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a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
b: Ta có: \(\left(x-3\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+9\left(x+2\right)^2\)
\(=x^3-9x^2+27x-27-x^3-8+9x^2+36x+36\)
\(=53x+1\)
a, (x+2)2+(x-3)2=2x(x+7)
x.2+2.2+x.2+(-3).2-2x=8
2x+4+2x-6-2x=8
(2x+2x)+(4-6)=8
4x-2=8
4x=8+2
4x=10
X=10:4
X=5/2
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+2\right)^3+2\left(x+2\right)\left(x^2-2x+4\right)+6x\left(x+2\right)\)
\(=x^3-27-x^3-6x^2-12x-27+2\left(x^3+8\right)+6x^2+12x\)
\(=-54+2x^3+16\)
\(=2x^3-38\)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
`a,3(x-2)^2+9(x-1)=3(x^2+x-3)`
`<=>3(x^2-4x+4)+9x-9=3x^2+3x-9`
`<=>3x^2-12x+12+9x-9=3x^2+3x-9`
`<=>3x^2-3x+3=3x^2+3x-9`
`<=>6x=12`
`<=>x=12`
`b,(x+3)^2-(x-3)=6x+18`
`<=>(x+3-x+3)(x+3+x-3)+6x+18`
`<=>6.2x=6(x+3)`
`<=>2x=x+3`
`<=>x=3`
`c,(2x+7)^2=9(x+2)^2`
`<=>(2x+7)^2=(3x+6)^2`
`<=>(3x+6-2x-7)(3x+6+2x+7)=0`
`<=>(x-1)(5x+13)=0`
`<=>` $\left[ \begin{array}{l}x-1=0\\5x+13=0\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\5x=-13\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=1\\x=-\dfrac{13}{5}\end{array} \right.$
a) Ta có: \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\)
\(\Leftrightarrow-6x+12=0\)
\(\Leftrightarrow-6x=-12\)
hay x=2
Vậy: x=2