Giúp mk câu hỏi này vs

\(2x\left(3x^2-7x+2\right)\)

\(=6x^3-14x^2+4x\)

\(\left(x-2\right)\left(3x^2+2x+4\right)\)

\(=3x^3+\left(-4x^2\right)+\left(-8\right)\)

\(\left(3x^2y^2+6x^2y^3-12xy\right)\div3xy\)

\(=xy+2xy-4\)

x^2/x-2+4-4x/x-2 ???

1) 2x(3x^2-7x+2)

<=> \(6x^3-14x^2+4x\)

2) (x-2)(3x^2+2x+4)

<=> \(3x^3-4x^2-8\)

3) (3x^2y^2+6x^2y^3-12xy)÷3xy

<=> xy + 2xy² -4 

4) X^2/x-2+4-4x/x-2

<=> \(\dfrac{x^2-4x+4}{x-2}=\dfrac{\left(x-2\right)^2}{x-2}=x-2\)

a: \(=-\left(10x^2+15x-8x-12\right)=-10x^2-7x+12\)

-10x^2-7x+12

= 12-7x-10x^2

Ta có:

\(\left(x-1\right).\left(x+1\right).\left(x+2\right)\)

\(=\left(x^2+x-x-1\right).\left(x+2\right)\)

\(=\left(x^2-1\right).\left(x+2\right)\)

\(=x^3+2x^2-x-2\)

1: ĐKXĐ:  \(x\notin\left\{0;3\right\}\)

\(\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\)

\(=\dfrac{x+3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\)

\(=\dfrac{\left(x+3\right)\left(x-3\right)-x^2+9}{x\left(x-3\right)}\)

\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)

=0

2: ĐKXĐ: \(x\notin\left\{0;1\right\}\)

\(\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x^2-x}\)

\(=\dfrac{3}{x}-\dfrac{5}{x-1}+\dfrac{3x+2}{x\left(x-1\right)}\)

\(=\dfrac{3x-3-5x+3x+2}{x\left(x-1\right)}\)

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

Chọn D

\(3^2\cdot5+2^3\cdot10-3^4:3\\ =9\cdot5+8\cdot10-3^4\\ =45+80-27\\ =98\)

\(3^2.5+2^3.10-3^4:3\)

\(=9.5+8.10-3^3\)

\(=45+80-27\)

\(=125-27\)

\(=98\)

\(#WendyDang\)

Anh ơi, dòng 2 ( \(3^4:3\) thì ra \(3^3\) -> a mới đổi thành được \(27\) chứ ạ? )