\(a,\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\left(-5< 0\right)\Leftrightarrow x>3\\ b,\dfrac{3-x}{x^2+1}\ge0\Leftrightarrow3-x\ge0\left(x^2+1>0\right)\Leftrightarrow x\le3\\ c,\dfrac{\left(x-1\right)^2}{x-2}< 0\Leftrightarrow x-2< 0\left[\left(x-1\right)^2\ge0\right]\Leftrightarrow x< 2\)

( x - 3/2 ) ( 2x + 1 ) > 0

TH1 : cả 2 thừa số đều lớn hơn 0

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}>0\\2x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x>\frac{3}{2}\\x>-\frac{1}{2}\end{cases}\Rightarrow}x>\frac{3}{2}}\)

TH2 : cả 2 thừa số đều bé hơn 0

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{2}< 0\\2x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< \frac{3}{2}\\x< -\frac{1}{2}\end{cases}\Rightarrow}x< -\frac{1}{2}}\)

Vậy,..........

a/ (x+1)(x-2) < 0 => \(\begin{cases}x+1>0\\x-2< 0\end{cases}\) hoặc \(\begin{cases}x+1< 0\\x-2>0\end{cases}\)

\(\Leftrightarrow-1< x< 2\) 

b/ (x+1/2)(x-2) > 0 => \(\begin{cases}x+\frac{1}{2}>0\\x-2>0\end{cases}\) hoặc \(\begin{cases}x+\frac{1}{2}< 0\\x-2< 0\end{cases}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x< -\frac{1}{2}\\x>2\end{array}\right.\)

a: \(\left|a-2b+3\right|^{2023}>=0\forall a,b\)

\(\left(b-1\right)^{2024}>=0\forall b\)

Do đó: \(\left|a-2b+3\right|^{2023}+\left(b-1\right)^{2024}>=0\forall a,b\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}a-2b+3=0\\b-1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}b=1\\a=2b-3=2\cdot1-3=-1\end{matrix}\right.\)

Thay a=-1 và b=1 vào P, ta được:

\(P=\left(-1\right)^{2023}\cdot1^{2024}+2024=2024-1=2023\)

a/ x.(x-1/3)<0

mà x > x-1/3

=> x>0 ; x-1/3 < 0 

=> x>0 ; x<1/3

=> 0<x<1/3, x thuộc Q

chọn ba số x là : 1/4 ; 1/5; 1/6

b/

x+y = x.y= x:y

x+y = x.y

=> x= x.y-y = y.[x-1]

=> x:y= x-1 [1]

=> x+y = x:y = x-1

=> y= -1 thay vào [1]

=> x: [-1] = x-1

=> -x = x-1

=> 2x = 1

=> x= 1/2

Vậy x= 1/2 ; y= -1

a)x(x-1/3)<0

Do x>x-1/3

=>x>0 x-1/3<0

<=>0<x<1/3

=>0<x<4/12

=>x={1/12;2/12;3/12;...}

Bạn bảo tìm 3 số nên mk tìm nấy chứ có vô số x

b)xy=x:y

=>y.y=x:x=1

=>y=1 hoặc y=-1

*)y=1

=>x+1=x

<=>x-x=1

<=>0=1(L)

*)y=-1

=>x-1=-x

<=>x+x=1

<=>2x=1

<=>x=1/2

Vậy y=-1 x=1/2

a/ x.(x-1/3)<0

mà x > x-1/3

=> x>0 ; x-1/3 < 0 

=> x>0 ; x<1/3

=> 0<x<1/3, x thuộc Q

chọn ba số x là : 1/4 ; 1/5; 1/6

b/

x+y = x.y= x:y

x+y = x.y

=> x= x.y-y = y.[x-1]

=> x:y= x-1 [1]

=> x+y = x:y = x-1

=> y= -1 thay vào [1]

=> x: [-1] = x-1

=> -x = x-1

=> 2x = 1

=> x= 1/2

Vậy x= 1/2 ; y= -1

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\)

\(\Rightarrow x^2-2x+1+9-x^2=0\)

\(\Rightarrow2x=10\Rightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\)

\(\Rightarrow\left(x-2-2x-1\right)\left(x-2+2x+1\right)=0\)

\(\Rightarrow-\left(x+3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)

a) \(\left(x-1\right)^2+\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow x^2-2x+1+9-x^2=0\\ \Leftrightarrow-2x=-10\\ \Leftrightarrow x=5\)

b) \(\left(x-2\right)^2-\left(2x+1\right)^2=0\\ \Leftrightarrow x^2-4x+4-4x^2-4x-1=0\\ \Leftrightarrow-3x^2-8x+3=0\\ \Leftrightarrow3x^2+8x-3=0\\ \Leftrightarrow\left(3x^2+9x\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)