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: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)

1:

a: =>28x-8=9x+3

=>19x=11

=>x=11/19

b: =>(3x-1)(x-1)=(2x+1)(x+1)

=>3x^2-4x+1=2x^2+3x+1

=>x^2-7x=0

=>x=0 hoặc x=7

\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)

a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)

\(=\dfrac{1-1-3x}{3x}\cdot\dfrac{2x}{x-1}\)

\(=\dfrac{-3x}{3x}\cdot\dfrac{2x}{x-1}=\dfrac{-2x}{x-1}\)

Không có dấu "=" hay như nào đâu giải tìm x được

a: \(P=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}-\dfrac{x}{x+1}\right):\dfrac{x-1+x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\left(\dfrac{x+1}{x-1}-\dfrac{x}{x+1}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}\)
\(=\dfrac{x^2+2x+1-x^2+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x}=\dfrac{3x+1}{2x}\)

b: 1/P-x+1/8>=1

=>\(\dfrac{2x}{3x+1}-\dfrac{x+1}{8}>=1\)

=>\(\dfrac{16x-3x^2-4x-1-24x-8}{8\left(3x+1\right)}>=0\)

=>\(\dfrac{-3x^2-24x-9}{8\left(3x+1\right)}>=0\)

=>\(\dfrac{x^2+8x+3}{3x+1}< =0\)

TH1: x^2+8x+3<=0 và 3x+1>0

=>x>-1/3 và \(-4-\sqrt{13}< =x< =-4+\sqrt{13}\)

=>Loại

TH2: x^2+8x+3>=0 và 3x+1<0

=>x<-1/3 và (x<=-4-căn 13 hoặc x>=-4+căn 13)

=>x<=-4-căn 13

Bạn xem lại \(a,b\) mình làm rồi nha.

\(c,P>0\Leftrightarrow\left(x+1\right)^2>0\) (luôn đúng \(\forall x\))

Vậy với mọi giá trị x thì \(P>0\).