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b, ĐK: \(x\ne8\)
\(A=\dfrac{x-5}{x-8}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5>0\\x-8>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5< 0\\x-8< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>5\\x>8\end{matrix}\right.\\\left\{{}\begin{matrix}x< 5\\x< 8\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>8\\x< 5\end{matrix}\right.\)
\(ĐKXĐ:x-3\ne0\Rightarrow x\ne3;x-1\ne0\Rightarrow x\ne1\\ \dfrac{1}{x-3}+2-1-\dfrac{5}{x-1}=0\\ \Leftrightarrow\dfrac{1}{x-3}+1-\dfrac{5}{x-1}=0\\ \Leftrightarrow\dfrac{1+x-3}{x-3}-\dfrac{5}{x-1}=0\\ \Leftrightarrow\dfrac{-2+x}{x-3}-\dfrac{5}{x-1}=0\\ \Leftrightarrow\dfrac{\left(-2+x\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{5\left(x-3\right)}{\left(x-3\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-1\right)}-\dfrac{5x-15}{\left(x-3\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{x^2-3x+2-5x+15}{\left(x-3\right)\left(x-1\right)}=0\\ \Rightarrow x^2-8x+17=0\\ \Leftrightarrow\left(x^2-8x+16\right)+1=0\\ \Leftrightarrow\left(x-4\right)^2=-1\left(vô lí\right)\)
suy ra pt vô nghiệm
\(\Leftrightarrow\dfrac{x+10}{2012}+1+\dfrac{x+8}{2014}+1+\dfrac{x+6}{2016}+1+\dfrac{x+4}{2018}+1=0\)
\(\Leftrightarrow\dfrac{x+2022}{2012}+\dfrac{x+2022}{2014}+\dfrac{x+2022}{2016}+\dfrac{x+2022}{2018}=0\Leftrightarrow x=-2022\)
do 2 pt tương đường nhau nên x = -2022 cũng là nghiệm của pt
\(\left(m-1\right)x+2020m-6=0\)
thay vào ta được : \(-2022\left(m-1\right)+2020m-6=0\)
\(\Leftrightarrow-2m+2022-6=0\Leftrightarrow-2m=-2016\Leftrightarrow m=1008\)
A, 3X+6>0
(=)3X>-6
(=)X>-2
VẬY ...
B,10-2X≥-4
(=)-2X≥-4-10
(=)-2X≥-14
(=)X≤7
VẬY....
C, 
(=)
(=) -15X+10>-3+3X
(=)-15X-3X>-3-10
(=)-18X>-13
(=)X<
\(Đkxđ:\\ \Leftrightarrow\left\{{}\begin{matrix}x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ \Rightarrow x\ne\pm4\)
ĐKXĐ: \(x\notin\left\{0;1\right\}\)
a) Thay m=1 vào phương trình, ta được:
\(\dfrac{2x+1}{x}=1+\dfrac{x+1}{x-1}\)
\(\Leftrightarrow\dfrac{2x+1}{x}=\dfrac{x-1+x+1}{x-1}\)
\(\Leftrightarrow\dfrac{2x+1}{x}=\dfrac{2x}{x-1}\)
\(\Leftrightarrow2x^2=\left(2x+1\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2=2x^2-2x+x-1\)
\(\Leftrightarrow2x^2-2x^2+2x-x-1=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(loại)
Vậy: Khi m=1 thì \(S=\varnothing\)
a) \(ĐKXĐ:\left\{{}\begin{matrix}x+1\ne0\\x\ne0\end{matrix}\right.< =>x\ne\left\{0;-1\right\}\)
b) \(\dfrac{3}{x+1}+\dfrac{5}{x}=0\\ < =>\dfrac{3x+5\left(x+1\right)}{x\left(x+1\right)}=0\\ =>3x+5\left(x+1\right)=0\\ < =>3x+5x+5=0\\ < =>8x=-5\\ < =>x=-\dfrac{5}{8}\left(TMDK\right)\)
Vậy tập nghiệm phương trình : \(S=\left\{-\dfrac{5}{8}\right\}\)