\(\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
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AP
1
22 tháng 2 2020
ĐKXĐ: \(\left\{{}\begin{matrix}4x-20\ne0\\50-2x^2\ne0\\6x+30\ne0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}4x-20\ne0\\x^2-25\ne0\\6x+30\ne0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
=> \(x\ne\pm5\)
Ta có : \(\frac{3}{4x-20}+\frac{15}{50-2x^2}+\frac{7}{6x+30}=0\)
=> \(\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}+\frac{7}{6\left(x+5\right)}=0\)
=> \(\frac{9\left(x+5\right)}{12\left(x^2-25\right)}-\frac{90}{12\left(x^2-25\right)}+\frac{14\left(x-5\right)}{12\left(x^2-25\right)}=0\)
=> \(9\left(x+5\right)-90+14\left(x-5\right)=0\)
=> \(9x+45-90+14x-70=0\)
=> \(23x=115\)
=> \(x=5\) ( KTM )
Vậy phương trình vô nghiệm .
TD
1
22 tháng 6 2022
\(\Leftrightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x-5\right)\left(x+5\right)}+\dfrac{7}{6\left(x+5\right)}=0\)
\(\Leftrightarrow3\cdot3\left(x+5\right)-15\cdot6+7\cdot2\cdot\left(x-5\right)=0\)
=>9x+45-90+14x-70=0
=>23x-115=0
hay x=5(loại)
17 tháng 3 2021
a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)
\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)
\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)
\(\Leftrightarrow9x-6=5x-20\)
\(\Leftrightarrow9x-5x=-20+6\)
\(\Leftrightarrow4x=-14\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
2 tháng 3 2018
\(\text{a) }\left|2-5x\right|=\left|3x+1\right|\\ \Leftrightarrow\left[{}\begin{matrix}2-5x=3x+1\\2-5x=-3x-1\end{matrix}\right. \Leftrightarrow\left[{}\begin{matrix}-5x-3x=1-2\\-5x+3x=-1-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-8x=-1\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{8}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{\dfrac{1}{8};\dfrac{3}{2}\right\}\)
\(\text{b) }\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\)
ĐXKĐ của phương trình \(:x\ne\pm5\)
\(\text{Ta có }:\dfrac{3}{4x-20}+\dfrac{15}{50-2x^2}+\dfrac{7}{6x+30}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{2\left(25-x^2\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x+5\right)\left(x-5\right)}+\dfrac{7}{6\left(x+5\right)}=0\\ \Rightarrow\dfrac{9\left(x+5\right)}{12\left(x+5\right)\left(x-5\right)}-\dfrac{90}{12\left(x+5\right)\left(x-5\right)}+\dfrac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\\ \Rightarrow9x+45-90+14x-70=0\\ \Leftrightarrow23x=115\\ \Leftrightarrow x=5\left(KTM\right)\)
Vậy phương trình vô nghiệm
\(\text{c) }\dfrac{x+29}{31}-\dfrac{x+27}{33}=\dfrac{x+17}{43}-\dfrac{x+15}{45}\\ \Leftrightarrow\left(\dfrac{x+29}{31}+1\right)-\left(\dfrac{x+27}{33}+1\right)=\left(\dfrac{x+17}{43}+1\right)-\left(\dfrac{x+15}{45}+1\right)\\ \Leftrightarrow\dfrac{x+60}{31}-\dfrac{x+60}{33}-\dfrac{x+60}{43}+\dfrac{x+60}{45}=0\\ \Leftrightarrow\left(x+60\right)\left(\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\right)=0\\ \Leftrightarrow x+60=0\left(\text{Vì }\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{43}+\dfrac{1}{45}\ne0\right)\\ \Leftrightarrow x=-60\)
Vậy \(x=-60\) là nghiệm của phương trình
23 tháng 7 2022
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
NH
1
8 tháng 8 2022
\(\Leftrightarrow\dfrac{-7}{x^2+3x-10}+\dfrac{x+4}{x+5}+\dfrac{x+3}{x-2}+3=0\)
\(\Leftrightarrow-7+x^2+2x-8+x^2+8x+15+3x^2+9x-30=0\)
\(\Leftrightarrow5x^2+19x-30=0\)
hay \(x\in\left\{\dfrac{6}{5}\right\}\)
24 tháng 9 2022
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
`3/(4x-20)+15/(50-2x^2)+7/(6x+30)=0(x ne +-5)`
`pt<=>3/(4(x-5))+15/(2(5-x)(5+x))+7/(6(x+5))=0`
`<=>3/(4(x-5))+7/(6(x+5))-15/(2(x-5)(x+5))=0`
`<=>9(x+5)+14(x-5)-90=0`
`<=>9x+45+14x-70-90=0`
`<=>23x=115`
`<=>x=5(ktm)`
Vậy PTVN