Lời giải:

a. ĐKXĐ: $x\neq 0;-1$

\(=\left(\frac{2x^2+3x}{(x+1)(x^2-x+1)}+\frac{x+1}{(x+1)(x^2-x+1)}\right).\frac{x^2-x+1}{x}\)

\(=\frac{2x^2+3x+x+1}{(x+1)(x^2-x+1)}.\frac{x^2-x+1}{x}=\frac{2x^2+4x+1}{x(x+1)}\)

b. ĐKXĐ: $x\neq 0; 1;2$

\(=\frac{x-(x-1)}{x(x-1)}:\frac{(x+1)(x-1)-(x-2)(x+2)}{(x-2)(x-1)}=\frac{1}{x(x-1)}:\frac{3}{(x-2)(x-1)}\)

\(=\frac{1}{x(x-1)}.\frac{(x-2)(x-1)}{3}=\frac{x-2}{3x}\)

c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)

1: Sửa đề: 2/x+2

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

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

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

=>-2x^2+x+1-2x^2+2x=0

=>-4x^2+3x+1=0

=>4x^2-3x-1=0

=>4x^2-4x+x-1=0

=>(x-1)(4x+1)=0

=>x=1(loại) hoặc x=-1/4(nhận)

b: \(\Leftrightarrow\dfrac{440}{x-2}-\dfrac{440}{x}=1\)

=>x(x-2)=440x-440x+880

=>x^2-2x-880=0

=>\(x=1\pm\sqrt{881}\)

c: \(\Leftrightarrow\dfrac{x+5+x}{x\left(x+5\right)}=\dfrac{1}{6}\)

=>x^2+5x=6(2x+5)

=>x^2+5x-12x-30=0

=>x^2-7x-30=0

=>(x-10)(x+3)=0

=>x=10 hoặc x=-3

d: =>(x-1)(x+1)-x=2x-1

=>x^2-1-x=2x-1

=>x^2-x-2x=0

=>x(x-3)=0

=>x=0(loại) hoặc x=3(nhận)

1/ \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\) (1)

Điều kiện: \(\left\{{}\begin{matrix}x-1\ne0\\3x+4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-\dfrac{4}{3}\end{matrix}\right.\)

(1) \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\\\Leftrightarrow12x^2+16x+21x+28=12x^2-12x+5x-5\\ \Leftrightarrow\left(16+21+12-5\right)x=-5-28\\ \Leftrightarrow44x=-33\\ \Leftrightarrow x=-\dfrac{3}{4}\) (Thỏa mãn)

Vậy \(x=-\dfrac{3}{4}\).

2/ \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\) (2)

Điều kiện: \(x\ne\pm1\)

(2)\(\Leftrightarrow\dfrac{x}{x-1}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+1\right)-2x}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow x\left(x+1\right)-2x=0\\ \Leftrightarrow x^2+x-2x=0\\ \Leftrightarrow x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

So sánh với điều kiện \(\Rightarrow x=0\) là nghiệm của PT.

3/ \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\) (3)

Điều kiện: \(x\ne\pm3\)

(3)\(\Leftrightarrow\dfrac{1}{3-x}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\\ \Leftrightarrow-\dfrac{\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\\ \Leftrightarrow-\left(x+3\right)-14=\left(x-3\right)\left(x+3\right)\\ \Leftrightarrow-x-17=x^2-9\Leftrightarrow x^2+x+8=0\) (Vô nghiệm do \(x^2+x+8>0\qquad\forall x\)).

Vậy PT vô nghiệm.

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

Điều kiện: \(x\ne\pm1\)

(4)\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x^2+2x+1\right)-\left(x^2-2x+1\right)=4\Leftrightarrow4x=4\Leftrightarrow x=1\) (loại)

Vậy PT vô nghiệm.

5/ \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\) (5)

Điều kiện: \(x\ne0\)

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

Đặt \(t=x+\dfrac{1}{x}\), ta có: \(t=t^2-2\\ \Leftrightarrow t^2-t-2=0\Leftrightarrow\left(t-2\right)\left(t+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=-1\end{matrix}\right.\)

Với \(t=2\) ta có: \(x+\dfrac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\) (thỏa mãn)

Với \(t=-1\) ta có: \(x+\dfrac{1}{x}=-1\Leftrightarrow x^2+1=-x\Leftrightarrow x^2+x+1=0\) (vô nghiệm).

