a, \(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)

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

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

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

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

b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)

a, + b, \(A=\frac{x+2}{x-3}+\frac{2x-1}{x-1}-\frac{2x-1}{2x+1}\)DKXD : \(x\ne3;1;-\frac{1}{2}\)

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

\(=\frac{2x^3+3x^2-3x-2+4x^3-12x^2-x+4-2x^3+9x^2-10x+3}{MTC}\)

\(=\frac{4x^3-14x+2x^3+5}{MTC}\)

Đề sai ko kiểm tra lại hộ nhé !!! 

a, \(\frac{x+1}{2x+6}=\frac{x+1}{2\left(x+3\right)}\)

b, \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)

\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)

c, \(\frac{x-x-2xy+x}{x+2y}+\frac{4xy}{4y^2-x^2}=\frac{x-2xy}{x+2y}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}\)

\(=\frac{\left(x-2xy\right)\left(2y-x\right)}{\left(x+2y\right)\left(2y-x\right)}+\frac{4xy}{\left(2y-x\right)\left(x+2y\right)}=\frac{2xy-x^2+4xy^2+2x^2y}{\left(2y-x\right)\left(x+2y\right)}\)

A = x² + x + 1 = ( x² + x + 1/4 ) + 3/4 = ( x + 1/2 )² + 3/4 ≥ 3/4 ∀ x

Dấu "=" xảy ra khi x = -1/2

=> MinA = 3/4 <=> x = -1/2

B = -x² - 4x + 12 = -( x² + 4x + 4 ) + 16 = -( x + 2 )² + 16 ≤ 16 ∀ x

Dấu "=" xảy ra khi x = -2

=> MaxB = 16 <=> x = -2

C = \(\frac{5}{x^2+6}\)

Ta có : x² + 6 ≥ 6 ∀ x

<=> \(\frac{1}{x^2+6}\le\frac{1}{6}\forall x\)

<=> \(\frac{5}{x^2+6}\le\frac{5}{6}\forall x\)

Dấu "=" xảy ra khi x = 0

=> MaxC = 5/6 <=> x = 0