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Ta có: \(x^3-5x^2+6x=0\)
\(\Leftrightarrow x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy: S={0;2;3}
\(x^3-5x^2+6x=0\)
\(\Leftrightarrow x^3-2x^2-3x^2+6x=0\)
\(\Leftrightarrow x^2\left(x-2\right)-3x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-3x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
\(S=\left\{0,2,3\right\}\)
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
6x^3 + x + 4 = 11x^2
<=>6x3-11x2+x+4=0
<=>6x3+3x2-14x2-7x+8x+4=0
<=>3x2(2x+1)-7x(2x+1)+4(2x+1)=0
<=>(2x+1)(3x2-7x+4)=0
<=>(2x+1)(3x2-3x-4x+4)=0
<=>(2x+1)(3x-4)(x-1)=0
<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0
<=>x\(\in\){-1/2;1;4/3}
b)x^6 - 14x^4 + 49x^2 = 36
<=>x6-14x4+49x2-36=0
<=>x6-x4-13x4+13x2+36x2-36=0
<=>x4(x2-1)-13x2(x2-1)+36(x2-1)=0
<=>(x2-1)(x4-13x2+36)=0
<=>(x+1)(x-1)(x4-9x2-4x2+36)=0
<=>(x+1)(x-1)[x2(x2-9)-4(x2-9)]=0
<=>(x-1)(x+1)(x2
-9)(x2-4)=0
<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0
<=>x\(\in\){-3;-2;-1;1;2;3}
p/s: kham khảo
6x^3 + x + 4 = 11x^2
<=>6x3-11x2+x+4=0
<=>6x3+3x2-14x2-7x+8x+4=0
<=>3x2(2x+1)-7x(2x+1)+4(2x+1)=0
<=>(2x+1)(3x2-7x+4)=0
<=>(2x+1)(3x2-3x-4x+4)=0
<=>(2x+1)(3x-4)(x-1)=0
<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0
<=>x\(\in\){-1/2;1;4/3}
b)x^6 - 14x^4 + 49x^2 = 36
<=>x6-14x4+49x2-36=0
<=>x6-x4-13x4+13x2+36x2-36=0
<=>x4(x2-1)-13x2(x2-1)+36(x2-1)=0
<=>(x2-1)(x4-13x2+36)=0
<=>(x+1)(x-1)(x4-9x2-4x2+36)=0
<=>(x+1)(x-1)[x2(x2-9)-4(x2-9)]=0
<=>(x-1)(x+1)(x2-9)(x2-4)=0
<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0
<=>x\(\in\){-3;-2;-1;1;2;3}
phù.mệt
a: ĐKXD: x<>0
\(\dfrac{14x^3+12x^2-14x}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(\dfrac{2x\left(7x^2+6x-7\right)}{2x}=\left(x+2\right)\left(3x-4\right)\)
=>\(7x^2+6x-7=3x^2-4x+6x-8\)
=>\(7x^2+6x-7=3x^2+2x-8\)
=>\(4x^2+4x+1=0\)
=>\(\left(2x+1\right)^2=0\)
=>2x+1=0
=>x=-1/2(nhận)
b: \(\left(4x-5\right)\left(6x+1\right)-\left(8x+3\right)\left(3x-4\right)=15\)
=>\(24x^2+4x-30x-5-\left(24x^2-32x+9x-12\right)=15\)
=>\(24x^2-26x-5-24x^2+23x+12=15\)
=>-3x+7=15
=>-3x=8
=>\(x=-\dfrac{8}{3}\)
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)(2x-1)^2-0,25=0`
`<=>(2x-1-0,5)(2x-1+0,5)=0`
`<=>(2x-1,5)(2x-0,5)=0`
`<=>[(x=0,75)(x=0,25):}`
`b)x^2+9=6x`
`<=>(x-3)^2=0`
`<=>x-3=0`
`<=>x=3`
`c)(x^2-4)-3x-6=0`
`<=>(x-2)(x+2)-3(x+2)=0`
`<=>(x+2)(x-2-3)=0`
`<=>(x+2)(x-5)=0`
`<=>[(x=-2),(x=5):}`
a: =>(2x-1-0,5)(2x-1+0,5)=0
=>(2x-1,5)(2x-0,5)=0
=>x=0,25 hoặc x=0,75
b: =>x^2-6x+9=0
=>(x-3)^2=0
=>x-3=0
=>x=3
c: =>(x-2)(x+2)-3(x+2)=0
=>(x+2)(x-5)=0
=>x=5 hoặc x=-2
Với \(x=0\) ko phải nghiệm
Với \(x\ne0\) chia 2 vế của pt cho \(x^2\) ta được:
\(x^2+3x-14-\dfrac{6}{x}+\dfrac{4}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\dfrac{4}{x^2}-4\right)+3\left(x-\dfrac{2}{x}\right)-10=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{x}\right)^2+3\left(x-\dfrac{2}{x}\right)-10=0\)
Đặt \(x-\dfrac{2}{x}=a\)
\(\Rightarrow a^2+3a-10=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{x}=2\\x-\dfrac{2}{x}=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x-2=0\\x^2+5x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\pm\sqrt{3}\\x=\dfrac{-5\pm\sqrt{33}}{2}\end{matrix}\right.\)