2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

a: Ta có: \(x^2+3x+4=0\)

\(\text{Δ}=3^2-4\cdot1\cdot4=9-16=-7< 0\)

Do đó: Phương trình vô nghiệm

Mình ko biết đặt biến phụ nên mình sẽ giải bừa :>

\(x^4+4x^3+6x^2+4x+1=0\)

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

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

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

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

Thấy ngay x= 0 không phải là nghiệm của pt. Chia 2 vế của pt cho x² ta được:

\(x^2+4x+6+4.\frac{1}{x}+\frac{1}{x^2}=0\)

\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+4\left(x+\frac{1}{x}\right)+6=0\left(1\right)\)

Đặt \(x+\frac{1}{x}=t\Rightarrow\left(x+\frac{1}{x}\right)^2=t^2\Rightarrow x^2+\frac{1}{x^2}=t^2-2\) Khi đó ta có:

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

\(\Leftrightarrow t=-2\Leftrightarrow x+\frac{1}{x}=-2\Leftrightarrow x^2+2x+1=0\Leftrightarrow x=-1\)

Vậy pt có 1 nghiệm x = -1

c)   \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)=40\)

\(\Leftrightarrow\)\(\left(x^2+6x+5\right)\left(x^2+6x+8\right)-40=0\)

Đặt      \(x^2+6x+5=t\)   ta có:

                       \(t\left(t+3\right)-40=0\)

          \(\Leftrightarrow\)\(t^2+3t-40=0\)

          \(\Leftrightarrow\)\(\left(t-5\right)\left(t+8\right)=0\)

        \(\Leftrightarrow\)\(\orbr{\begin{cases}t-5=0\\t+8=0\end{cases}}\)

Thay trở lại ta có:      \(\orbr{\begin{cases}x^2+6x=0\\x^2+6x+13=0\end{cases}}\)

(*)     \(x^2+6x=0\)

 \(\Leftrightarrow\)\(x\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x+6=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

(*)   \(x^2+6x+13=0\)

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

Vậy......

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

\(x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\)

Ta có: \(x^2+2x+17=(x^2+2x+1)+16=\left(x+1\right)^2+16\ge16\)

\(\Rightarrow\sqrt{x^2+2x+17}\ge\sqrt{16}=4\)

\(\Rightarrow x^4+4x^3+6x^2+4x+\sqrt{x^2+2x+17}=3\ge x^4+4x^3+6x^2+4x+4\)

\(\Leftrightarrow x^4+4x^3+6x^2+4x+1\le0\)

\(\Leftrightarrow\left(x+1\right)^4\le0\)

Mà \(\left(x+1\right)^4\ge0\Rightarrow(x+1)^4=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Thử lại ta thấy x=-1 thỏa mãn bài toán

Vậy, pt có nghiệm duy nhất là x=-1

\(4x^4+4x^3+x^2+3x\ge0\)

\(4x^4+4x^2+1-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(\Leftrightarrow\left(2x^2+1\right)^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)\sqrt{\left(x^2-x+1\right)\left(2x^2+1\right)+2x^4+6x^3-2x^3+4x-1}\)

\(2x^2+1=u;\sqrt{4x^4+4x^3+x^2+3x}=v\left(u>0;v>0\right)\)

\(\hept{\begin{cases}u^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)v\\v^2-\left(2x^4+6x^3-2x^2+4x-1\right)=\left(x^2-x+1\right)u\end{cases}\Rightarrow u^2-v^2=\left(x^2-x+1\right)\left(v-u\right)\Leftrightarrow\orbr{\begin{cases}u=v\\u+v+x^2-x+1=0\end{cases}}}\)

Chọn đáp án D

c) (x+1)(x+2)(x+4)(x+5)=40

<=> (x+1)(x+5)(x+2)(x+4)=40

<=>(x^2+6x+5)(x^2+6x+8)=40

Đặt x^2+6x+5=y

=>y(y+3)=40

=>y^2+3y=40<=>y^2+2.\(\frac{3}{2}\)y+\(\frac{9}{4}\)=40+\(\frac{9}{4}\)<=> (y+\(\frac{3}{2}\))²=42,25<=> y+\(\frac{3}{2}\)=6,5 hoặc -6,5

Bạn tự làm tiếp nha :333

a)x⁴ - 4x³ - 19x² +106x - 120 = 0

=>x⁴ -2x³ -2x³+4x² -23x² +46x +60x - 120 = 0

=>x³(x-2) -2x²(x-2) -23x(x-2) +60(x-2)= 0

=>(x³- 2x² -23x+ 60)(x-2) =0

=>(x³ - 3x² +x² -3x -20x+60)(x -2) = 0

=>(x² +x -20)(x-3)(x-2) = 0

=>(x² -4x +5x -20)(x-3)(x-2) = 0

=>(x+5)(x-4)(x-3)(x-2) =0

=>x= -5; 4; 3; 2

b)=>4x⁴ -4x³ +16x³ -16x² +21x² -21x +15x -15= 0

=>(x-1)(4x³ +16x² +21x+15)= 0

=>...bạn tự làm phần tiếp theo nhé

c)Làm giống nguyễn thị ngọc linh