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

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

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

x²  - 1 + 3(6x + 6)=0

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

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

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

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

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

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

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

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

\(\Leftrightarrow\)   \(x+2=0\)  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  \(S=\left\{-2;-3;\frac{1}{2};\frac{1}{3}\right\}\)

đây là pt đỗi xứng bậc chẵn bạn ơi
cos cachs giải đó bạn

(+) Kiểm tra x = 0 , sau đó chia cả hai vế cho x^2

(+) đặt x- 1/x = a => x^2 + 1/x^2 = a^2 + 2 

Thay vô giải pt bậc hai 

à ha , tự nhiên tui quên mất ^^! , thks nhá , k cần giải nữa đâu , bik giải gòi 

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

Bạn dùng phương pháp phân tích đa thức thành nhân tử sẽ đc :

(2x-1).(x+3).(x+2).(3x-1) = 0

<=> x=1/2 hoặc x=-3 hoặc x=-2 hoặc x=1/3

Vậy .............

Tk mk nha

giải ra

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

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

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

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

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

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

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

P/S:lm tiếp nha   x = -2;  x = -3;  x = 1/2;  x = 1/3

\(b.6x^4+25x^3+12x^2-25x+6=0\\\Leftrightarrow 6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\\\Leftrightarrow 6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\\\Leftrightarrow \left(6x^3+13x^2-14x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^3+18x^2-5x^2-15x+x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)\right]\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-5x+1\right)\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-3x-2x+1\right)\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[3x\left(2x-1\right)-\left(2x-1\right)\right]\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(2x-1\right)\left(x+3\right)\left(x+2\right)=0\)

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

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

\(2x^4-9x^3+14x^2-9x+2=0\\\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\\\Leftrightarrow 2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\\\Leftrightarrow \left(2x^3-7x^2+7x-2\right)\left(x-1\right)=0\\\Leftrightarrow \left[2\left(x^3-1\right)-7x\left(x-1\right)\right]\left(x-1\right)=0\\\Leftrightarrow \left(x-1\right)^2\left[2\left(x^2+x+1\right)-7x\right]=0\\\Leftrightarrow \left(2x^2+2x+2-7x\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-5x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-x-4x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x-1\right)^2=0\\\Leftrightarrow \left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)

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

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

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 \(S=\left\{0;-1\right\}\)

b. sửa đề

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

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

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

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

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

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

Vậy........

Bài 1 : Giải phương trình

a) (x + 3)⁴ + (x + 5)⁴ = 16

Đặt : x + 3 = t

=> x + 5 = x + 3 + 2 = t + 2

Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :

t⁴ + (t + 2)⁴ = 16

<=> 2t⁴ + 8t³ + 24t² + 32t + 16 = 16

<=> 2(t⁴ + 4t³ + 12t² + 16t) = 0

<=> t⁴ + 4t³ + 12t² + 16t = 0

<=> (t + 2) . t . (t² + 2y + 4) = 0

TH1 : t = 0

TH2 : t + 2 = 0 <=> t = -2

TH3 : t² + 2y + 4 = 0 (vô nghiệm => loại)

Nên t = 0 hoặc t = -2

hay x + 3 = -2 hoặc x + 3 = 0

<=> x = -5 hoặc x = -3

\(S=\left\{-5;-3\right\}\)

b) 6x⁴ + 25x³ + 12x² - 25x + 6 = 0

<=> 6x⁴ + 12x³ + 13x³ + 26x² - 14x² - 28x + 3x + 6 = 0

<=> 6x³ (x + 2) + 13x² (x + 2) - 14x (x + 2) + 3(x + 2) = 0

<=> (x + 2)(6x³ + 13x² - 14x + 3) = 0

<=> (x + 2)(6x³ + 18x² - 5x² - 15x + x + 3) = 0

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

<=> (x + 2)(x + 3) (6x² - 5x + 1) = 0

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

TH1 : x + 2 = 0 <=> x = -2

TH2 : x + 3 = 0 <=> x = -3

TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)

TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)

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

\(\text{a) }\left(x+3\right)^4+\left(x+5\right)^4=16\\ \Leftrightarrow\left(x^2+6x+9\right)^2+\left(x^2+10x+25\right)^2=16\\ \Leftrightarrow x^4+36x^2+81+12x^3+18x^2+108x+x^4+100x^2+625+20x^3+50x^2+500x=16\\ \Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\\ \Leftrightarrow x^4+16x^3+102x^2+304x+345=0\\ \Leftrightarrow x^4+5x^3+11x^3+55x^2+47x^2+235x+373x+69x+345=0\\ \Leftrightarrow\left(x^4+5x^3\right)+\left(11x^3+55x^2\right)+\left(47x^2+235x\right)+\left(69x+345\right)=0\\ \Leftrightarrow x^3\left(x+5\right)+11x^2\left(x+5\right)+47x\left(x+5\right)+69\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+11x^2+47x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x^2+24x+23x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x^3+3x^2\right)+\left(8x^2+24x\right)+\left(23x+69\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left[x^2\left(x+3\right)+8x\left(x+3\right)+23\left(x+3\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left(x^2+8x+23\right)\left(x+3\right)\left(x+5\right)=0\)\(\Leftrightarrow\left(x^2+8x+16+7\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x+4\right)^2+7\right]\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+5\right)=0\left(\text{Vì }\left(x+4\right)^2+7\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy tập nghiệm phương trình là \(S=\left\{-3;-5\right\}\)