Vậy \(x=1\) là nghiệm PT.

6/ \(\dfrac{x-1}{x^2+4}=\dfrac{x-1}{x+1}\) (6)

Điều kiện: \(x\ne-1\)

(6)\(\Leftrightarrow\dfrac{x-1}{x^2+4}-\dfrac{x-1}{x+1}=0\\ \Leftrightarrow\left(x-1\right)\left(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\end{matrix}\right.\)

\(x-1=0\Leftrightarrow x=1\) (Thỏa mãn)

\(\dfrac{1}{x^2+4}-\dfrac{1}{x+1}=0\Leftrightarrow\dfrac{1}{x^2+4}=\dfrac{1}{x+1}\Leftrightarrow x^2+4=x+1\\ \Leftrightarrow x^2-x+3=0\) (vô nghiệm).

Vậy \(x=1\) là nghiệm PT.

1) ĐKXĐ: \(x\notin\left\{1;-\dfrac{4}{3}\right\}\)

Ta có: \(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)

\(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)

\(\Leftrightarrow12x^2+16x+21x+28=12x^2+12x+5x-5\)

\(\Leftrightarrow12x^2+37x+28-12x^2-17x+5=0\)

\(\Leftrightarrow20x+33=0\)

\(\Leftrightarrow20x=-33\)

\(\Leftrightarrow x=-\dfrac{33}{20}\)(nhận)

Vậy: \(S=\left\{-\dfrac{33}{20}\right\}\)

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

Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

Suy ra: \(x^2+x-2x=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Vậy: S={0}

3) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{1}{3-x}-\dfrac{14}{x^2-9}=1\)

\(\Leftrightarrow\dfrac{-1}{x-3}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=1\)

\(\Leftrightarrow\dfrac{-\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{14}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(-x-3-14=x^2-9\)

\(\Leftrightarrow x^2-9=-x-17\)

\(\Leftrightarrow x^2-9+x+17=0\)

\(\Leftrightarrow x^2+x+8=0\)

\(\Leftrightarrow x^2+2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{31}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{31}{4}=0\)(vô lý)

Vậy: \(S=\varnothing\)

4) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

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

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

5) ĐKXĐ: \(x\ne0\)

Ta có: \(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)

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

\(\Leftrightarrow x^2\left(x^2+1\right)=x\left(x^4+1\right)\)

\(\Leftrightarrow x^4+x^2=x^5+x\)

\(\Leftrightarrow x^5+x-x^4-x^2=0\)

\(\Leftrightarrow x\left(x^4-x^3-x+1\right)=0\)

\(\Leftrightarrow x\left[x^3\left(x-1\right)-\left(x-1\right)\right]=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x^3-1\right)=0\)

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

mà \(x^2+x+1>0\)

nên \(x\cdot\left(x-1\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)

Vậy: S={1}

6) ĐKXĐ: \(x\in R\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x^2+4\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x^2+4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1-x^2-4\right)=0\)

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

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

mà \(x^2-x+3>0\)

nên x-1=0

hay x=1(nhận)

Vậy: S={1}

\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)

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

\(a,=\left(\dfrac{1-x}{x}+\dfrac{x^3-x}{x}\right)\times\dfrac{x}{x-1}\\ =\dfrac{1-x+x^3-x}{x}\times\dfrac{x}{x-1}\\ =\dfrac{1-2x+x^3}{x-1}\\ b,=\left(\dfrac{x-x^2}{x.x^2}\right).\dfrac{x^2}{y}+\dfrac{x}{y}\\ =\dfrac{x-x^2}{xy}+\dfrac{x}{y}\\ =\dfrac{x-x^2+x^2}{xy}=\dfrac{x}{xy}=\dfrac{1}{y}\)

\(c,=\dfrac{3}{x}-\dfrac{2}{x}\times x+\dfrac{x}{3}\\ =\dfrac{3}{x}-2+\dfrac{x}{3}\\ =\dfrac{3-2x+x^2}{3x}\)

help me pls!!!

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

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

\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)

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

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

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

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

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

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

\(a,=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\\ b,=\dfrac{6x-3+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2}{2x\left(2x-1\right)}\\ =\dfrac{2\left(2x-1\right)\left(2x+1\right)}{2x\left(2x-1\right)}=\dfrac{2x+1}{x}\\ c,=\dfrac{x^3+x^2+x+2x-2+4x^2-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^3+5x^2+3x-3}{x^3-1}\)