x \(\dfrac{2}{3}\)=\(\dfrac{1}{5}\)
\(\dfrac{2}{x}\)=\(\dfrac{12}{5}\)
\(\dfrac{1}{2}\)x-\(\dfrac{3}{5}\)=\(\dfrac{-4}{5}\)
3x+2x=-5,05
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Những câu hỏi liên quan
11 tháng 8 2021
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
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: \(S=\left\{\dfrac{20}{7}\right\}\)
23 tháng 2 2019
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 \(S=\left\{-1\right\}\)
19 tháng 1 2022
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
B
19 tháng 1 2022
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
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
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) 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: \(S=\varnothing\)
Y
17 tháng 1 2023
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)
26 tháng 9 2021
\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
26 tháng 9 2021
e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
1.\(x\times\dfrac{2}{3}=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}:\dfrac{2}{3}\)
\(x=\dfrac{3}{10}\)
2.\(\dfrac{2}{x}=\dfrac{12}{5}\)
\(12x=2.5\)
\(x=\dfrac{10}{12}=\dfrac{5}{6}\)
3.\(\dfrac{1}{2}x-\dfrac{3}{5}=-\dfrac{4}{5}\)
\(\dfrac{1}{2}x=\dfrac{-4}{5}+\dfrac{3}{5}\)
\(\dfrac{1}{2}x=-\dfrac{1}{5}\)
\(x=-\dfrac{1}{5}:\dfrac{1}{2}\)
\(x=-\dfrac{2}{5}\)
4.\(3x+2x=-5,05\)
\(5x=-5,05\)
\(x=-\dfrac{5,05}{5}\)
\(x=-1,01\)
\(x\cdot\dfrac{2}{3}=\dfrac{1}{5}\)
\(x=\dfrac{1}{5}:\dfrac{2}{3}\)
\(x=\dfrac{1}{5}\cdot\dfrac{3}{2}\)
\(x=\dfrac{3}{10}\)
\(\dfrac{2}{x}=\dfrac{12}{5}\)
\(x=\dfrac{\left(2\cdot5\right)}{12}\)
\(x=\dfrac{10}{12}\)
\(x=\dfrac{5}{6}\)
\(\dfrac{1}{2}x-\dfrac{3}{5}=\dfrac{-4}{5}\)
\(\dfrac{1}{2}x=\left(\dfrac{-4}{5}+\dfrac{3}{5}\right)\)
\(\dfrac{1}{2}x=-\dfrac{1}{5}\)
\(x=\left(-\dfrac{1}{5}\right):\dfrac{1}{2}\)
\(x=\left(-\dfrac{1}{5}\right)\cdot2\)
\(x=-\dfrac{2}{5}\)
\(3x+2x=-5,05\)
\(\left(3+2\right)x=-5,05\)
\(5x=-5,05\)
\(x=\left(-5,05\right):5\)
\(x=-1,01\)