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\(a,=x^4+6x^3+8x^2\\ b,=x^2+3x-28\\ c,=x^2-3x-x^2+6x-9+9=3x\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+2\right)^3+2\left(x+2\right)\left(x^2-2x+4\right)+6x\left(x+2\right)\)
\(=x^3-27-x^3-6x^2-12x-27+2\left(x^3+8\right)+6x^2+12x\)
\(=-54+2x^3+16\)
\(=2x^3-38\)
`a,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
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(3-2x\right)\left(4x^2+6x+9\right)\)
\(M=\left(x^3+3^3\right)-\left[3^3-\left(2x\right)^3\right]\)
\(M=x^3+27-27+8x^3\)
\(M=9x^3\)
Thay x=20 vào M ta có:
\(M=9\cdot20^3=72000\)
Vậy: ...
\(N=\left(x-2y\right)\left(x^2+2xy+4y^2\right)+16y^3\)
\(N=x^3-\left(2y\right)^3+16y^3\)
\(N=x^3-8y^3+16y^3\)
\(N=x^3+8y^3\)
\(N=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Thay \(x+2y=0\) vào N ta có:
\(N=0\cdot\left(x^2-2xy+4y^2\right)=0\)
Vậy: ...
\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)
\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)
\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)
\(=-\left(x-3\right)^3-7\left(x-3\right)^2-4\)
\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)