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Cái này t dùng máy tính
\(\left(x-2\right)\left(x+3\right)\left(2x+1\right)\left(3x-1\right)=0\)
Đến đây thì pt có 4 nghiệm:\(x=2;-3;-\frac{1}{2};\frac{1}{3}\)
Vậy....
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
a, \(x^4-6x^3+11x^2-6x+1=0\)
=> \(x^4-6x^3+9x^2+2x^2-6x+1=0\)
=> \(x^2+3x+1=0\)
=> \(\Delta\) =\(b^2-4c\)
=\(3^2.4=5\)
Nên \(\sqrt{\Delta}=5\)
x= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{5}}{2}\)
hoặc x= \(\dfrac{b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{5}}{2}\)
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
\(6x^4+5x^3-38x^2+5x+6=0\)
\(6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6=0\)
\(6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\left(x-2\right)\left(6x^3+17x^2-4x-3\right)=0\)
\(\left(x-2\right)\left[6x^3-3x^2+20x^2-10x+6x-3\right]=0\)
\(\left(x-2\right)\left[6x^2\left(x-\dfrac{1}{2}\right)+20x\left(x-\dfrac{1}{2}\right)+6\left(x-\dfrac{1}{2}\right)\right]=0\)
\(\left(x-2\right)\left(x-\dfrac{1}{2}\right)\left(6x^2+20x+6\right)=0\)
=> \(\left[{}\begin{matrix}x-2=0\\x-\dfrac{1}{2}=0\\6x^2+20x+6=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\\left(3x+1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\\x=-3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
Ta có : \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x-2\right)-24=0\)
Đặt t = x2 + 5x - 1
Khi đó : (x2 + 5x) = t + 1 ; (x2 + 5x - 2) = t - 1
Ta có : C = (x2 + 5x - 2)2 (x2 + 5x - 2) - 24 = 0
=> (x2 + 5x - 2)3 = 24
MK chỉ giả được đến đây thôi
Đang rảnh, buồn ngủ nên giải cho tỉnh táo :D
Ta nhận thấy x=0 không phải là nghiệm của phương trình, vậy ta chia cả 2 vế của phương trình cho x2 khác 0, ta được:
\(6x^2+5x-38+\dfrac{5}{x}+\dfrac{6}{x^2}=0\)
\(\Leftrightarrow6\left(x^2+\dfrac{1}{x}\right)+5\left(x+\dfrac{1}{x}\right)-38=0\)
Đặt \(x+\dfrac{1}{x}=y\Rightarrow x^2+\dfrac{1}{x^2}=y^2-2\)
Ta được: \(6\left(y^2-2\right)+5y-38=0\)
Do đó: y1=2,5;y2=-10/3
Với y=2,5\(\Rightarrow x+\dfrac{1}{x}=2,5\Rightarrow x_1=2;x_2=0,5\)
Với y=-10/3
\(\Rightarrow x+\dfrac{1}{x}=-\dfrac{10}{3}\Rightarrow x_3=-\dfrac{1}{3};x_4=-3\)
Vậy: \(S=\left\{2;0,5;-\dfrac{1}{3};-3\right\}\)
Bài a tự giải
Bài b thì biến đổi xong rồi đặt ẩn phụ \(y=x+\dfrac{1}{x}\)
Bài c:
Đặt x-1=y
Phương trình trở thành: \(\left(y+2\right)^4+\left(y-2\right)^4=82\)
Rút gọn ta được: \(2y^4+48y^2-50=0\)
Đặt y2=z ( \(z\ge0\) )
Phương trình này cho z1=1, z2=-25(Loại)
\(z=1\Rightarrow y^2=1\Rightarrow y=\pm1\)
\(\Rightarrow x_1=2;x_2=0\)
X=-2,3,1/3
\(6x^4-5x^3-38x^2-5x+6=0\)
\(\Leftrightarrow6x^4-12x^3+17x^3-34^2-4x^2+8x-3x+6=0\)
\(\Leftrightarrow6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-4x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-x^2-3x-x-3=0\right)\)
\(\Leftrightarrow\left(x-2\right)\left[6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left[6x\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{2}\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-\frac{1}{2}\right)\left(6x+2\right)=0\)