A=-(3x+7)+(5x-2)+(2x-10)

=-3x-7+5x-2+2x-10

=(-3x+5x+2x)-(7+2+10)

=4x-19

B = (6x+8)-(4x-5)-3x

= 6x+8-4x+5-3x

= (6x-4x-3x) + (8+5)

= -x + 13

= 13-x

C = 2(5x+3) - (2x-1) + 12

= 10x+6 - 2x + 1 + 12

= (10x-2x) + (6+1+12)

= 8x + 19

D = (x+7)-3(x+1)+2x-5

= x+7-3x-3+2x-5

= (x-3x+2x) + (7-3-5)

= -1

:) bóc lột !

Câu 1: 

a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x

b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)

\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)

\(=2x^2+6x+17\)

c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)

Câu 2: 

\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x^2-2.x.3+3^2\right)=2x\left(x-3\right)^2\)

a) (3x-1).(2x+7)-(x+1).(6x-5)-(18x-12)

=6x²+19x-7-(6x²+x-5)-18x+12

=6x²+19x-7-6x²-x+5-18x+12

=10

vậy giá trị của biểu thức (3x-1).(2x+7)-(x+1).(6x-5)-(18x-12) ko phụ thuộc vào giá trị của biến;

a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)

\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)

\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)

\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)

b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)

\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)

\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)

\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)

\(\Leftrightarrow\frac{1}{3x^2-y^2}\)

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

(3x−1)(2x+7)−(x+1)(6x−5)−(18x−12)

=6x^2+21x−2x−7−6x^2+5x−6x+5−18x+12(3x−1)(2x+7)−(x+1)(6x−5)−(18x−12)

=6x^2+21x−2x−7−6x2+5x−6x+5−18x+12

=(6x^2−6x^2)+(21x−2x+5x−6x−18x)+(12+5−7)

=10

b) (3x-1)(2x+3)-(x-5)(6x-1)-38x

= 6x^2+9x-2x-3-6x^2+x+30x-5-38x

=-8

Suy ra (3x-1)(2x+3)-(x-5)(6x-1)-38x ko phụ thuộc vào giá trị biến của x

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)