`#Sad`

`\text{·Bài}` `9:`

`d)`

`Q= x(x-y)^2-y(x-y)^2+xy^2-x^2y`

`= [x(x-y)^2-y(x-y)^2]+xy(y-x)`

`= (x-y)^3-xy(x-y)`

`\text{→Thay}` `x-y=7; x.y=9` `\text{vào ta có:}`

`= 7^3-9. 7`

`= 343-63`

`= 280`

`\text{·Bài}` `10:`

`a)`

`2-x = 2(x-2)^3`

`⇔ 2-x-2(x-2)^3 = 0`

`⇔ (-x+2)-2(x-2)^3 = 0`

`⇔ (x-2)-2(x-2)^3 = 0`

`⇔ (x-2)[1-2(x-2)^2] = 0`

`\text{→Vì}` `(x-2)^2 >= 0`

`⇒` `1-2(x-2)^2 >= 1`

`\text{→Mà:}` `(x-2)^2 = 1/2` `\text{(loại)}`

`⇔ x-2 = 0`

`⇔ x = 2`

`\text{Vậy S=}` `{2}`

`b)`

`8x^3-72x = 0`

`⇔ 8x(x^2-9) = 0`

`⇔ 8x(x-3)(x+3) = 0`

`⇔` \(\left[ \begin{array}{l}8x=0\\x-3=0\\x+3=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=0\\x=3\\x=-3\end{array} \right.\) 

`\text{Vậy S=}` `{0; 3; -3}`

`c)`

`(x-1,5)^6+2(1,5-x)^2 = 0`

`⇔ (x-1,5)^6+2(x-1,5)^2 = 0`

`⇔ (x-1,5)^2[(x-1,5)^4+2] = 0`

`\text{→Vì:}` `(x-1,5)^4 >= 0`

`\text{→Mà:}` `(x-1,5)^4 = -2` `\text{(loại)}`

`⇔ (x-1,5)^2 = 0`

`⇔ x-1,5 = 0`

`⇔ x = 1,5`

`\text{Vậy S=}` `{1,5}`

`d)`

`2x^3+3x^2+3+2x = 0`

`⇔ (2x^3+2x)+(3x^2+3) = 0`

`⇔ 2x(x^2+1)+3(x^2+1) = 0`

`⇔ (x^2+1)(2x+3) = 0`

`\text{→Vì:}` `x^2 >= 0`

`⇒ x^2+1 >= 1`

`\text{→Mà:}` `x^2 = -1` `\text{(loại)}`

`⇔ 2x+3 = 0`

`⇔ 2x = -3`

`⇔ x = -3/2`

`\text{Vậy S=}` `{-3/2}`

`e)`

`x^2(x+1)-x(x+1)+x(x-1) = 0`

`⇔ [x^2(x+1)-x(x+1)]+x(x-1) = 0`

`⇔ (x^2-x)(x+1)+x(x-1) = 0`

`⇔ x(x-1)(x+1)+x(x-1) = 0`

`⇔ x(x-1)(x+1+1) = 0`

`⇔ x(x-1)(x+2) = 0`

`⇔` \(\left[ \begin{array}{l}x=0\\x-1=0\\x+2=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=0\\x=1\\x=-2\end{array} \right.\) 

`\text{Vậy S=}` `{0; 1; -2}`

`f)`

`x^3-4x-14x(x-2) = 0`

`⇔ (x^3-4x)-14x(x-2) = 0`

`⇔ x(x^2-4)-14x(x-2) = 0`

`⇔ x(x-2)(x+2)-14x(x-2) = 0`

`⇔ x(x-2)(x+2-14) = 0`

`⇔ x(x-2)(x-12) = 0`

`⇔` \(\left[ \begin{array}{l}x=0\\x-2=0\\x-12=0\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x=0\\x=2\\x=12\end{array} \right.\) 

`\text{Vậy S=}` `{0; 2; 12}`

Đáp án:

$\begin{array}{l}
9)\\
d)\\
Q = x{\left( {x - y} \right)^2} - y{\left( {x - y} \right)^2} + x{y^2} - {x^2}y\\
 = x{\left( {x - y} \right)^2} - y{\left( {x - y} \right)^2} + xy\left( {y - x} \right)\\
 = {\left( {x - y} \right)^2}.\left( {x - y} \right) - xy.\left( {x - y} \right)\\
 = {\left( {x - y} \right)^3} - xy\left( {x - y} \right)\\
 = {7^3} - 9.7\\
 = 280\\
B10)\\
a)2 - x = 2{\left( {x - 2} \right)^3}\\
 \Leftrightarrow 2{\left( {x - 2} \right)^3} + \left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right).\left( {2{{\left( {x - 2} \right)}^2} + 1} \right) = 0\\
 \Leftrightarrow x - 2 = 0\\
 \Leftrightarrow x = 2\\
Vậy\,x = 2\\
b)8{x^3} - 72x = 0\\
 \Leftrightarrow 8x\left( {{x^2} - 9} \right) = 0\\
 \Leftrightarrow 8x\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
 \Leftrightarrow x = 0;x = 3;x =  - 3\\
Vậy\,x = 0;x = 3;x =  - 3\\
c){\left( {x - 1,5} \right)^6} + 2.{\left( {1,5 - x} \right)^2} = 0\\
 \Leftrightarrow {\left( {x - 1,5} \right)^2}.\left[ {{{\left( {x - 1,5} \right)}^4} + 2} \right] = 0\\
 \Leftrightarrow x - 1,5 = 0\\
 \Leftrightarrow x = 1,5\\
Vậy\,x = 1,5\\
d)2{x^3} + 3{x^2} + 3 + 2x = 0\\
 \Leftrightarrow \left( {2x + 3} \right).{x^2} + 2x + 3 = 0\\
 \Leftrightarrow \left( {2x + 3} \right)\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow 2x + 3 = 0\\
 \Leftrightarrow x =  - \dfrac{3}{2}\\
Vậy\,x = \dfrac{{ - 3}}{2}\\
e){x^2}\left( {x + 1} \right) - x\left( {x + 1} \right) + x\left( {x - 1} \right) = 0\\
 \Leftrightarrow x\left( {x + 1} \right)\left( {x - 1} \right) + x\left( {x - 1} \right) = 0\\
 \Leftrightarrow x.\left( {x - 1} \right).\left( {x + 1 + 1} \right) = 0\\
 \Leftrightarrow x\left( {x - 1} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow x = 0;x = 1;x =  - 2\\
Vậy\,x = 0;x = 1;x =  - 2\\
f){x^3} - 4x - 14x\left( {x - 2} \right) = 0\\
 \Leftrightarrow x.\left( {{x^2} - 4 - 14\left( {x - 2} \right)} \right) = 0\\
 \Leftrightarrow x\left( {x - 2} \right)\left( {x + 2 - 14} \right) = 0\\
 \Leftrightarrow x\left( {x - 2} \right)\left( {x - 12} \right) = 0\\
 \Leftrightarrow x = 0;x = 2;x = 12\\
Vậy\,x = 0;x = 2;x = 12
\end{array}$