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6 tháng 10 2018

\(a.x^3-6x=x^3-4^3=\left(x-4\right)\left(x^2+4x+16\right)\)

\(b.x^4+6x^3+11x^2+6x+1=x^4+6x^3+9x^2+2x^2+6x+1\)

\(=\left(x^2+3x+1\right)^2\)

\(c.x^2+3x+2=x^2+x+2x+2=x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(x+2\right)\)

\(d.x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)

\(=x\left(x+3\right)\left(x+1\right)\left(x+2\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)

Đặt \(x^2+3x=y\Rightarrow y\left(y+2\right)+1=y^2+2y+1=\left(y+1\right)^2\)

Thay \(y=x^2+3x\) ta được: \(\left(y+1\right)^2=\left(x^2+3x+1\right)^2\)

\(e.x^3+9x^2+27x+27=\left(x+3\right)^3\)

\(f.\left(x+1\right)\left(x+7\right)\left(x^2+8x+15\right)+15=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(a=x^2+8x+11\Rightarrow\left(a-4\right)\left(a+4\right)+15=a^2-16+15=a^2-1=\left(a+1\right)\left(a-1\right)\)

Thay \(a=x^2+8x+11\) ta được: \(\left(a+1\right)\left(a-1\right)=\left(x^2+8x+12\right)\left(x^2+8x+10\right)\)

6 tháng 10 2018

\(x^3-6x=x^3-64\) ??? . Căng nhể -.-

7 tháng 10 2019

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) \(1-9x+27x^2-27x^3\)

\(=-\left(27x^3-27x^2+9x-1\right)\)

\(=-\left(3x-1\right)^3\)

7 tháng 10 2019

d) \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

\(=\left(x+\frac{1}{2}\right)^3\)

e) \(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x-2y\right)^3\)

23 tháng 7 2017

\(a,x^3-3x^2+3x-1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

\(b,\left(x-2\right)^3+6\left(x+1\right)^2-x+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x+12=0\)\(\Leftrightarrow x^3+23x+10=0\) (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= \(\sqrt[3]{-0,2355375386+\sqrt{-0,2355375386^2+1}}\)

\(\alpha=\dfrac{1}{2}\left(a-\dfrac{1}{a}\right)\) , ta được:

\(4\alpha^3+3\alpha=-0,2355375386\) , vậy \(t=\alpha\) là nghiệm của pt

Vậy t= \(\dfrac{1}{2}\left(\sqrt[3]{-0,2355375386}+\sqrt{-0,2355375386^2+1}\right)\) \(\left(\sqrt[3]{-0,2355375386-\sqrt{-0,2355375386^2+1}}\right)\)\(=-0,07788262891\)

\(\Rightarrow x=\dfrac{2\sqrt{69}}{3}.t=-0,4312944692\)

\(c,x^3+6x^2+12x+8=0\)

\(\Leftrightarrow\left(x+2\right)^3=0\)

\(\Leftrightarrow x+2=0\Rightarrow x=-2\)

\(d,x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

\(e,8x^3-12x^2+6x-1=0\)

\(\Leftrightarrow\left(2x-1\right)^3=0\)

\(\Rightarrow2x-1=0\Rightarrow x=\dfrac{1}{2}\)

\(f,x^3+9x^2+27x+27=0\)

\(\Leftrightarrow\left(x+3\right)^3=0\)

\(\Rightarrow x+3=0\Rightarrow x=-3\)

15 tháng 10 2020

e, \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)

\(=x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x^2+4x+4\right)=\left(x+1\right)\left(x+2\right)^2\)

d, \(27x^3-27x^2+18x-4=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

12 tháng 9 2020

a. \(\left(x+y\right)^3+\left(x-y\right)^3\)

\(=x^3+3x^2y+3xy^2+y^3+x^3-3x^2y+3xy^2-y^3\)

\(=2x^3+6xy^2\)

\(=2x\left(x^2+6y^2\right)\)

b. \(x^3-y^3+2x^2-2y^2\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x+y\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)

c. \(x^3-y^3-3x^2+3x-1\)

\(=\left(x^3-3x^2+3x-1\right)-y^3\)

\(=\left(x-1\right)^3-y^3\)

\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)

\(=\left(x-y-1\right)\left(x^2+y^2+xy-2x-y+1\right)\)

1 tháng 12 2015

câu d nè bạn

\(x^3+9x^2+23x+15=x^3+5x^2+4x^2+20x+3x+15\)

=\(x^2\left(x+5\right)+4x\left(x+5\right)+3\left(x+5\right)\)

=\(\left(x^2+4x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)

câu c nè

\(x^3-6x^2-x+30=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=\left(x^2-x-6\right)\left(x-5\right)\)

=\(\left(x+2\right)\left(x-3\right)\left(x-5\right)\)

tick rui minh làm tiếp cho