bài 15 :tìm x∈Q, biết.
a) ( x - 3).( x +4 ) > 0 b)( x + 2).( x - 5) < 0
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c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
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a: =>3x+10-2x=0
hay x=-10
c: \(\Leftrightarrow3x^2-3x^2+6x=36\)
=>6x=36
hay x=6
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Bài 1:
a: \(=6x^3-10x^2+6x\)
b: \(=-2x^3-10x^2-6x\)
Bài 4:
a: =>3x+10-2x=0
=>x=-10
c: =>3x2-3x2+6x=36
=>6x=36
hay x=6
Bài 1:
\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)
Bài 4:
\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)
Bài 1:
\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)
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a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dễ mak
nhưng mik nhìn đề thấy dài quá nên ko muốn làm
hihi^_$
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a) \(\left(x+2\right)^3-x^2\left(x+6\right)=0\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=0\)
\(\Leftrightarrow12x+8=0\)
\(\Leftrightarrow12x=-8\)
\(\Leftrightarrow x=-\dfrac{8}{12}\)
\(\Leftrightarrow x=-\dfrac{2}{3}\)
b) \(\left(2x+3\right)^3-8x\left(x+1\right)\left(x-1\right)=9x\left(4x-3\right)\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x\left(x^2-1\right)=36x^2-27x\)
\(\Leftrightarrow8x^3+36x^2+54x+27-8x^3+8x=36x^2-27x\)
\(\Leftrightarrow8x^3-8x^3+36x^2-36x^2+54x+27x+8x+27=0\)
\(\Leftrightarrow89x+27=0\)
\(\Leftrightarrow x=-\dfrac{27}{89}\)
c) \(\left(2-x\right)^3+\left(2+x\right)^3-12x\left(x+1\right)=0\)
\(\Leftrightarrow8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x^2-12x=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(6x^2+6x^2-12x^2\right)-\left(12x-12x\right)+12x+\left(8+8\right)=0\)
\(\Leftrightarrow12x+16=0\)
\(\Leftrightarrow x=-\dfrac{16}{12}\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
`#040911`
`a)`
`(x + 2)^3 - x^2(x + 6) = 0`
`<=> x^3 + 6x^2 + 12x + 8 - x^3 - 6x^2 = 0`
`<=> (x^3 - x^3) + (6x^2 - 6x^2) + 12x = 0`
`<=> 12x = 0`
`<=> x = 0`
Vậy, `x = 0.`
`b)`
`(2x + 3)^3 - 8x(x - 1)(x + 1) = 9x(4x - 3)`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x(x^2 - 1) = 36x^2 - 27x`
`<=> 8x^3 + 36x^2 + 54x + 27 - 8x^3 + 8x - 36x^2 + 27x = 0`
`<=> (8x^3 - 8x^3) + (36x^2 - 36x^2) + (54x + 8x + 27x) + 27 = 0`
`<=> 89x + 27 = 0`
`<=> 89x = -27`
`<=> x = -27/89`
Vậy, `x = -27/89`
`c)`
`(2 - x)^3 + (2 + x)^3 - 12x(x + 1) = 0`
`<=> 8 - 12x + 6x^2 - x^3 + 8 + 12x + 6x^2 + x^3 - 12x^2 - 12x = 0`
`<=> (-x^3 + x^3) + (12x - 12x - 12x) + (6x^2 + 6x^2 - 12x^2) + (8 + 8)=0`
`<=> -12x + 16 = 0`
`<=> -12x = -16`
`<=> 12x = 16`
`<=> x=4/3`
Vậy, `x = 4/3.`
a: \(\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -4\end{matrix}\right.\)
b: -2<x<5
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