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1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
a) \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x-1\right)\left(x+1\right)+3x=2\)
\(\Leftrightarrow x^3+8-x\left(x^2-1\right)+3x-2=0\)
\(\Leftrightarrow x^3-x^3+x+3x+6=0\)
\(\Leftrightarrow4x+6=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
Vậy....
b) \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x^2=25\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
Vậy....
1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
\(a.x^4-16x^2=0\Leftrightarrow\left(x^2+4x\right)\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
\(b.\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
a) x4 - 16x2 = 0
<=> x2 ( x2 - 16 ) = 0
<=> \(\left[{}\begin{matrix}x^2=0\\x^2-16=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
Vậy...
b) ( x - 5)3 - x + 5 = 0
<=> ( x - 5)3 - (x - 5) = 0
<=> (x - 5) [ (x - 5)2 - 1] =0
<=> \(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
c) 5(x - 2) = x2 - 4
<=> 5(x - 2) - (x2 - 4) = 0
<=> (x - 2)( 5 - x - 2) = 0
<=> (x - 2)( 3 - x ) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy...
d) x - 3 = (3 - x)2
<=> x - 3 - (x - 3)2 = 0
<=> (x - 3)(1 - x + 3) = 0
<=> (x - 3)( 4 - x ) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy...
e) x2 (x - 5) + 5 - x = 0
<=> x2 (x - 5) - (x - 5) = 0
<=> (x2 - 1)( x - 5) = 0
<=> \(\left[{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
,
a) 2x(x-3)+5(x-3)=0
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy: phương trình đã cho có tập nghiệm S=\(\left\{3;-\frac{5}{2}\right\}\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)