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\(3x^4+2x^3-10x^2+2x+3=0\)
\(\Leftrightarrow3x^4-6x^3+3x^2+8x^3-16x^2+8x+3x^2-6x+3=0\)
\(\Leftrightarrow3x^2\left(x^2-2x+1\right)+8x\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(3x^2+8x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(3x^2+8x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3\left(x+\dfrac{4}{3}\right)^2-\dfrac{7}{3}=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-8\pm\sqrt{28}}{6}\end{matrix}\right.\)
(3x+4)2-(3x-1).(3x+1)=49
<=> 9x2+24x+16-(9x2-1)=49
<=>9x2+24x+16-9x2+1=49
<=>24x+17=49
<=>24x =32
<=>x =4/3
Vậy ...
(x+2).(x^2-2x+4)-x.(x+3).(x-3)
=x3+8-x(x2-9)
=x3+8-x3+9x
=9x+8
(3x+4)2-(3x-1).(3x+1)=49
<=> 9x2+24x+16-(9x2-1)=49
<=>9x2+24x+16-9x2+1=49
<=>24x+17=49
<=>24x =32
<=>x =4/3
Vậy ...
(x+2).(x^2-2x+4)-x.(x+3).(x-3)
=x3+8-x(x2-9)
=x3+8-x3+9x
=9x+8
\(x^4-2x^3+4x^2-3x+2=0\)
\(\Leftrightarrow x^4-2x^3+x^2+3x^2-3x+\dfrac{9}{4}-1=0\)
\(\Leftrightarrow\left(x^2-x\right)^2+3\left(x^2-x\right)+\dfrac{9}{4}-1=0\)
\(\Leftrightarrow\left(x^2-x+\dfrac{3}{2}\right)^2-1=0\)
\(\Leftrightarrow\left(x^2-x+\dfrac{3}{2}\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+\dfrac{3}{2}=1\\x^2-x+\dfrac{3}{2}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=1\\x^2-x+\dfrac{1}{4}+\dfrac{5}{4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=1\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2=-\dfrac{1}{4}\\\left(x-\dfrac{1}{2}\right)^2=-\dfrac{9}{4}\end{matrix}\right.\)
\(\Rightarrow\) Vô lý ( vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\) )
\(\Rightarrow PT\) vô nghiệm .
(x^2+2x+1)(x+2)-x^2(x-3)-7x(x-1)=3x-9
<=>x3+4x2+5x+2-x3+3x2-7x2+7x=3x-9
<=>14x+2=3x-9
<=>14x-3x=-9-2
<=>11x=-11
<=>x=-1
vậy S={-1}
ĐKXĐ: ...
\(\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)
\(\Leftrightarrow\frac{2}{x+\frac{12}{x}-3}+\frac{6}{x+\frac{12}{x}+2}=1\)
Đặt \(x+\frac{12}{x}-3=t\)
\(\Rightarrow\frac{2}{t}+\frac{6}{t+5}=1\Leftrightarrow2\left(t+5\right)+6t=t\left(t+5\right)\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{12}{x}-3=-2\\x+\frac{12}{x}-3=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+12=0\\x^2-8x+12=0\end{matrix}\right.\) (casio)
=> \(\left(2x-5\right)^2-\left(3x+4\right)^2=0\)
=> (2x - 5 + 3x +4).(2x - 5 - 3x - 4) = 0
=> (5x - 1).(-x - 9) = 0
=> 5x - 1 = 0 hoặc - x - 9 = 0
+) 5x - 1 = 0 => x = 1/5
-x - 9 = 0 => x = -9
Vậy..............
3X- 2 =2X -3
<=> X=-1
học tốt
\(3x-2x=2-3\)
\(\Leftrightarrow x=-1\)