Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 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: \(B=\left(a+b-a+b\right)\left(a+b+a-b\right)=4ab=4\cdot2\cdot5=40\)
\(a,x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (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= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)
\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)
\(c,x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Leftrightarrow x+2=0\Rightarrow x=-2\)
\(d,x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
\(e,8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)
\(f,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Rightarrow x+3=0\Rightarrow x=-3\)
a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)
b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)
c. \(8x^3-12x^2+6x-1=0\)
\(\Rightarrow\left(2x-1\right)^3=0\)
\(\Rightarrow x=\frac{1}{2}\)
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) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
\(2x^4+3x^3-9x^2-3x+2\)
\(=2x^4+5x^3-2x^2-2x^3-5x^2+2x-2x^2-5x+2\)
\(=x^2\left(2x^2+5x-2\right)-x\left(2x^2+5x-2\right)-\left(2x^2+5x-2\right)\)
\(=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
b/
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4-4x^3-x^2+x^3-4x^2-x-x^2+4x+1\)
\(=x^2\left(x^2-4x-1\right)+x\left(x^2-4x-1\right)-\left(x^2-4x-1\right)\)
\(=\left(x^2+x-1\right)\left(x^2-4x-1\right)\)
c/
\(x^4-6x^3+12x^2-14x+3\)
\(=x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3\)
\(=x^2\left(x^2-4x+1\right)-2x\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)\)
\(=\left(x^2-2x+3\right)\left(x^2-4x+1\right)\)
e/
Đề sai, sao có 2 hạng tử chứa \(x^4\) thế kia?
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)