\)
Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)
\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)
TH4: \(x>7\)
\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)
\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)
\(\Leftrightarrow x^2-5x+22\le0\)
\(\Rightarrow\) vô nghiệm
Vậy ...
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