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26 tháng 12 2021

\(\left(6x^4-3x^3+x^2\right):3x^2\\ =\left[3x^2\left(2x^2-x+\dfrac{1}{3}\right)\right]:3x^2\\ =2x^2-x+\dfrac{1}{3}\)

17 tháng 8 2023

Chịu

a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)

\(=x^4-16-x^4+9=-7\)

b: \(=27x^3-8-27x^3+6=-2\)

c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

26 tháng 10 2023

6:

a: ĐKXĐ: x<>0

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

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

b: ĐKXĐ: x<>1

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

\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)

c: ĐKXĐ: x<>-2

\(\dfrac{x^2+4x+4}{2x+4}\)

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

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)

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

e: ĐKXĐ: x<>-y

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

g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

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

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

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

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

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

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

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

23 tháng 11 2017

a. 6x3-x2-486x+81

= 6x3-54x2+53x2-477x-9x+81

= 6x2.(x-9)+53x.(x-9)-9.(x-9)

= (x-9).(6x2+53x-9)

= (x-9)(6x2+54x-x-9)

=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)

b. x3-5x2+3x+9

= x3+x2-6x2-6x+9x+9

=x2.(x+1)-6x.(x+1)+9.(x+1)

=(x+1)(x2-6x+9)=(x+1)(x-3)2

c. x3+3x2+6x+4

= x3+x2+2x2+2x+4x+4

= x2.(x+1)+2x.(x+1)+4.(x+1)

= (x+1)(x2+2x+4)

d. 

15 tháng 11 2021

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

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

\(\dfrac{2x^3\left(x-1\right)-x^2\left(x-1\right)-4x\left(x-1\right)+2\left(x-1\right)}{x^2-2}\)

\(\dfrac{\left(2x^3-x^2-4x+2\right)\left(x-1\right)}{x^2-2}\)

\(\dfrac{\left[x^2\left(2x-1\right)-2\left(2x-1\right)\left(x-1\right)\right]}{x^2-2}\)

\(\dfrac{\left(x^2-2\right)\left(2x-1\right)\left(x-1\right)}{x^2-2}\)

\(\left(2x-1\right)\left(x-1\right)\)

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

15 tháng 11 2021

\(\dfrac{2x^4-3x^3-3x^2+6x-2}{x^2-2}\)

\(\dfrac{2x^4-2x^3-x^3+x^2-4x^2+4x+2x-2}{x^2-2}\)

\(\dfrac{2x^3\left(x-1\right)-x^2\left(x-1\right)-4x\left(x-1\right)+2\left(x-1\right)}{x^2-2}\)

\(\dfrac{\left(2x^3-x^2-4x+2\right)\left(x-1\right)}{x^2-2}\)

\(\dfrac{\left[x^2\left(2x-1\right)-2\left(2x-1\right)\left(x-1\right)\right]}{x^2-2}\)

\(\dfrac{\left(x^2-2\right)\left(2x-1\right)\left(x-1\right)}{x^2-2}\)

\(\left(2x-1\right)\left(x-1\right)\)

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

10 tháng 11 2019

Có ai làm dc câu này ko thầy mk cho đề hack não quá

10 tháng 11 2019

\(\frac{12x^4-6x^3-9x^2}{-3x^2}-\left(2-3x\right)\left(2+3x\right)=-\left(3x+1\right)\)\(Dk:-3x^2\ne0\)\(< =>x\ne0\)

<=> \(-4x^2+2x+3-\left(2-3x\right).\left(2+3x\right)=-\left(3x+1\right)\)

<=> \(-4x^2+2x+3-4-6x+6x+9x^2=-3x-1\)

<=>\(5x^2+5x=0\)

<=> \(\orbr{\begin{cases}x=-1\left(n\right)\\x=0\left(l\right)\end{cases}}\)