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a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
a, x2-x+1/4=(x-1/2)2
b, (x+1)3
c,(2x+1)3
d, (2-3x03
e, (10x)2-(x2+25)2=:[10x+(x2+25)][10x-(x2+25)]=(10x+x2+25)(10x-x2-25)
Bài 2:
a: \(A=a^2+b^2+c^2+2ab-2ac-2bc+a^2+b^2+c^2-2ab-2bc+2ac\)
\(=2a^2+2b^2+2c^2-4bc\)
\(=2+2\cdot9+2\cdot1-4\cdot3\cdot\left(-1\right)=22+12=34\)
b: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)
a)
A = \(\left(2x\right)^3+3.\left(2x\right)^2.y+3.\left(2x\right).y+y^3\)
= \(\left(2x+y\right)^3\)
b)
\(B=x^3-3.x^2.1+3.x.1-1^3\)
= \(\left(x-1\right)^3\)
a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
Bài 1 : Phân tích các đa thức sau thành nhân tử :
a) 8x3 - 64
=(2x)3 + 43
=(2x+4)(4x2 - 8x + 16)
c) 125x3 + 1
=5x3 + 13
=(5x+1)(25x2 +5x+1)
d) 8x3 - 27
=(2x)3 - 33
=(2x - 3)(2x2 + 6x + 9)
e) 1 + 8x6y3
=1 + (2x2y)3
=(1 + 2x2y)(4x4y2 -2x2y + 1)
f) 125x3 + 27y3
=(5x)3 + (3y3)
=(5x + 3y)(25x2 - 15xy + 9y2)
Bài 1
a) \(8x^3-64\)
\(=\left(2x\right)^3-4^3\)
\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)
c) \(125x^3+1\)
\(=\left(5x\right)^3+1^3\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)
\(=\left(2x\right)^3-3^3\)
\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)
e) \(1+8x^6x^3\)
\(=1^3+\left(2x^2y\right)^3\)
\(=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)
f) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
a) Ta có: \(3x^2-6xy+3y^2\)
\(=3\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)^2\)
b) Ta có: \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
c) Ta có: \(64xy-96x^2y+48x^3y-8x^4y\)
\(=8xy\left(8-12x+6x^2-x^3\right)\)
\(=8xy\left(2-x\right)^3\)
d) Ta có: \(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
\(A=2^3-3.2^2.x+3.2.x^2-x^3\)
\(A=\left(2-x\right)^3\)
\(B=\left(2x\right)^3-2.\left(2x\right)^2.y+3.2x.y^2-y^3\)
\(B=\left(2x-y\right)^3\)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (1)
Đặt \(t=\dfrac{x}{\dfrac{2\sqrt{69}}{3}}\Leftrightarrow x=\dfrac{2\sqrt{69}}{3}t\)
Khi đó: (1) \(\Leftrightarrow4t^3+3t=-0,2355375386\)
Đặt a= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)
Và \(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:
\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt
Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
\(1,x^3-3x^2+3x-1\\ =x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\\ =\left(x-1\right)^3\\ 2,x^3+3x^2+3x+1\\ =x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\\ =\left(x+1\right)^3\\ 3,8x^3+12x^2+6x+1\\ =\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3\\ =\left(2x+1\right)^3\\ 4,8x^3-12x^2+6x-1\\ =\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\\ =\left(2x-1\right)^3\\ 5,8-36x+54x^2-27x^3\\ =2^3-3\cdot2^2\cdot3x+3\cdot2\cdot\left(3x\right)^2-\left(3x\right)^3\\ =\left(2-3x\right)^3\\ 6,8+36x+54x^2+27x^3\\ =2^3+3\cdot2^2\cdot3x+3\cdot2\cdot\left(3x\right)^2+\left(3x\right)^3\\ =\left(2+3x\right)^3\\ 7,8x^3-12x^2y+6xy^2-y^3\\ =\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3\\ =\left(2x-y\right)^3\\ 8,8x^3+12x^2y+6xy^2+y^3\\ =\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\\ =\left(2x+y\right)^3\)
\(1,x^3-3x^2+3x-1=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=\left(x-1\right)^3\\ 2,x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\\ 3,8x^3+12x^2+6x+1=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2+1^3=\left(2x+1\right)^3\\ 4,8x^3-12x^2+6x-1=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1-1^3=\left(2x-1\right)^3\\ 5,8-36x+54x^2-27x^3=2^3-3\cdot2^2\cdot3x+3\cdot2\cdot\left(3x\right)^2-\left(3x\right)^3=\left(2-3x\right)^3\\ 6,8+36x+54x^2+27x^3=2^3+3\cdot2^2\cdot3x+3\cdot2\cdot\left(3x\right)^2+\left(3x\right)^3=\left(2+3x\right)^3\\ 7,8x^3-12x^2y+6xy^2-y^3=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\\ 8,8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot2x\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)