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\(x^3-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)\)
\(=\left(x-1\right)\left(x-2\right)^2\)
a) 5x ( x - 2000 ) - x + 2000 = 0
5x ( x - 2000 ) - ( x - 2000 ) = 0
5x ( x - 2000 ) = 0
\(\Rightarrow\orbr{\begin{cases}5x=0\\x-2000=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2000\end{cases}}\)
Vậy ....
b) x3 - 13x = 0
x ( x2 - 13 ) = 0
x ( x - \(\sqrt{13}\)) - ( x + \(\sqrt{13}\)) = 0
\(\Rightarrow\hept{\begin{cases}x=0\\x-\sqrt{13}\\x+\sqrt{13}\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\sqrt{13}\\x=\sqrt{-13}\end{cases}}\)
Vậy ....
a) x2 + 6 + 9
= x2 + 2 . 3 . x + 32
= ( x + 3 )2
b) 10x - 25 - x2
= - ( x2 - 10x + 25 )
= - ( x - 5 )2
c) 8x3 - 1/8
= ( 2x )3 - ( 1/2 )3
= ( 2x - 1/2 ) ( 4x2 + x + 1/4 )
d) 1/25 x2 - 64x2
= ( 1/5x )2 - ( 8x )2
= ( 1/5x + 8x ) ( 1/5 - 8x )
\(x^3-13x=0\)
<=> \(x\left(x^2-13\right)=0\)
<=> \(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
<=> \(x=0\)
hoặc \(x-\sqrt{13}=0\)
hoặc \(x+\sqrt{13}=0\)
<=> .....
a) \(x^3-x^2-4\)
\(=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+x+2\right)\left(x-2\right)\)
b) \(x^3+x^2-10x+8\)
\(=x^3+4x^2-3x^2-12x+2x+8\)
\(=x^2\left(x+4\right)-3x\left(x+4\right)+2\left(x+4\right)\)
\(=\left(x^2-2x-x+2\right)\left(x+4\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+4\right)\)
c) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x^2-4x+3x-12\right)\left(x+1\right)\)
\(=\left(x-4\right)\left(x+3\right)\left(x+1\right)\)
f) \(x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+2\right)\)
\(=\left(x^2+4x+4\right)\left(x+1\right)\)
\(=\left(x+2\right)^2\left(x+1\right)\)
\(x^3-x^2-4\)
\(=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)
b, \(x^3+x^2-10x+8\)
\(=x^3-x^2+2x^2-2x-8x+8\)
\(=x^2\left(x-1\right)+2x\left(x-1\right)-8\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+2x-8\right)\)
\(=\left(x-1\right).\left[x^2+4x-2x-8\right]\)
\(=\left(x-1\right).\left[x\left(x+4\right)-2\left(x+4\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x-2\right)\)
c, \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x-4\right)\)
d, \(x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
Chúc bạn học tốt.
a.x3-x2-4
=x2(x-1)-22
=(x\(\sqrt{x-1}\))2-22
=(x\(\sqrt{x-1}\)-2)(x\(\sqrt{x-1}\)+2)
b.x3+x2-10x+8
=x3+23+x2-10x
=(x+2)(x2+2x+4)+x(x-10)
=(x+2)[x(x+2)+4]+x(x-10)
=x(x+2)2+4(x+2)+x(x-10)
=x(x2+4x+4+x-10)+4x-8
=x(x2+5x-6)+4x-8
=x(x2+9x-6)+23
=[\(\sqrt[3]{x\left(x^2+9x-6\right)}\)]3+23
áp dung hằng đẳng thức số 6 làm tiếp
c.x3-13x-12
=x3-13x-13+1
=-13(x+1)+(x3+1)
=-13(x+1)+(x+1)(x2+x+1)
=(x+1)(x2+x+1-13)
=(x+1)(x2+x-12)
d.x3+5x2+8x+4
=5x2+22+x3+8x
=(\(\sqrt{5x}\)+2)(\(\sqrt{5x}\)-2)+x(x2+8)
tương tự như những câu trên dùng hằng đẳng thức làm
\(x^3-x^2-4=\left(x^3-2x^2\right)+\left(x^2-4\right)=x^2\left(x-2\right)+\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
\(x^3+x^2-10x+8=\left(x^3-x^2\right)+\left(2x^2-2x\right)-\left(8x-8\right)=x^2\left(x-1\right)+2x\left(x-1\right)-8\left(x-1\right)=\left(x-1\right)\left(x^2+2x-8\right)\)\(x^3-13x-12=\left(x^3-x\right)-\left(12x+12\right)=x.\left(x^2-1\right)-12.\left(x+1\right)=x\left(x-1\right)\left(x+1\right)-12.\left(x+1\right)=\left(x+1\right).\left[x\left(x-1\right)-12\right]\)\(x^3+5x^2+8x+4=\left(x^3+x^2\right)+\left(4x^2+4x\right)+\left(4x+4\right)=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)Tham khảo nhé~
cảm ơn ạ