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NH
1
NT
28 tháng 6 2019
Đây là phương trình đối xứng, cách giải những bài phương trình đối xứng khác cũng giống vậy nhé!
Xét x = 0 không phải là nghiệm của phương trình
Chia cả hai vế của phương trình cho x2, ta được:
\(2x^2-21x+74-\frac{105}{x}+\frac{50}{x^2}=0\\ \Rightarrow\left(2x^2+\frac{50}{x^2}\right)-\left(21x+\frac{105}{x}\right)+74=0\\ \Rightarrow2\left(x^2+\frac{25}{x^2}\right)-21\left(x+\frac{5}{x}\right)+74=0\)
Đặt \(x+\frac{5}{x}=y\Rightarrow x^2+\frac{25}{x^2}=y^2-10\)
Thay vào phương trình, ta được:
\(2\left(y^2-10\right)-21y+74=0\\ \Rightarrow2y^2-20-21y+74=0\\ \Rightarrow2y^2-21y+54=0\\ \Rightarrow\left(2y^2-12y\right)-\left(9y-54\right)=0\\ \Rightarrow2y\left(y-6\right)-9\left(y-6\right)=0\\ \Rightarrow\left(y-6\right)\left(2y-9\right)=0\\ \Rightarrow\left(x+\frac{5}{x}-6\right)\left(2x+\frac{10}{x}-9\right)=0\\ \Rightarrow x=1;x=2\)
T
28 tháng 6 2019
Nhận thấy x = 0 không phải là nghiệm.
Xét x khác 0.Chia hai vế của pt cho x2 ta được:
\(x^2-3x-6+\frac{3}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-6=0\)
Đặt \(x-\frac{1}{x}=a\). PT trở thành:
\(a^2-3a-4=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-1\end{matrix}\right.\)
Với a = 4 thì \(x=4+\frac{1}{x}=\frac{4x+1}{x}\Leftrightarrow x^2-4x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{5}\\x=2-\sqrt{5}\end{matrix}\right.\) (nghiệm xấu chút nhưng dễ giải lắm ạ)
Với a = -1 thì \(x=\frac{1}{x}-1=\frac{1-x}{x}\Leftrightarrow x^2+x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1+\sqrt{5}}{2}\\x=\frac{-1-\sqrt{5}}{2}\end{matrix}\right.\) (cái này thì max xấu rồi ;( )
T
28 tháng 6 2019
\(PT\Leftrightarrow\left(x^4-x^3\right)-\left(6x^3-6x^2\right)+\left(12x^2-12x\right)-\left(9x-9\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)-6x^2\left(x-1\right)+12x\left(x-1\right)-9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)-\left(3x^2-9x\right)+\left(3x-9\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)-3x\left(x-3\right)+3\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2-3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\) (do \(x^2-3x+3>0\forall x\))
Vậy..
30 tháng 6 2019
Xét thấy x = 0 không thỏa mãn pt
Ta có : \(6x^4+7x^3-36x^2+7x+6=0\)
\(\Leftrightarrow x^2\left(6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}\right)=0\)
\(\Leftrightarrow6x^2+7x-36+\frac{7}{x}+\frac{6}{x^2}=0\)
\(\Leftrightarrow6\left(x^2+\frac{1}{x^2}\right)+7\left(x+\frac{1}{x}\right)-36=0\)
\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-36-12=0\)
\(\Leftrightarrow6\left(x+\frac{1}{x}\right)^2-7\left(x+\frac{1}{x}\right)-48=0\)
Đặt \(x+\frac{1}{x}=a\)
\(pt\Leftrightarrow6a^2-7a-48=0\)
\(\Leftrightarrow6\left(a^2-\frac{7}{6}a-8\right)=0\)
\(\Leftrightarrow a^2-\frac{7}{6}a-8=0\)
\(\Leftrightarrow a^2-2\cdot a\cdot\frac{7}{12}+\frac{49}{144}-\frac{1201}{144}=0\)
\(\Leftrightarrow\left(a-\frac{7}{12}\right)^2=\left(\frac{\pm\sqrt{1201}}{12}\right)^2\)
\(\Leftrightarrow a=\frac{\pm\sqrt{1201}+7}{12}\)
\(\Leftrightarrow x+\frac{1}{x}=\frac{\pm\sqrt{1201}+7}{12}\)
Giải nốt nha bạn. Nghiệm hơi xấu
27 tháng 6 2019
\(\Leftrightarrow\left(x^2-x-20\right)\left(x^2-x-6\right)+24=0\)
\(\Leftrightarrow\left(x^2-x-13-7\right)\left(x^2-x-13+7\right)+24=0\)
\(\Leftrightarrow\left(x^2-x-13\right)^2-7^2+24=0\)
\(\Leftrightarrow\left(x^2-x-13\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-13=5\\x^2-x-13=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-x-18=0\\x^2-x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x\cdot\frac{1}{2}+\frac{1}{4}=18+\frac{1}{4}\\x^2-2x\cdot\frac{1}{2}+\frac{1}{4}=8+\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{2}\right)^2=\frac{73}{4}\\\left(x-\frac{1}{2}\right)^2=\frac{33}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{73}}{2}\\x=\frac{1-\sqrt{73}}{2}\\x=\frac{1+\sqrt{33}}{2}\\x=\frac{1-\sqrt{33}}{2}\end{matrix}\right.\) ( TM )
NH
1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 6 2019
\(\Leftrightarrow\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)-4=0\)
\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4=0\)
Đặt \(12x^2+11x-1=a\)
\(\left(a+3\right)a-4=0\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}12x^2+11x-1=1\\12x^2+11x-1=-4\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}12x^2+11x-2=0\\12x^2+11x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
1 tháng 1 2018
\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)
\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=-2\\3x=1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\\x=\dfrac{1}{3}\\x=3\end{matrix}\right.\\ \text{Vậy }S=\left\{-\dfrac{1}{2};-2;\dfrac{1}{3};3\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
1.
a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)
\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)
\(\Leftrightarrow x^3+3x^2+2x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)
b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 3 2020
1c/
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x-1\right)^2+3x^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn
Vậy pt có nghiệm duy nhất \(x=-1\)
MP
1
18 tháng 4 2017
A. \(\left(x+6\right)\left(3x-1\right)+x+6=0\)
\(\Leftrightarrow\left(x+6\right)\left(3x-1+1\right)=0\\ \Leftrightarrow\left(x+6\right)\cdot3x=0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\3x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=0\end{matrix}\right.\)
Vậy.................................
B. \(\left(x+4\right)\left(5x+9\right)-x-4=0\)
\(\Leftrightarrow\left(x+4\right)\left(5x+9\right)-\left(x+4\right)=0\\ \Leftrightarrow\left(x+4\right)\left(5x+9-1\right)=0\\ \Leftrightarrow\left(x+4\right)\left(5x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+4=0\\5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{-8}{5}\end{matrix}\right.\)
Vậy.......................................
(x5 - 2x3 ) - (2x2 - 4) =0
x3 (x2 - 2) - 2 (x2 - 2) =0
(x2 - 2)(x3 - 2) =0
=> x2 - 2 =0 => x=\(\sqrt{2}\)
=> x3 - 2 =0 => x=\(\sqrt[3]{2}\)