Bài 1:

a: ĐKXĐ: \(x\notin\left\{0;-1;\dfrac{1}{2}\right\}\)

\(P=\left(\dfrac{x+1}{3x^2+3x}+\dfrac{1-2x}{6x^2-3x}-1\right):\dfrac{1-x}{2x}\)

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

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

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

b: Để P nguyên thì \(2x⋮x-1\)

=>\(2x-2+2⋮x-1\)

=>\(2⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2\right\}\)

=>\(x\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{2;3\right\}\)

c: P<1

=>P-1<0

=>\(\dfrac{2x}{x-1}-1< 0\)

=>\(\dfrac{2x-x+1}{x-1}< 0\)

=>\(\dfrac{x+1}{x-1}< 0\)

=>-1<x<1

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}-1< x< 1\\x\ne0\end{matrix}\right.\)

a) đặt mẫu chứng là x-2

`a,` Với `x=3`

\(B=\dfrac{x^2-x}{2x+1}\\ \Rightarrow\dfrac{3^2-3}{2\cdot3+1}\\ =\dfrac{9-3}{6+1}\\ =\dfrac{6}{7}\)

`b,` Ta có `M=A*B`

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

`c,` Để `M=1/2`

`=> x/(x+1)=1/3`

`<=> (3x)/(3(x+1))= (x+1)/(3(x+1))`

`<=> 3x=x+1`

`<=>3x-x=1`

`<=>2x=1`

`<=>x=1/2`

các học bá đâu rùi

a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)

\(\Leftrightarrow1=0\)(vô lý)

Vậy: \(S=\varnothing\)

Ai giúp vs

a:

ĐKXĐ: x<>-1

 \(\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\)

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

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

b: \(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)

\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)

\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)

\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\dfrac{1}{x}+\dfrac{1}{x+2}\)

\(=\left(\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}+\dfrac{1}{x+2}\right)-\dfrac{1}{x}\)

\(=\dfrac{x+2-x-4+x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x}\)

\(=\dfrac{x-4}{x^2-4}-\dfrac{1}{x}\)

\(=\dfrac{x^2-4x-x^2+4}{x\left(x^2-4\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

c: \(\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\)

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

\(=\dfrac{-x-1-3x+3+4x}{2\left(x-1\right)\left(x+1\right)}\)

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

d:

\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)

b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)

c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)

\(P-2=\dfrac{2x-2x-2}{x+1}=\dfrac{-2}{x+1}\)

P<=2

=>x+1>0

=>x>-1

`P <= 1` là `P` ở đâu cậu nhỉ cộng `A` với `B` lại với nhau à?