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

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

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

chiu roi 

ban oi $$

sao cung duoc nhe Vy Do

tk nhe@@@@@@@@@@__@

a) (3x + 1)^2 - 2(3x + 1)(3x - 5) + (3x - 5)^2 

= 9x^2 + 6x + 1 - 18x^2 + 24x + 10 + 9x^2 - 30x + 25

= 36

b) (3x^2 - y)^2

= 9x^4 - 6x^2y + y^2

c) (3x + 5)^2 + (3x - 5)^2 - (3x + 2)(3x - 2)

= 9x^2 + 30x + 25 + 9x^2 - 30x + 25 - 9x^2 + 4

= 9x^2 + 54

d) 2x(2x - 1)^2 - 3x(x + 3)(x - 3) - 4x(x + 1)^2

= 8x^3 - 8x^2 + 2x - 3x^2 + 27x - 4x^3 - 8x^2 - 4x

= x^3 - 16x^2 + 25x

e) (x - 2)(x^2 + 2x + 4) - (x + 1)^2 + 3(x - 1)(x + 1)

= x^3 - 8 - x^2 - 2x - 1 + 3x^2 - 2

= x^3 + 2x^2 - 2x - 12

f) (x^4 - 5x^2 + 25)(x^2 + 5) - (2 + x^2)^2 + 3(1 + x^2)^2

= x^6 + 125 - 4 - 4x^2 - x^2 + 3 + 6x^2 + 3x^4

= x^6 + 2x^4 + 2x^2 + 124

   R(x) =           2x² + 3x - 1

^-  M(x) =   -x³ + x² 

                x³ + x² + 3x - 1

Vậy R(x) - M(x) = x³ + x² + 3x - 1

1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)

=-27x^3-18x^2+4x+10

2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27

=7x^3+37x^2+46x+33

5:

\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)

\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)

=7x^3-48x^2+8x-35

mình nghĩ để sai. biểu thức ở giữa phải là (3x-1)^2 mới đúng

\(1,\dfrac{x-2}{2}=3.\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{x-2}{2}=\dfrac{1-3x}{2}\\ \Leftrightarrow x-2=1-3x\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)

2, mik có sửa đề vì đề của bn sai

ĐKXĐ:\(x\ne\pm\dfrac{1}{3}\)

\(\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}=\dfrac{5}{1-9x^2}\\ \Leftrightarrow\dfrac{\left(1-3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}-\dfrac{5}{\left(1-3x\right)\left(1+3x\right)}=0\\ \Leftrightarrow\dfrac{1-6x+9x^2-1-6x-9x^2-5}{\left(1+3x\right)\left(1-3x\right)}=0\\ \Rightarrow-12x-5=0\\ \Leftrightarrow x=-\dfrac{5}{12}\left(tm\right)\)