giúp tôi với

1) 2x⁴ - 9x³ + 14x² - 9x + 2 = 0

<=> (2x⁴ - 4x³) - (5x³ - 10x²) + (4x² - 8x) - (x - 2) = 0

<=> 2x³(x - 2) - 5x²(x - 2) + 4x(x - 2) - (x - 2) = 0

<=> (2x³ - 5x² + 4x - 1)(x - 2) = 0

<=> [(2x³ - 2x²) - (3x² - 3x) + (x - 1)](x - 2) = 0

<=> [2x²(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0

<=> (2x² - 2x - x + 1)(x - 1)(x - 2) = 0

<=> (2x - 1)(x - 1)²(x - 2) = 0

<=> 2x - 1=0

hoặc x - 1 = 0

hoặc x - 2 = 0

<=> x = 1/2

hoặc x = 1

hoặc x = 2

Vậy S = {1/2; 1; 2}

1) \(2x^4-9x^3+14x^2-9x+2=0\)

\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)

\(\Leftrightarrow2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[2\left(x^3-1\right)-7x\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[2\left(x-1\right)\left(x^2+x+1\right)-7x\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(2x^2+2x+2-7x\right)=0\)

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

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

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

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

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

hoặc \(2x-1=0\)

hoặc \(x-2=0\)

\(\Leftrightarrow\)\(x=1\)hoặc \(x=\frac{1}{2}\)hoặc \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{1;\frac{1}{2};2\right\}\)

2) \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4-3x^3+28x^3-14x^2+26x^2-13x-12x+6=0\)

\(\Leftrightarrow3x^3\left(2x-1\right)+14x^2\left(2x-1\right)+13x\left(2x-1\right)-6\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3+14x^2+13x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x^3-x^2+15^2-5x+18x-6\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left[x^2\left(3x-1\right)+5x\left(3x-1\right)+6\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-1\right)\left(x^2+5x+6\right)=0\)

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

\(\Leftrightarrow\)\(2x-1=0\)

hoặc \(3x-1=0\)

hoặc \(x+2=0\)

hoặc \(x+3=0\)

\(\Leftrightarrow\)\(x=\frac{1}{2}\)hoặc \(x=\frac{1}{3}\)hoặc \(x=-2\)hoặc \(x=-3\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{2};\frac{1}{3};-2;-3\right\}\)

3) Ktra lại đề nhé :D

4) \(x^3-3x^2+3x+8=0\)

\(\Leftrightarrow2x^3+2x^2-5x^2-5x+8x+8=0\)

\(\Leftrightarrow2x^2\left(x+1\right)-5x\left(x+1\right)+8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x^2-5x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x^2-5x+8=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(TM\right)\\2\left(x-\frac{5}{4}\right)^2+\frac{39}{8}=0\left(L\right)\end{cases}}\)

Vậy x = -1

5) \(x^4+2x^3+x^2=0\)

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

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

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

Vậy tập nghiệm của phương trình là \(S=\left\{0;-1\right\}\)

a) Gần giống cho nó giống luôn.

cần thêm (-x^3+2x^2-x) là giống

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

\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)

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

Nghiệm duy nhất: x=1

câu d

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

\(6\left(x^2+\frac{1}{x^2}\right)+25\left(x-\frac{1}{x}\right)+12=0\)

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

\(\Rightarrow6\left(t^2+2\right)+25t+12=0\)

\(\Leftrightarrow6t^2+25t+24=0\Rightarrow\left[{}\begin{matrix}t=-\frac{3}{2}\\t=-\frac{8}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{x}=-\frac{3}{2}\\x-\frac{1}{x}=-\frac{8}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-2=0\\3x^2+8x-3=0\end{matrix}\right.\)

Mình giải cho bạn rồi, bạn vào xem lại lời giải nhé:

http://olm.vn/hoi-dap/question/430226.html

a: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)

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

hay \(x\in\left\{-2;-3;\dfrac{1}{3};\dfrac{1}{2}\right\}\)

b: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)

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

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

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

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

=>x+1=0

hay x=-1

c: \(x^2\left(x^2+2\right)-x^2-2=0\)

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

=>x=1 hoặc x=-1

Ta có: \(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(6x^3-3x^2+16x^2-8x-6x+3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[3x^2\left(2x-1\right)+8x\left(2x-1\right)-3\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+8x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(3x^2+9x-x-3\right)=0\)

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

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

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

Vậy: \(S=\left\{-2;\dfrac{1}{2};-3;\dfrac{1}{3}\right\}\)

\(6x^4+25x^3+12x^2-25x+6=0\)

\(\Leftrightarrow\)  \(6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)

\(\Leftrightarrow\)  \(6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left[6x^2\left(x+3\right)-5x\left(x+3\right)+x+3\right]=0\)

\(\Leftrightarrow\)  \(\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)

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

\(\Leftrightarrow\)   \(x+2=0\)  hoặc  \(x+3=0\)  hoặc  \(2x-1=0\)  hoặc  \(3x-1=0\)

\(\Leftrightarrow\)   \(x=-2\)  hoặc \(x=-3\)  hoặc  \(x=\frac{1}{2}\)  hoặc  \(x=\frac{1}{3}\)

Vậy, tập nghiệm của pt là  \(S=\left\{-2;-3;\frac{1}{2};\frac{1}{3}\right\}\)