i: \(=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=3\)

Bài 1:

\(i,\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}-\dfrac{x+2}{5-x}=\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}=\dfrac{x+1+x-18+x+2}{x-5}=\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)

\(j,\dfrac{3x\left(x-2\right)}{3x-2}+\dfrac{6x^2}{3x-2}-\dfrac{2\left(2-3x\right)}{2-3x}=\dfrac{3x^2-6x}{3x-2}+\dfrac{6x^2}{3x-2}+\dfrac{4-6x}{3x-2}=\dfrac{3x^2-6x+6x^2+4-6x}{3x-2}=\dfrac{9x^2-12x+4}{3x-2}=\dfrac{\left(3x-2\right)^2}{3x-2}=3x-2\)

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

Bài 2:

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

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

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

giúp mình với ạ câu nào cũng được

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

Đây là bài giải pt chứ có phải biểu thức đâu mà thu gọn hả bạn?

Lời giải:

a. ĐKXĐ: $x\neq 1$

PT $\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}+\frac{2x(x-1)}{(x-1)(x^2+x+1)}=\frac{3x^2}{(x-1)(x^2+x+1)}$

$\Leftrightarrow x^2+x+1+2x(x-1)=3x^2$

$\Leftrightarrow 3x^2-x+1=3x^2$

$\Leftrightarrow x=1$ (không thỏa đkxđ)

Vậy pt vô nghiệm.

b. ĐKXĐ: $x\neq \pm 3$

PT $\Leftrightarrow \frac{(x+2)(x+3)}{(x-3)(x+3)}=\frac{x^2+3x}{(x-3)(x+3)}$

$\Leftrightarrow (x+2)(x+3)=x^2+3x$

$\Leftrightarrow x^2+5x+6=x^2+3x$

$\Leftrightarrow 2x+6=0$

$\Leftrightarrow x=-3$ (không thỏa mãn đkxđ)

Do đó pt vô nghiệm.

c. ĐKXĐ: $x\neq \pm 2$

PT $\Leftrightarrow \frac{(x-2)^2-(x+2)^2}{(x+2)(x-2)}=\frac{-16}{(x-2)(x+2)}$

$\Leftrightarrow (x-2)^2-(x+2)^2=-16$

$\Leftrightarrow -8x=-16$

$\Leftrightarrow x=2$ (vi phạm đkxđ)

Do đó pt vô nghiệm.

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)

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

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

c: \(=\dfrac{2x^2-3x-9-x^2+3x+x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)

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

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

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

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

\(\dfrac{2}{36a^2b^2-1}=\dfrac{2}{\left(6ab-1\right)\left(6ab+1\right)}\\ \dfrac{1}{6ab+1}=\dfrac{6ab-1}{\left(6ab-1\right)\left(6ab+1\right)};\dfrac{1}{6ab-1}=\dfrac{6ab+1}{\left(6ab-1\right)\left(6ab+1\right)}\)

\(\dfrac{x}{x^3-27}=\dfrac{x\left(x-3\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{2x}{x^2-6x+9}=\dfrac{2x\left(x^2+3x+9\right)}{\left(x-3\right)^2\left(x^2+3x+9\right)}\\ \dfrac{1}{x^2+3x+9}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)^2\left(x^2+3x+9\right)}\)

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

a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)

\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)

c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)

\(=\dfrac{-x+7x-4}{3x-2}\)

\(=\dfrac{6x-4}{3x-2}=2\)