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

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

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

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

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

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

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

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

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

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

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

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

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

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

26 tháng 12 2018

1) 4x\(^2\).(5x3+2x-1)

= 20x\(^5\)+8x\(^3\)-4x\(^2\).

2) 4x\(^3\): x2

= 4x

3) ( 15x2y3-10x3y3+6xy): 5xy

= 3xy2-2x2y2+\(\dfrac{6}{5}\)

4) (5x3+14x2+12x+8 ): (x+2)

= 5x2+4x+4

5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)

=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)

6) \(\dfrac{x}{x+2}\) +\(\dfrac{3}{\left(x+2\right)\left(4x-7\right)}\)

7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)

= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).

8)\(\dfrac{1}{2}\)x2y2.(2x+y)(2x-y)

= \(\dfrac{1}{2}\)x2y2.(4x2-2xy+2xy-y2)

= \(\dfrac{1}{2}\)x2y2.(4x2-y2)

= 2x4y2-\(\dfrac{1}{2}\)x2y4

9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)

= x2.(4x-1)

= 4x3-x2

10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)

= \(\dfrac{3x}{2\left(x+3\right)}\)+\(\dfrac{6-x}{2x\left(x+3\right)}\)= \(\dfrac{3x^2+6-x}{2x\left(x+3\right)}\)=\(\dfrac{3-x}{3}\)= -x

11) x2-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)

12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)

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

26 tháng 12 2018

cảm ơn bạn nhé ^^

x2 - 6x + 9 

= (x -3)2 (hàng đẳng thức đáng nhớ số 2)

x2 + x + 1/4 

= x2 + 2.x.1/2 + 1/4

= (x +1/2)2 (hàng đẳng thức 1)

12 tháng 9 2018

x2-6x+9=(x+3)2

x2+x+\(\frac{1}{4}\)=\(\left(x+\frac{1}{2}\right)^2\)

Học tốt!

\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)

\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

\(TH2:x+6=0\Leftrightarrow x=-6\)

\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha ) 

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

\(2x+5=3x-1\)

\(2x-3x=-1-5\)

\(-1x=-6\)

\(x=6\)

AH
Akai Haruma
Giáo viên
7 tháng 8 2019

1.

\(x^2-22x+12\) : biểu thức không phân tích được thành nhân tử nữa.

2.

\(9x^2+6x+1=(3x)^2+2.3x.1+1^2=(3x+1)^2\)

3.

\(x^2-10x+2\): không p. tích được thành nhân tử.

4.

\(x^3+1=x^3+1^3=(x+1)(x^2-x+1)\)

5.

\(8x^3-27y^3=(2x)^3-(3y)^3=(2x-3y)[(2x)^2+(2x)(3y)+(3y)^2]\)

\(=(2x-3y)(4x^2+6xy+9y^2)\)

AH
Akai Haruma
Giáo viên
7 tháng 8 2019

6.

\((x+3y)^2-(3y+1)^2=[(x+3y)-(3y+1)][(x+3y)+(3y+1)]\)

\(=(x-1)(x+6y+1)\)

7.

\(4y^2-36x^2=(2y)^2-(6x)^2=(2y-6x)(2y+6x)=4(y-3x)(y+3x)\)

8.

\(27-(x+4)^3=3^3-(x+4)^3=[3-(x+4)][3^2+3(x+4)+(x+4)^2]\)

\(=-(x+1)(37+x^2+11x)\)

9.

\(25x^2-10xy+y^2=(5x)^2-2.5x.y+y^2=(5x-y)^2\)

10.

\(9x^6-12x^7+4x^8=x^6(9-12x+4x^2)=x^6[3^2-2.3.2x+(2x)^2]\)

\(=x^6(3-2x)^2\)

23 tháng 2 2020

Bài 1 : Khai triển :

a, \(\left(x+5\right)^2=x^2+10x+25\)

b, \(\left(x-3y\right)^2=x^2-6xy+9y^2\)

c, \(\left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\)

d, \(\left(x+3y\right)^3=x^3+9x^2y+27xy^2+27y^3\)

e, \(27x^3-9y^2+y-\frac{1}{27}=\left(3x-\frac{1}{3}\right)^3\)

g, \(8x^6+12x^4y+6x^2y^2+y^3=\left(2x^2+y\right)\)

h, \(4x^2+12x^4y+6x^22y^2+y^3=\left(\sqrt[3]{4x^2}+y\right)\)