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2) 2x4-21x3+74x2-105x+50=0
<=>(2x4-2x3)+(-19x3+19x2)+(55x2-55x)+(-50x+50)=0
<=>2x3.(x-1)-19x2.(x-1)+55x.(x-1)-50.(x-1)=0
<=>(x-1)(2x3-19x2+55x-50)=0
<=>(x-1)[(2x3-20x2+50x)+(x2+5x-50)]=0
<=>(x-1)[2x.(x-5)2+(x2-5x+10x-50)]=0
<=>(x-1){2x.(x-5)2+[x.(x-5)+10.(x-5)]}=0
<=>(x-1)[2x.(x-5)2+(x-5)(x+10)]=0
<=>(x-1)(x-5)(2x2-10x+x+10)=0
<=>(x-1)(x-5)(2x2-5x-4x+10)=0
<=>(x-1)(x-5)[x.(2x-5)-2.(2x-5)]=0
<=>(x-1)(x-5)(x-2)(2x-5)=0
<=>x=1 hoặc x=5 hoặc x=2 hoặc x=5/2
a) x^4 - 3x^3 + 3x - 1 = 0
<=> (x^3 - 2x^2 - 2x + 1)(x - 1) = 0
<=> (x^3 - 3x + 1)(x + 1)(x - 1) = 0
<=> x^3 - 3x + 1 khác 0 hoặc x + 1 = 0 hoặc x - 1 = 0
<=> x + 1 = 0 hoặc x - 1 = 0
<=> x = -1 hoặc x = 1
a, \(x\left(x-3\right)-x^2+2=0\)
\(\Leftrightarrow x^2-3x-x^2+2=0\\ \Leftrightarrow-3x+2=0\)
\(\Leftrightarrow-3x=-2\\ \Rightarrow x=\frac{2}{3}\)
b, \(x^2-2x+1=0\\ \Leftrightarrow\left(x-1\right)^2=0^2\)
\(\Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
c, x(x-1)-(x+3)(x+4)=5x
\(\Leftrightarrow x^2-x-x^2-4x-3x-12=5x\)
\(\Leftrightarrow x^2-x-x^2-4x-3x-5x=12\\ \Leftrightarrow-13x=12\\ \Rightarrow x=\frac{-12}{13}\)
d, ko có vế phải ạ
e, \(x^2+2x=15\)
\(\Leftrightarrow\left(x^2+2x+1\right)-16=0\\ \Leftrightarrow\left(x+1\right)^2-4^2=0\)
\(\Leftrightarrow\left(x+1-4\right)\left(x+1+4\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
f, \(x^4-5x^3+4x^2=0\)
\(\Leftrightarrow x^4-x^3-4x^3+4x^2=0\\ \Leftrightarrow x^3\left(x-1\right)-4x^2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^3-4x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right).x^2\left(x-4\right)=0\)
\(\left[{}\begin{matrix}x^2=0\\x-1=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=4\end{matrix}\right.\)
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
\(2x^4-10x^2+17=2\left(x^4-5x^2+\frac{25}{4}\right)+\frac{9}{2}=2\left(x^2-\frac{5}{2}\right)^2+\frac{9}{2}>0\left(vl\right)\)
=> PT vô nghiệm
\(x^4-x^3+2x^2-x+1=x^2\left(x^2-x+1\right)+x^2-x+1=\left(x^2-x+1\right)\left(x^2+1\right)=\left(x^2+1\right)\left(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right)>0\forall x\)=> Pt vô nghiệm
b) \(\Leftrightarrow x^2\left(x^2+2x-2\right)-3x\left(x^2+2x-2\right)-2\left(x^2+2x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x-2\right)\left(x^2+2x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x-2=0\\x^2+2x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{3}{2}\right)^2=\frac{17}{4}\\\left(x+1\right)^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\frac{\sqrt{17}}{2}\\x-\frac{3}{2}=-\frac{\sqrt{17}}{2}\\x+1=\sqrt{3}\\x+1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{17}}{2}\\x=\frac{3-\sqrt{17}}{2}\\x=\sqrt{3}-1\\x=-1-\sqrt{3}\end{matrix}\right.\) ( TM )
a) Dễ thấy x = 0 không là nghỉ=ệm của pt đã cho
Chia cả 2 vế của pt cho \(x^2\ne0\) ta đc :
\(2x^2-21x+74-\frac{105}{x}+\frac{50}{x^2}=0\)
\(\Leftrightarrow2\left(x^2+\frac{25}{x^2}+10\right)-21\left(x+\frac{5}{x}\right)+54=0\)
\(\Leftrightarrow2\left(x+\frac{5}{x}\right)^2-21\left(x+\frac{5}{x}\right)+54=0\)
\(\Leftrightarrow2t^2-21t+54=0\) ( với \(t=x+\frac{5}{x}\) )
\(\Leftrightarrow\left(2t-9\right)\left(t-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{9}{2}\\t=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{x}=\frac{9}{2}\\x+\frac{5}{x}=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-\frac{9}{2}x+5=0\\x^2-6x+5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{9}{4}\right)^2=\frac{1}{16}\\\left(x-1\right)\left(x-5\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{9}{4}=\frac{1}{4}\\x-\frac{9}{4}=-\frac{1}{4}\\x-1=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=2\\x=1\\x=5\end{matrix}\right.\) ( TM )
Vậy tập nghiệm của pt đã cho là \(S=\left\{\frac{5}{2};2;1;5\right\}\)