a.

câu 1 là (x-y).(6x^2-4y^2+\(\dfrac{1}{2}\)xy) nha 

1. \(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)\)

\(=6x^3-4xy^2+\dfrac{1}{2}x^2y-6x^2y+4y^3-\dfrac{1}{2}xy^2\)

\(=6x^3+4y^3-\dfrac{11}{2}x^2y-\dfrac{9}{2}xy^2\)

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2. \(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)\)

\(=18x+6x^2-3-x-6x^2-15x\)

\(=2x-3\)

Chúc bạn học tốt!

`1)(x-y)(6x^2-4y^2+1/2xy)`

`=6x^3-4xy^2+1/2x^2y-6x^2y+4y^3-1/2xy^2`

`=6x^3-11/2x^2y-9/2xy^2+4y^3`

`2)(6x-1)(3+x)+(2x+5)(-3x)`

`=18+6x^2-x-3-6x^2-15x`

`=-15x+15`

1)\(\left(x-y\right)\left(6x^2-4y^2+\dfrac{1}{2}xy\right)=6x^3+4y^3-\dfrac{9}{2}xy^2-\dfrac{11}{2}x^2y\)

2)\(\left(6x-1\right)\left(3+x\right)+\left(2x+5\right)\left(-3x\right)=\left(6x^2+17x-3\right)+\left(-6x^2-15x\right)=2x-3\)

tick mik nha

2: Ta có: \(\left(6x-1\right)\left(x+3\right)+\left(-3x\right)\left(2x+5\right)\)

\(=6x^2+18x-x-3-6x^2-15x\)

\(=2x-3\)

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

=(3x^3-6x)(3x^2+x-2)

=9x^5+3x^4-6x^3-18x^3-6x^2+12x

=9x^5+3x^4-12x^3-6x^2+12x

2x(x^2-1)=2x^3-2x

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

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

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

a: \(=\dfrac{4x-2+6x^2-6x+2x^2+1}{2x\left(2x-1\right)}=\dfrac{8x^2-2x-1}{2x\left(2x-1\right)}\)

a) \(\dfrac{2x}{x^2-6x+9}+\dfrac{x-2}{x-3}\) (ĐK: \(x\ne3\))

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

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

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

\(=\dfrac{x^2-3x+6}{x^2-6x+9}\)

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

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

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

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

\(=\dfrac{1}{x^2+x+1}\)

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