a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)

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

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

\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)

b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)

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

\(=\dfrac{x-3}{3x}\)

\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)

c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

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

\(=\dfrac{x-2}{2x}\)

\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)

Bài 1: 

a: \(\left(\dfrac{1}{3}x+2\right)\left(3x-6\right)\)

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

\(=x^2+3x-12\)

b: \(\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)

c: \(\left(-2xy+3\right)\left(xy+1\right)\)

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

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

d: \(x\left(xy-1\right)\left(xy+1\right)\)

\(=x\left(x^2y^2-1\right)\)

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

Bài 2: 

a: Ta có: \(M=\left(3x+2\right)\left(9x^2-6x+4\right)\)

\(=27x^3+8\)

\(=27\cdot\dfrac{1}{27}+8=9\)

b: Ta có: \(N=\left(5x-2y\right)\left(25x^2+10xy+4y^2\right)\)

\(=125x^3-8y^3\)

\(=125\cdot\dfrac{1}{125}-8\cdot\dfrac{1}{8}\)

=0

Bài 3: 

Ta có: \(A=\left(x+2\right)\left(3x-1\right)-x\left(3x+3\right)-2x+7\)

\(=3x^2-x+6x-2-3x^2-9x-2x+7\)

=5

a: Khi x=-2 thì \(A=3\cdot\left(-2\right)^2+5\cdot\left(-2\right)-1=12-10-1=1\)

b: \(B=6xyz^4=6\cdot3\cdot2\cdot1^4=36\)

a) thay x = -2 và biểu thúc ta đc

\(3.\left(-2\right)^2+5.\left(-2\right)-1=12-10-1=1\)

b)thay x = 3 ; y=2 ,z=1 và biểu thúc ta đc

\(9.3.2.1^4+5.3.2.1^4-8.3.2.1^4=3.2.1^4\left(9+5-8\right)=6.6=36\)

a)=1
b)=36

=> (8 - x)/(x - 5) ∈ Z
=> 8 - x chia hết cho x - 5
=> 3 - (x - 5) chia hết cho x - 5
=> 3 chia hết cho x - 5
=> x - 5 ∈ Ư(3) = (-3 ; -1 ; 1 ; 3)
=> x ∈ (2 ; 4 ; 6 ; 8)
vậy x ∈ (2 ; 4 ; 6 ; 8) mik ko chắc đâu 

A=x(y+2)+2(y+2)=(x+2)(y+2)=100.100=10000

B=9\(^2\)(9+1) +11(-9-1)=\(9^2\).10+11.(-10)=810-110=700

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`

b) Ta có: \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)

\(=5x^2-20xy-4y^2+20xy\)

\(=5x^2-4y^2\)

\(=5\cdot\left(-\dfrac{1}{5}\right)^2-4\cdot\left(-\dfrac{1}{2}\right)^2\)

\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}\)

\(=\dfrac{1}{5}-1=\dfrac{-4}{5}\)