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![](https://rs.olm.vn/images/avt/0.png?1311)
a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)
Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)
Suy ra: \(9-3x+10x-2=4\)
\(\Leftrightarrow7x+7=4\)
\(\Leftrightarrow7x=-3\)
hay \(x=-\dfrac{3}{7}\)
Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
1/
Ta có: 6x4 -x3-7x2+x+1=0
<=> 6x4-6x3+5x3-5x2-2x2+2x-x+1=0
<=> 6x3(x-1)+5x2(x-1)-2x(x-1)-(x-1)=0
<=> (x-1) ( 6x3+5x2-2x-1)=0
<=> ( x-1) ( 6x3-3x2+8x2-4x+2x-1)=0
<=> (x-1)\(\left[3x^2\left(2x-1\right)+4x\left(2x-1\right)+\left(2x-1\right)\right]\)=0
<=> (x-1) ( 2x-1) ( 3x2+4x+1)=0
<=> (x-1) ( 2x-1) (3x2+3x+x+1)=0
<=> (x-1) (2x-1) \(\left[3x\left(x+1\right)+\left(x+1\right)\right]\)=0
<=> (x-1)(2x-1)(x+1)(3x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\\x+1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=1\\x=-1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\x=-1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
vậy \(S=\left\{\pm1;\dfrac{1}{2};\dfrac{-1}{3}\right\}\)
\(6x^4-x^3-7x^2+x+1=0\)
\(\Leftrightarrow6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)
\(\Leftrightarrow6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)
\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)
Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1
b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)
\(\Leftrightarrow x< 5\)
c) \(\dfrac{-1}{2x+3}< 0\)
dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp
a) \(15x-3\left(3x-2\right)=45-5\left(2x-5\right)\)
\(\Leftrightarrow15x-9x+6=45-10x+25\)
\(\Leftrightarrow15x-9x+10x=45+25-6\)
\(\Leftrightarrow16x=64\)
\(\Leftrightarrow x=4\)
b) \(x^2-9+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)+4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\x+7=0\Leftrightarrow x=-7\end{matrix}\right.\)
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow x+4+x^2-4x+2x-8=5x-4\)
\(\Leftrightarrow x^2+x-4x+2x-5x=-4+8-4\)
\(\Leftrightarrow x^2-6x=0\)
\(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-6=0\Leftrightarrow x=6\end{matrix}\right.\)
a) 15x - 3(3x - 2) = 45 - 5(2x - 5)
\(\Leftrightarrow\) 15x - 9x + 6 = 45 - 10x + 25
\(\Leftrightarrow\) 6x + 10x = 70 - 6
\(\Leftrightarrow\) 16x = 64
\(\Leftrightarrow\) x = 4
Vậy.......................
b) x2 - 9 + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3) + 4(x - 3) = 0
\(\Leftrightarrow\) (x - 3)(x + 3 + 4) = 0
\(\Leftrightarrow\) (x - 3)(x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=3\end{matrix}\right.\)
Vậy........................
c) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{x^2-16}\)
\(\Leftrightarrow\) \(\dfrac{1}{x-4}+\dfrac{x+2}{x+4}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\) (đk: x\(\ne\pm\)4)
\(\Leftrightarrow\) \(\dfrac{x+4}{\left(x+4\right)\left(x-4\right)}+\dfrac{\left(x+2\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5x-4}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow\) x + 4 + x2 - 4x + 2x - 8 = 5x - 4
\(\Leftrightarrow\) x2 - x - 5x - 4 + 4 = 0
\(\Leftrightarrow\) x2 - 6x = 0
\(\Leftrightarrow\) x(x - 6) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=6\left(tmđk\right)\end{matrix}\right.\)
Vậy...............