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Sửa lại câu d) là `25y^2`
`a)x^3-1`
`=(x-1)(x^2+x+1)`
`b)8x^3-y^3`
`=(2x)^3-y^3`
`=(2x-y)(4x^2+2xy+y^2)`
`c)x^2-8x+16`
`=x^2-2.x.4+4^2`
`=(x-4)^2`
`d)25y^2-1`
`=(5y)^2-1`
`=(5y-1)(5y+1(`
`e)27-8y^3`
`=3^3-(2y)^3`
`=(3-2y)(9+6y+4y^2)`
`f)2x^2-8x+8`
`=2(x^2-4x+4)`
`=2(x-2)^2`
1) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x\right)^2-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
2) \(x^3-6x^2+12x-8=27\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot2+3\cdot2^2\cdot x-2^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=27\)
\(\Leftrightarrow\left(x-2\right)^3=3^3\)
\(\Leftrightarrow x-2=3\)
\(\Leftrightarrow x=3+2\)
\(\Leftrightarrow x=5\)
3) \(x^2-8x+16=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(x-4\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow\left(4-x\right)^2=5\left(4-x\right)^3\)
\(\Leftrightarrow5\left(4-x\right)=1\)
\(\Leftrightarrow4-x=\dfrac{1}{5}\)
\(\Leftrightarrow x=4-\dfrac{1}{5}\)
\(\Leftrightarrow x=\dfrac{19}{5}\)
4) \(\left(2-x\right)^3=6x\left(x-2\right)\)
\(\Leftrightarrow8-12x+6x^2-x^3=6x^2-12x\)
\(\Leftrightarrow-12x+6x^2-6x^2+12x=8-x^3\)
\(\Leftrightarrow8-x^3=0\)
\(\Leftrightarrow x^3=8\)
\(\Leftrightarrow x^3=2^3\)
\(\Leftrightarrow x=2\)
5) \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(3x-3x\right)+\left(3x^2+3x^2\right)+\left(1+1\right)-6x^2+12x-6=-10\)
\(\Leftrightarrow0+0+0+\left(6x^2-6x^2\right)+12x-4=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-10+4\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=\dfrac{-6}{12}\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
6) \(\left(3-x\right)^3-\left(x+3\right)^3=36x^2-54x\)
\(\Leftrightarrow27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x\)
\(\Leftrightarrow-54x-2x^3=36x^2-54x\)
\(\Leftrightarrow-2x^3=36x^2\)
\(\Leftrightarrow-2x^3-36x^2=0\)
\(\Leftrightarrow-2x^2\left(x+18\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x^2=0\\x+18=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-18\end{matrix}\right.\)
\(x^3-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
a, Ta có: 2x2-8x+8=2(x2-4+4)=2(x2-2.x.x+22)=2(x-2)2 . Vậy 2x2-8x+8=2(x2-2.x.x+22)
b, Ta có: 4x3 + 9x2 - 4x - 9=4x2(x-1)+9x(x-1)=(x-1)x(4x+9). Vậy 4x3 + 9x2 - 4x - 9=(x-1)x(4x+9)
c, Ta có: (x+1) ( x+3) (x+5) (x+7) =16
=> [(x+1)(x+7)][(x+3)(x+5)]=16
=> (x2+8x+7)(x2+8x+15)=16
=> (x2+8x+7)(x2+8x+7+8)=16
=> (x2+8x+7)2+8(x2+8x+7)-16=0
=>-[(x2+8x+7)2-8(x2+8x+7)+16]=0
=> -(x2+8x+7-4)2=0
=> -(x2+8x+3)2=0
=> x2+8x+3=0
=> x(x+8)=-3
Vì -3=1.(-3)=(-1).3 nên {(x=1);(x+8=-3)}hoặc {(x=-1);(x+8=3)}
Không có trường hợp nào thỏa mãn đề bài ( hơi nghi câu này nhưng bạn có thể dựa theo cách mik làm câu này để làm nha tại mình chưa chắc chắn) Chúc bạn học tốt. Kick đúng cho tui nka
a)\(2x^2-8x+8\)
\(=2\left(x-2\right)^2\)
b)\(4x^3+9x^2-4x-9\)
\(=4x^3-4x^2+13x^2-13x+9x-9\)
\(=\left(x-1\right)4x^2+13x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(4x^2+13x+9\right)\left(x-1\right)\)