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 \(S=\left\{-3;-5\right\}\)
\(\text{b) }6x^4+25x^3+12x^2-25x+6=0\\ \Leftrightarrow x^2\left(6x^2+25x+12-\dfrac{25}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left(6x^2-12+\dfrac{6}{x^2}+25x-\dfrac{25}{x}+24\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2-12+\dfrac{6}{x^2}\right)+\left(25x-\dfrac{25}{x}\right)+24\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2-2+\dfrac{1}{x^2}\right)+25\left(x-\dfrac{1}{x}\right)+24\right]=0\\ \Leftrightarrow x^2\left[6\left(x-\dfrac{1}{x}\right)^2+25\left(x-\dfrac{1}{x}\right)+24\right]=0\)
Đặt \(x-\dfrac{1}{x}=t\)
\(\Leftrightarrow x^2\left(6t^2+25t+24\right)=0\\ \Leftrightarrow x^2\left(6t^2+9t+16t+24\right)=0\\ \Leftrightarrow x^2\left[3t\left(2t+3\right)+8\left(2t+3\right)\right]=0\\ \Leftrightarrow x^2\left(3t+8\right)\left(2t+3\right)=0\\ \Leftrightarrow x^2\left(3x-\dfrac{3}{x}+8\right)\left(2x-\dfrac{2}{x}+3\right)=0\\ \Leftrightarrow\left(3x^2-3+8x\right)\left(2x^2-2+3x\right)=0\\ \Leftrightarrow\left(3x^2+9x-x-3\right)\left(2x^2+4x-x-2\right)=0\\ \Leftrightarrow\left[3x\left(x+3\right)-\left(x+3\right)\right]\left[2x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x+3\right)\left(2x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+3=0\\2x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=-3\\2x=1\\x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-3\\x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{-\dfrac{1}{3};-3;\dfrac{1}{2};-2\right\}\)
