\(\dfrac{x+1}{3}=\dfrac{1-3x}{-5}\)
Trong đó, x=
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22 tháng 3 2021
a) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
Suy ra: x+2=0
hay x=-2(thỏa ĐK)
Vậy: S={-2}
22 tháng 3 2021
d)
ĐKXĐ: \(x\notin\left\{1;3\right\}\)
Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)
Suy ra: \(x^2-3x+5x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+9=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3(loại)
Vậy: \(S=\varnothing\)
3 tháng 2 2022
a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)
\(\Rightarrow-x^2=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{6;-6\right\}\)
b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)
\(\Rightarrow5-9x=-3\)
\(\Rightarrow-9x=-8\)
\(\Rightarrow x=\dfrac{8}{9}\)
Vậy: \(x=\dfrac{8}{9}\)
c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)
\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)
\(\Rightarrow x=\dfrac{24}{5}\)
Vậy: \(x=\dfrac{24}{5}\)
d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)
\(\Rightarrow\left(3x-1\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)
8 tháng 12 2021
i: \(=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)
8 tháng 12 2021
Bài 1:
\(i,\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}-\dfrac{x+2}{5-x}=\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
\(j,\dfrac{3x\left(x-2\right)}{3x-2}+\dfrac{6x^2}{3x-2}-\dfrac{2\left(2-3x\right)}{2-3x}=\dfrac{3x^2-6x}{3x-2}+\dfrac{6x^2}{3x-2}+\dfrac{4-6x}{3x-2}=\dfrac{3x^2-6x+6x^2+4-6x}{3x-2}=\dfrac{9x^2-12x+4}{3x-2}=\dfrac{\left(3x-2\right)^2}{3x-2}=3x-2\)
\(n,\dfrac{2}{x}+\dfrac{3}{x-1}+\dfrac{1-4x}{x^2-x}=\dfrac{2\left(x-1\right)+3x+1-4x}{x\left(x-1\right)}=\dfrac{2x-2+3x+1-4x}{x\left(x-1\right)}=\dfrac{x-1}{x\left(x-1\right)}=\dfrac{1}{x}\)
Bài 2:
\(j,\dfrac{2}{3x}-\dfrac{1}{2x-2}-\dfrac{x-4}{6x-6x^2}=\dfrac{4\left(x-1\right)}{6x\left(x-1\right)}-\dfrac{3x}{6x\left(x-1\right)}-\dfrac{x-4}{6x\left(1-x\right)}=\dfrac{4x-4-3x+x-4}{6x\left(x-1\right)}=\dfrac{2x-8}{6x\left(x-1\right)}=\dfrac{2\left(x-4\right)}{6x\left(x-1\right)}=\dfrac{x-4}{3x\left(x-1\right)}\)
11 tháng 7 2023
1: Sửa đề: 2/x+2
\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)
=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
=>4x-3=-3x-6
=>7x=-3
=>x=-3/7(nhận)
2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)
=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)
=>-6x^2+6=2(3x^2-10x+3)
=>-6x^2+6=6x^2-20x+6
=>-12x^2+20x=0
=>-4x(3x-5)=0
=>x=5/3(nhận) hoặc x=0(nhận)
3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)
=>x*19/6=35/12
=>x=35/38
5 tháng 2 2022
e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)
\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)
\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)
\(\Leftrightarrow x=-1\left(TM\right)\)
22 tháng 3 2021
a) Ta có: \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)
\(\Leftrightarrow\dfrac{2\left(x+5\right)}{6\left(x-2\right)}-\dfrac{3\left(x-2\right)}{6\left(x-2\right)}=\dfrac{3\left(2x-3\right)}{6\left(x-2\right)}\)
Suy ra: \(2x+5-3x+6=6x-9\)
\(\Leftrightarrow-x+11-6x+9=0\)
\(\Leftrightarrow20-7x=0\)
\(\Leftrightarrow7x=20\)
hay \(x=\dfrac{20}{7}\)(thỏa ĐK)
Vậy: \(S=\left\{\dfrac{20}{7}\right\}\)
11 tháng 8 2021
\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)
\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)
\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)
\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)
\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)
Tick nha
11 tháng 8 2021
2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)
\(\Leftrightarrow6x-18=15-5x\)
\(\Leftrightarrow11x=33\)
hay x=3
9 tháng 3 2021
a) \(\dfrac{x+5}{3\left(x-1\right)}+1=\dfrac{3x+7}{5\left(x-1\right)}\) ( đk: \(x\ne1\))
\(\Leftrightarrow\dfrac{5\left(x+5\right)}{15\left(x-1\right)}+\dfrac{15\left(x-1\right)}{15\left(x-1\right)}=\dfrac{3\left(3x+7\right)}{15\left(x-1\right)}\)
\(\Rightarrow5\left(x+5\right)+15\left(x-1\right)=3\left(3x+7\right)\)
\(\Leftrightarrow5x+25+15x-15=9x+21\)
\(\Leftrightarrow5x+15x-9x=21-25+15\)
\(\Leftrightarrow11x=11\Leftrightarrow x=1\) (loại)
Vậy tập nghiệm: \(S=\varnothing\)
b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}-\dfrac{8}{x^2+2x-3}=1\) (đk: \(x\ne1,x\ne-3\))
\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{x^2+2x-3}-\dfrac{\left(2x+5\right)\left(x-1\right)}{x^2+2x-3}-\dfrac{8}{x^2+2x-3}=\dfrac{x^2+2x-3}{x^2+2x-3}\)
\(\Rightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-8=x^2+2x-3\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5-8=x^2+2x-3\)
\(\Leftrightarrow3x=3\Leftrightarrow x=1\) (loại)
Vậy tập nghiệm: \(S=\varnothing\)
\(\dfrac{x+1}{3}=\dfrac{1-3x}{-5}\)
=>\(-5\left(x+1\right)=3\left(1-3x\right)\)
=>\(5\left(x+1\right)=3\left(3x-1\right)\)
=>5x+5=9x-3
=>5x-9x=-3-5
=>-4x=-8
=>x=2