\(\frac{1}{a}\)<\(\frac{1}{b}\)

Ta có P = xyz - xy - yz - zx + x + y + z - 1 

= (xyz - xy) - z(x + y) + (x + y) + (z - 1)

= xy(z - 1) - (x + y)(z - 1) + (z - 1) 

= (z - 1)(xy - x - y + 1) 

= (z - 1)[x(y - 1) - (y - 1)]

= (x - 1)(y - 1)(z - 1) 

`Answer:`

`xyz-(xy+yz+zx)+(x+y+z)-1`

`=xyz-xy-yz-zx+x+y+z-1`

`=(xyz-xy)-(yz-y)-(zx-x)+(z-1)`

`=xy(z-1)-y(z-1)-x(z-1)+(z-1)`

`=(z-1)(xy-y-x+1)`

`=(z-1)[(xy-y)-(x-1)]`

`=(z-1)[(y(x-1)-(x-1)]`

`=(z-1)(x-1)(y-1)`

`=(x-1)(y-1)(z-1)`

\(A = n 3 + ( n + 1 ) 3 + ( n + 2 ) 3\)

\(= n 3 + n 3 + 3 n 2 + 3 n + 1 + n 3 + 6 n 2 + 12 n + 8\)

\(= 3 n 3 + 9 n 2 + 15 n + 9\)

\(= 3 n 2 ( n + 1 ) + 6 n ( n + 1 ) + 9 ( n + 1 )\)

\(= 3 ( n + 1 ) ( n 2 + 2 n + 3 )\)

\(= 3 ( n + 1 ) [ n ( n + 2 ) + 3 ]= 3 ( n + 1 ) [ n ( n + 2 ) + 3 ]\)

\(= 3 n ( n + 1 ) ( n + 2 ) + 9 ( n + 1 )\)

Do \(n , n + 1 , n + 2 \) là 3 số tự nhiên liên tiếp

\(⇒ 3 n ( n + 1 ) ( n + 2 ) ⋮ 9\)

\(⇒ A = 3 n ( n + 1 ) ( n + 2 ) + 9 ( n + 1 ) ⋮ 9 ( đ p c m )\)

P/s : Bài này bạn có thể sử dụng phương pháp quy nạp

làm như vậy sẽ nhanh hơn

tíc cho tui 

Xét hằng đẳng thức sau:
x^3 + y^3 + z^3 - 3xyz
= (x + y)^3 - 3xy(x + y) + z^3 - 3xyz
= [(x + y)^3 + z^3] - 3xy(x + y + z)
= (x + y + z)[(x + y)^2 - z(x + y) + z^2) - 3xy(x + y + z)
= (x + y + z)(x^2 + y^2 + z^2 + 2xy - xz - yz) - 3xy(x + y + z)
= (x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz)
---> x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - xz) + 3xyz

Áp dụng hằng đẳng thức trên, ta có:
n^3 + (n + 1)^3 + (n + 2)^3
= (n + n + 1 + n + 2)[ n^2 + (n + 1)^2 + (n + 2)^2 -n(n + 1) - (n + 1)(n + 2) - n(n + 2)] - 3n(n + 1)(n + 2)
= (3n + 3)(n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 - n^2 - n - n^2 - 3n - 2 - n^2 - 2n) - 3n(n + 1)(n + 2)
= 9(n + 1) - 3n(n + 1)(n + 2)
Vì n(n + 1)(n + 2) là tích 3 số nguyên liên tiếp nên chia hết 6
--> 3n(n + 1)(n + 2) chia hết 3.6 = 18 chia hết 9
--> 9(n + 1) - 3n(n + 1)(n + 2) chia hết 9
--> n^3 + (n + 1)^3 + (n + 2)^3 chia hết cho 9

\(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy.\left(x+y\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy.\left(x+y+z\right)\)

\(=\left(x+y+z\right).\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy.\left(x+y+z\right)\)

\(=\left(x+y+z.\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)

\(=\left(x+y+z\right).\left(x^2+y^2+z^2-xy-yz-zx\right)\)

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= ( x^2+7x +10 )(x^2+7x+12)-24

= (x^2+7x)^2+22(x^2+7x)+96

=(x^2+7x+6)(x^2+7x+16)

=(x+1)(x+6)(x^2+7x+16)

k nha nếu sai k sai 

ht

tui hơm ít

\(\left(x^2+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1;1\\x=-1;1\end{cases}}\)

\(\Rightarrow S=\left\{-1;1\right\}\)

Ta có :(x²+1)(x-1)=0

x²+1=0 hoặc x-1=0

x²=-1               x=-1

\(\Rightarrow x\in\varnothing\)

Vây x=-1

_HT_

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