\(x^2+x=6\)

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

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12 tháng 10 2018

\(x^2+x=6\)

<=>  \(x^2+x-6=0\)

<=>   \(\left(x-2\right)\left(x+3\right)=0\)

tự lm tiếp

b)  \(6x^3+x^2=2x\)

<=>  \(6x^3+x^2-2x=0\)

<=>  \(x\left(6x^2+x-2\right)=0\)

<=>   \(x\left(2x-1\right)\left(3x+2\right)=0\)

tự giải ra

12 tháng 10 2018

a/\(x^2+x=6\)

\(x\left(x+1\right)=6\)

=> TH1 :x =0

     TH2 : x+1 =0  nên x = ( -1 )

b/\(6x^3+x^2=2x\)

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

\(2x\left(x-1\right)\left(x+1\right)=0\)

TH1 : 2x =0  nên x =0

TH2 : x-1 =0 nên x =1

TH2 : x+1 =0 nên x = (-1)

14 tháng 12 2018

a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)

14 tháng 12 2018

thanks

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

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

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

b: \(=\dfrac{x+3}{x\left(x-6\right)}-\dfrac{x+9}{\left(x-6\right)\left(x+4\right)}+1\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)-x\left(x+9\right)+x\left(x-6\right)\left(x+4\right)}{x\left(x-6\right)\left(x+4\right)}\)

\(=\dfrac{x^2+7x+12-x^2-9x+x\left(x^2-2x-24\right)}{x\left(x-6\right)\left(x+4\right)}\)

\(=\dfrac{-2x+12+x^3-2x^2-24x}{x\left(x-6\right)\left(x+4\right)}\)

\(=\dfrac{x^3-2x^2-26x+12}{x\left(x-6\right)\left(x+4\right)}\)

\(=\dfrac{x^3-6x^2+4x^2-24x-2x+12}{x\left(x-6\right)\left(x+4\right)}\)

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

13 tháng 12 2021

\(\text{A.}\)\(\text{x3+6x2+3x−10}\)

20 tháng 12 2018

a.

\(\dfrac{x+3}{x-2}+\dfrac{4+x}{2-x}\\ =\dfrac{x+3}{x-2}-\dfrac{4+x}{x-2}\\ =\dfrac{x+3-4-x}{x-2}\\ =-\dfrac{1}{x-2}\)

b. \(\dfrac{x+1}{2x+6}+\dfrac{2x+3}{x^2+3x}\)

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

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

\(=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x^2+3x+2x+6}{2x\left(x+3\right)}\)

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

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

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

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

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

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

\(=\dfrac{1}{x}\)

d. \(\dfrac{2x+6}{3x^2-x}:\dfrac{x^2+3x}{1-3x}\)

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

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

\(=-\dfrac{2}{x^2}\)

20 tháng 12 2018

bị mỏi tay nhỉ

27 tháng 8 2017

\(a,\)\(x^4-4x^3+4x^2=0\)

\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b,\)\(x^2+5x+4=0\)

\(\Leftrightarrow x^2+x+4x+4=0\)

\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

\(c,\)\(9x-6x^2-3=0\)

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

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow2x^2-2x-x+1=0\)

\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)

\(d,\)\(2x^2+5x+2=0\)

\(\Leftrightarrow2x^2+4x+x+2=0\)

\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)

19 tháng 12 2016

a/ S=\(\left(\frac{x}{\left(x+6\right)\left(x-6\right)}-\frac{x-6}{x\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2x-6}-\frac{x}{x-6}\)

S=\(\frac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}.\frac{x\left(x+6\right)}{2x-6}-\frac{x}{x-6}\)=\(\frac{\left(x-x+6\right)\left(x+x-6\right)}{\left(x-6\right)\left(2x-6\right)}-\frac{x}{x-6}\)

\(\frac{6\left(2x-6\right)}{\left(x-6\right)\left(2x-6\right)}-\frac{x}{x-6}\)\(\frac{6}{\left(x-6\right)}-\frac{x}{x-6}\)\(\frac{6-x}{x-6}=-1\)

b/ S luôn =-1 với mọi x

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

a)

\(x^3+6x^2+11x+6=(x^3-x)+(6x^2+12x+6)\)

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

\(=x(x-1)(x+1)+6(x+1)^2\)

\(=(x+1)[x(x-1)+6(x+1)]=(x+1)(x^2+5x+6)\)

\(=(x+1)(x^2+2x+3x+6)\)

\(=(x+1)[x(x+2)+3(x+2)]=(x+1)(x+2)(x+3)\)

b) \(x^3+6x^2-13x-42\)

\(=x^3+2x^2+4x^2+8x-21x-42\)

\(=x^2(x+2)+4x(x+2)-21(x+2)\)

\(=(x+2)(x^2+4x-21)\)

\(=(x+2)[x^2-3x+7x-21)\)

\(=(x+2)(x+7)(x-3)\)

AH
Akai Haruma
Giáo viên
15 tháng 7 2018

c)

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

\(=x^2(x-1)-4(x^2-2x+1)\)

\(=x^2(x-1)-4(x-1)^2\)

\(=(x-1)[x^2-4(x-1)]=(x-1)(x^2-4x+4)\)

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

d) \(2x^3-x^2+3x+6\)

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

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

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

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

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

\(=(x+1)(2x^2-3x+6)\)

27 tháng 11 2017

a) Tớ làm luôn nhé , không chép lại đề đâu

P = \(\left[\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right].\dfrac{x\left(x+6\right)}{2x-6}\)

ĐKXĐ : x # -6 ; x # 6 ; x # 0 ; x # 3 . Khi đó , ta có :

P = \(\left[\dfrac{x^2-\left(x-6\right)^2}{x\left(x-6\right)\left(x+6\right)}\right]\).\(\dfrac{x\left(x+6\right)}{2x-6}\)

P = \(\dfrac{x^2-x^2+12x-36}{x-6}.\dfrac{1}{2x-6}\)

P = \(\dfrac{6\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}=\dfrac{6}{x-6}\)

b) Tương tự