Bài 6:

M= 2.2 - 2.3+3.2.3

M= 4 - 6 + 18

M= 20

Bài 7: 

P= 1.2 - 5.-1.-2 + 8.-2.2

P = 2 -10 -32

P= -44

Bài 8:

A (thiếu dữ kiện bn ơi)

B= -1.2 . 3.2 + -1.3 +3.3 +-1.3

B= -2 . 6 + -3 + 9 +-3

B= -2 . 6 - 3 + 9 - 3

B= -12 - 3 + 9 - 3

B= -9

a: x và y là hai đại lượng tỉ lệ thuận

=>\(\frac{x_1}{y_1}=\frac{x_2}{y_2}\)

=>\(\frac{x_1}{y_1}=-\frac13\)

=>\(y_1=-3\cdot x_1\)

\(y_1-2x_1=9\)

=>\(-3x_1-2x_1=9\)

=>\(-5x_1=9\)

=>\(x_1=-1,8\)

=>\(y_1=-3\cdot1,8=-5,4\)

b: \(\frac{x}{y}=-\frac13\)

=>y=-3x

c: y=-10

=>-3x=-10

=>\(x=\frac{10}{3}\)

y=12

=>-3x=12

=>x=-4

d: Khi x=-5 thì \(y=\left(-3\right)\cdot\left(-5\right)=15\)

Khi x=7 thì \(y=-3\cdot7=-21\)

a) \(P=x\left(x-y\right)+y\left(x-y\right)=\left(x-y\right)\left(x+y\right)=x^2-y^2=5^2-4^2=9\)

b) \(Q=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=0\)

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

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

=3

=>Đúng

b: \(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)

\(=x^3+x_{}^2+x-x^3-x^2-x+5\)

=5

=>Đúng

c: \(3x\left(x-2\right)-5x\left(x-1\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x_{}^2+5x-8x^3+24\)

=-x+24

=>Sai

d: \(2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)

\(=2y^3+2y^2+2y-2y^3-2y^2-2y-20\)

=-20

=>Đúng

b: \(A=\dfrac{2-1}{3\cdot2}=\dfrac{1}{6}\)

$\textbf{a)}$

Điều kiện: $x\ne0,\ x\ne-1,\ x\ne1.$

Ta có $B=\left(\dfrac{x+1}{2(x-1)}+\dfrac{3x-1}{(x-1)(x+1)}-\dfrac{x+3}{2(x+1)}\right):\dfrac3{x+1}.$

Quy đồng các phân thức trong ngoặc:

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

$\dfrac{x+3}{2(x+1)}=\dfrac{(x+3)(x-1)}{2(x-1)(x+1)}.$

Do đó \[\begin{aligned}&\dfrac{(x+1)^2+2(3x-1)-(x+3)(x-1)}{2(x-1)(x+1)}\\&=\dfrac{x^2+2x+1+6x-2-(x^2+2x-3)}{2(x-1)(x+1)}\\&=\dfrac{6x+2}{2(x-1)(x+1)}=\dfrac{3x+1}{(x-1)(x+1)}.\end{aligned}\]

Suy ra $B=\dfrac{3x+1}{(x-1)(x+1)}\cdot\dfrac{x+1}{3}=\dfrac{3x+1}{3(x-1)}.$

\(A=\dfrac{-x^2-2x+2014}{x^2}=\dfrac{2014}{x^2}-\dfrac{2}{x}-1=2014\left(\dfrac{1}{x}-\dfrac{1}{2014}\right)^2-\dfrac{2015}{2014}\ge-\dfrac{2015}{2014}\)

\(A_{min}=-\dfrac{2015}{2014}\) khi \(x=2014\)

cứuuuuuuuuuuu