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: \(\dfrac{x^2-4x+4}{2x^2-4x}\)

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

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

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

a) x² + 4x + 4 = x² + 2 . x . 2 + 2² = (x+ 2)²

Với x = 98: (98+ 2)² =100² = 10000

b) x³ + 3x² + 3x + 1 = x³ + 3 . 1 . x² + 3 . x .1²+ 1³ = (x + 1)³ 

Với x = 99: (99+ 1)³ = 100³ = 1000000

áp dụng hằng đẳng thức thứ 1

a) \(\left(x+2\right)^2\)

Thay x = 98 :

\(\left(98+2\right)^2\)\(=100^2=10000\)

Áp dụng hằng đẳng thức thứ 4

\(\left(x+1\right)^3\)

Thay x = 99

\(\left(99+1\right)^2\)\(=100^2=10000\)

\(a,x^2+4x+4=x^2+2.x.2+2^2=\left(x+2\right)^2.\)

Thay x=98 vào biểu thức : \(\left(98+2\right)^2=100^2=10000\)

\(b,x^3+3x^2+3x+1=x^3+3.x^2.1+3.x.1^2+1^3=\left(x+1\right)^3\)

Thay x=99 vào biểu thức :  \(\left(99+1\right)^3=100^3=1000000\)

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

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

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

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

a: A=2/3x^2y+4x^2y=14/3x^2y

=14/3*9*7=294

b: B=xy^2(1/2+1/3+1/6)=xy^2=3/4*1/4=3/16

c: C=x^3y^3(2+10-20)=-8x^3y^3

=-8*1^3(-1)^3=8

d: D=xy^2(2018+16-2016)

=18xy^2

=18(-2)*1/9=-4

a ) \(x^2+4x+4\)

\(=x^2+2.x.2+2^2\)

\(=\left(x+2\right)^2\)

Khi \(x=98\) , ta có : 

\(\left(98+2\right)^2\)

\(=100^2=10000\)

\(x^3+3x^2+3x+1\)

\(=x^3+3.x^2.1+3.x.1^2+1^3\)

\(=\left(x+1\right)^3\)

Khi \(x=99\) , ta có : 

\(\left(99+1\right)^2\)

\(=100^2=10000\)

phân tích ra thành 7 hằng đẳng thức đáng nhớ ý 

x^2+4x+4=(x+2)^2

thay x =98 vào biểu thức (x+2)^2 ta có:

(98+2)^2=10 000

1, a) 

Ta có:

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

Thay x=99 vào ta có:

\(\left(99+1\right)^2=100^2=10000\)

b) Ta có:

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

Thay x=101 vào ta có:

\(\left(101-1\right)^3=100^3=1000000\)