a, \(x^4-x^2-2=0\Leftrightarrow x^4-2x^2+x^2-2=0\)

\(\Leftrightarrow x^2\left(x^2-2\right)+\left(x^2-2\right)=0\Leftrightarrow\left(x^2+1>0\right)\left(x^2-2\right)=0\Leftrightarrow x=\pm\sqrt{2}\)

b, \(\Leftrightarrow x^2\left(x^2+2x+1\right)=0\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow x=0;x=-1\)

c, \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1>0\right)=0\Leftrightarrow x=1\)

d, \(\Leftrightarrow6x^2-3x-4x+2=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow x=\dfrac{2}{3};x=\dfrac{1}{2}\)

a) 

/ \(x^4+x^2-2=0\)

\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

x 4   +   2 x 3   –   8 x   –   16   =   0     ⇔   ( x 4   +   2 x 3 )   –   ( 8 x   +   16 )   =   0     ⇔   x 3 ( x   +   2 )   –   8 ( x   +   2 )   =   0     ⇔   ( x 3   –   8 ) ( x   +   2 )   =   0

Mà x 0 < 0 nên x 0 = -2 suy ra -3< x 0 < -1

Đáp án cần chọn là: A

d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

Thu gọn:

`P(x)=`\(5x^4 + 3x^2 - 3x^5 + 2x - x^2 - 4 +2x^5\)

`= (-3x^5 + 2x^5) + 5x^4 + (3x^2 - x^2) + 2x - 4`

`= -x^5 + 5x^4 + 2x^2 + 2x - 4`

`Q(x) =`\(x^5 - 4x^4 + 7x - 2 + x^2 - x^3 + 3x^4 - 2x^2\)

`= x^5 + (-4x^4 + 3x^4) - x^3 + (x^2 - 2x^2) + 7x - 2`

`= x^5 - x^4 - x^3 - x^2 + 7x - 2`

`@` Tổng:

`P(x)+Q(x)=`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) + (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 + x^5 - x^4 - x^3 - x^2 + 7x - 2`

`= (-x^5 + x^5) - x^3 + (5x^4 - x^4) + (2x^2 - x^2) + (2x + 7x) + (-4-2)`

`= 4x^4 - x^3 + x^2 + 9x - 6`

`@` Hiệu:

`P(x) - Q(x) =`\((-x^5 + 5x^4 + 2x^2 + 2x - 4) - (x^5 - x^4 - x^3 - x^2 + 7x - 2)\)

`= -x^5 + 5x^4 + 2x^2 + 2x - 4 - x^5 + x^4 + x^3 + x^2 - 7x + 2`

`= (-x^5 - x^5) + (5x^4 + x^4) + x^3 + (2x^2 + x^2) + (2x - 7x) + (-4+2)`

`= -2x^5 + 6x^4 + x^3 + 3x^2 - 5x - 2`

`b)`

`@` Thu gọn:

\(H (x) = ( 3x^5 - 2x^3 + 8x + 9) - ( 3x^5 - x^4 + 1 - x^2 + 7x)\)

`= 3x^5 - 2x^3 + 8x + 9 - 3x^5 + x^4 - 1 + x^2 - 7x`

`= (3x^5 - 3x^5) + x^4 - 2x^3 - x^2 + (8x + 7x) + (9+1)`

`= x^4 - 2x^3 - x^2 + 15x + 10`

\(R( x) = x^4 + 7x^3 - 4 - 4x ( x^2 + 1) + 6x\)

`= x^4 + 7x^3 - 4 - 4x^3 - 4x + 6x`

`= x^4 + (7x^3 - 4x^3) + (-4x + 6x) - 4`

`= x^4 + 3x^3 + 2x - 4`

`@` Tổng:

`H(x)+R(x)=` \((x^4 - 2x^3 - x^2 + 15x + 10)+(x^4 + 3x^3 + 2x - 4)\)

`= x^4 - 2x^3 - x^2 + 15x + 10+x^4 + 3x^3 + 2x - 4`

`= (x^4 + x^4) + (-2x^3 + 3x^3) - x^2 + (15x + 2x) + (10-4)`

`= 2x^4 + x^3 - x^2 + 17x + 6`

`@` Hiệu: 

`H(x) - R(x) =`\((x^4 - 2x^3 - x^2 + 15x + 10)-(x^4 + 3x^3 + 2x - 4)\)

`=x^4 - 2x^3 - x^2 + 15x + 10-x^4 - 3x^3 - 2x + 4`

`= (x^4 - x^4) + (-2x^3 - 3x^3) - x^2 + (15x - 2x) + (10+4)`

`= -5x^3 - x^2 + 13x + 14`

`@` `\text {# Kaizuu lv u.}`

a:

Sửa đề: A=x^4-9x^3+21x^2+x+a

A chia hết cho B

=>x^4-2x^3-7x^3+14x^2+7x^2-14x+15x-30+a+30 chia hết cho x-2

=>a+30=0

=>a=-30

b: A chia hết cho B

=>x^4+2x^3-12x^3-24x^2+45x^2+90x-82x-164+a+164 chia hết cho x+2

=>a+164=0

=>a=-164

a: A=x^5-32

Khi x=3 thì A=3^5-32=243-32=211

b: B=x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+x^7-x^6+x^5-x^4+x^3-x^2+x-1

=x^8-1

=2^8-1=255

a) Ta có: \(x^3+x^2+x+1=0\)

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

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

mà \(x^2+1>0\forall x\)

nên x+1=0

hay x=-1

Vậy: S={-1}

b) Ta có: \(x^3-6x^2+11x-6=0\) 

\(\Leftrightarrow x^3-x^2-5x^2+5x+6x-6=0\)

\(\Leftrightarrow x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)

Vậy: S={1;2;3}

c) Ta có: \(x^3-x^2-21x+45=0\)

\(\Leftrightarrow x^3-3x^2+2x^2-6x-15x+45=0\)

\(\Leftrightarrow x^2\left(x-3\right)+2x\left(x-3\right)-15\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+2x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2+5x-3x-15\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2\cdot\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: S={3;-5}

d) Ta có: \(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6=0\)

\(\Leftrightarrow x^3\left(x-2\right)+4x^2\cdot\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+4x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^3+3x^2+x^2+4x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+3\right)+\left(x+1\right)\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1>0\forall x\)

nên (x-2)(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy: S={2;-3}

1.

a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)

\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)

\(\Leftrightarrow x^3+3x^2+2x-6=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)

b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

1c/

\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)

\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

Vậy pt có nghiệm duy nhất \(x=-1\)

2.

a. \(x^4-x^3+x^2+x^2-x+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

b.

\(x^4+x^3+x^2+x+1=0\)

\(\Leftrightarrow x\left(x^3+1\right)+x^3+1+x^2=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+1\right)+x^2=0\)

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

Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn