Ta có : \(M=\frac{8x^6-27}{4x^4+6x^2+9}=\frac{\left(2x^2\right)^3-3^3}{\left(2x^2\right)^2+\left(2x^2\right).3+3^2}\)

\(=\frac{\left(2x^2-3\right)\left[\left(2x^2\right)^2+2x^2.3+3^2\right]}{\left(2x^2\right)^2+2x^2.3+3^2}=2x^2-3\)

\(N=\frac{y^4-1}{y^3+y^2+y+1}=\frac{\left(y-1\right)\left(y^3+y^2+y+1\right)}{y^3+y^2+y+1}=y-1\)

Vậy \(\frac{M}{N}=\frac{2x^3-3}{y-1}\)

Khi \(x=8,y=251\) , ta có :

\(\frac{M}{N}=\frac{2.8^3-3}{251-1}=\frac{1}{2}\)

M = (8x⁶ - 27) : (4x⁴ + 6x² + 9)

= (2x² - 3)(4x⁴ + 6x² + 9) : (4x⁴ + 6x² + 9)

= 2x² - 3

N = (y⁴ - 1) : (y³ + y² + y + 1)

= (y² + 1)(y - 1)(y + 1) : [(y + 1)(y² + 1)]

= y - 1

\(\frac{M}{N}=\frac{2x^2-3}{y-1}=\frac{2\cdot8^2-3}{251-1}=\frac{125}{250}=\frac{1}{2}\)

2 có nghĩa là câu 2 nhé

\(1,H=a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab[\left(a+b\right)^2-2ab]+6a^2b^2\left(a+b\right)\)

\(=\left(a+b\right)[\left(a+b\right)^2-3ab]+3ab[\left(a+b\right)^2-2ab]+6a^2b^2\left(a+b\right)\)

\(=1-ab+3ab\left(1-2ab+6a^2b^2\right)\)

\(=1-3ab+3ab-6a^2b^2+6a^2b^2\)

\(=1\)

6) c) x³ - x² + x = 1

<=> x³ - x² + x - 1 = 0

<=> (x³ - x²) + (x - 1) = 0

<=> x² (x - 1) + (x - 1) = 0

<=> (x - 1) (x² + 1) = 0

=> x - 1 = 0 hoặc x² + 1 = 0

* x - 1 = 0 => x = 1

* x² + 1 = 0 => x² = -1 => x = -1

Vậy x = 1 hoặc x = -1

Bài 5: 

a) Đặt   \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow8A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{8}\)

b) (7x+6)² + (5-6x)² - (10-12x)(7x+6)

=(7x+6)² + (5-6x)² - 2(5-6x)(7x+6)

\(=\left(7x+6-5+6x\right)^2\)

\(=\left(13x+1\right)^2\)

\(2\left(a^2+b^2+ab-b-ac\right)+\left(c^2+1\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2ab-2b-2ac+c^2+1=0\)

\(\Leftrightarrow\left(a^2+2ab+b^2\right)+\left(a^2-2ab+c^2\right)+\left(b^2-2b+1\right)=0\)

\(\Leftrightarrow\left(a+b\right)^2+\left(a-c\right)^2+\left(b-1\right)^2=0\)

Ta có: \(VT\ge0\Rightarrow\hept{\begin{cases}a+b=0\\a-c=0\\b-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=-1\\b=1\\c=-1\end{cases}}\)

`a, = 3x^2y - 3xy + 6x^2y + 5xy - 9x^2y`

`= 2xy`.

Thay `x = 2/3; y = -3/4` vào BT:

`2 . 2/3 . -3/4 = -1.`

`b, x(x-2y) - y(y^2-2x)`

`= x^2 - 2xy - y^3 + 2xy`

`= x^2 - y^3`

Thay `x = 5; y =3` vào BT:

`= 5^2 - 3^3 = 25 - 27 = -2`

a) \(3x^2y-\left(3xy-6x^2y\right)+\left(5xy-9x^2y\right)\)

\(=3x^2y-3xy+6x^2y+5xy-9x^2y\)

\(=2xy\)

Thay \(x=\dfrac{2}{3},y=-\dfrac{3}{4}\) vào Bt ta có:

\(2\cdot\dfrac{2}{3}\cdot-\dfrac{3}{4}=-1\)

b) \(x\left(x-2y\right)-y\left(y^2-2x\right)\)

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

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

Thay \(x=5,y=3\) vào Bt ta có:
\(5^2-3^3=-3\)