a) ĐKXĐ: x≠ \(\dfrac{1}{2}\); x≠ \(\dfrac{-1}{2}\); x≠0

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

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

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

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

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

       = \(\dfrac{-4}{2x-1}\)

b) Tại x= -2 ta có A= \(\dfrac{-4}{2.\left(-2\right)-1}\)= \(\dfrac{4}{5}\)

c)  A= 4 ta có \(\dfrac{-4}{2x-1}\)=4

                  ⇔ -4 = 4(2x-1)

                  ⇔ -4 = 8x-4 

                   ⇔ x = 0

d)  A=1 ta có \(\dfrac{-4}{2x-1}\)=1

                   ⇔  -4 = 2x-1

                    ⇔ x= \(\dfrac{-3}{2}\)

Các bn giúp mk nhanh nhanh nha câu b thôi câu a mk bt rồi nếu ko hiểu bảo mk gửi lại cho

a, a x 6 = 3 x 6 = 18

b, a + b = 4 + 2 = 6

c, b + a = 2 + 4 = 6

d, a - b = 8 - 5 = 3

e, m x n = 5 x 9 = 45

a. 18

b. 6

c. 6

d. 3

e. 45

chờ xíu

a) ĐKXĐ: \(x\notin\left\{5;-5\right\}\)

b) Ta có: \(A=\dfrac{2x}{x^2-25}+\dfrac{5}{5-x}-\dfrac{1}{x+5}\)

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

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

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

\(=\dfrac{-4x-20}{\left(x-5\right)\left(x+5\right)}\)

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

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

Để A nguyên thì \(-4⋮x-5\)

\(\Leftrightarrow x-5\inƯ\left(-4\right)\)

\(\Leftrightarrow x-5\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{6;4;7;3;9;1\right\}\)(nhận)

Vậy: Để A nguyên thì \(x\in\left\{6;4;7;3;9;1\right\}\)

a) \(A=\dfrac{3}{x-1}\)

Điều kiện \(|x-1|\ge0\)

\(\Rightarrow A=\dfrac{3}{x-1}\ge0\)

\(GTNN\left(A\right)=0\) \(\Rightarrow x-1=+\infty\Rightarrow x\rightarrow+\infty\)

b) \(GTLN\left(A\right)\) không có \(\left(A=\dfrac{3}{x-1}\ge0\right)\)

Câu 2: D

Câu 1: C