a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

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

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

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

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

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

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

a) \(\dfrac{4x-1}{3x^2y}-\dfrac{7x-1}{3x^2y}\)

\(=\dfrac{\left(4x-1\right)-\left(7x-1\right)}{3x^2y}\)

\(=\dfrac{4x-1-7x+1}{3x^2y}\)

\(=\dfrac{-3x}{3x^2y}\)

\(=\dfrac{-1}{xy}\)

b) \(\dfrac{4x+5}{2x-1}-\dfrac{5-9x}{2x-1}\)

\(=\dfrac{\left(4x+5\right)-\left(5-9x\right)}{2x-1}\)

\(=\dfrac{4x+5-5+9x}{2x-1}\)

\(=\dfrac{13x}{2x-1}\)

c) \(\dfrac{11x}{2x-3}-\dfrac{x-18}{3-2x}\)

\(=\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\)

\(=\dfrac{11x+\left(x-18\right)}{2x-3}\)

\(=\dfrac{11x+x-18}{2x-3}\)

\(=\dfrac{12x-18}{2x-3}\)

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

\(=\dfrac{6}{1}\)

\(=6\)

d) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)

\(=\dfrac{2x-7}{10x-4}+\dfrac{3x+5}{10x-4}\)

\(=\dfrac{\left(2x-7\right)+\left(3x+5\right)}{10x-4}\)

\(=\dfrac{2x-7+3x+5}{10x-4}\)

\(=\dfrac{5x-2}{10x-4}\)

\(=\dfrac{5x-2}{2\left(5x-2\right)}\)

\(=\dfrac{1}{2}\)

a)

b)

c)

d)

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

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

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

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

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

\(=\dfrac{5\cdot4x}{2x\cdot\left(x+1\right)}=\dfrac{10}{x+1}\)

\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)

\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)

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

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

a) (2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x(2x+12x−1−2x−12x+1):4x10x−5=(2x+1)2−(2x−1)2(2x−1)(2x+1).10x+54x

=4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x4x2+4x+1−4x2+4x−1(2x−1)(2x+1).5(2x+1)4x

=8x.5(2x+1)(2x−1)(2

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

c) nè

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

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

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

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

\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)\cdot2}{\left(x^2+1\right)\left(x-1\right)^2\left(x+1\right)}\)

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

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

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

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

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

a) Không chép lại đề bài , tớ làm luôn nhé

P = \(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{10x-5}{4x}\)

ĐKXĐ :* ) 2x - 1 # 0 => x # \(\dfrac{1}{2}\)

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

*) x # 0

P = \(\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{10x-5}{4x}\)

P = \(\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)

P = \(\dfrac{10}{2x+1}\)

b) / x / = 2

=> x = 2 ; x = -2

*) Thay : x = 2 vào phân thức P đã rút gọn tại phần a , ta có :

\(P=\dfrac{10}{2.2+1}=\dfrac{10}{5}=2\)

Vậy , phân thức P = 2 tại x = 2

*) Thay : x = -2 vào phân thức P đã rút gọn tại phần a , ta có :

\(P=\dfrac{10}{2.\left(-2\right)+1}=\dfrac{10}{-4+1}=-\dfrac{10}{3}\)

Vậy , phân thức P = \(-\dfrac{10}{3}\) tại x = -2