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21 tháng 6 2018

mk sửa đề 1 chút

\(x^6-x^5+3x^4-16x^2+16x-48=0\)

\(\Leftrightarrow\left(x^6-16x^2\right)-\left(x^5-16x\right)+\left(3x^4-48\right)=0\)

\(\Leftrightarrow x^2\left(x^4-16\right)-x\left(x^4-16\right)+3\left(x^4-16\right)=0\)

\(\Leftrightarrow\left(x^4-16\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\left(x^2-x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x^2+4>0\\x^2-x+3=0\left(vonghiem\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy.............

19 tháng 9 2023

hết cứu đi mà làm

22 tháng 12 2020

Rảnh rỗi thật sự .-.

undefined

a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0

=>-5x-4=0

=>x=-4/5

b: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81

=>83x=83

=>x=1

23 tháng 6 2023

Cảm ơn nhìu ạ :3

19 tháng 6 2019

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

19 tháng 6 2019

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

\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)

\(\Leftrightarrow3x\left(x^2+x-2x-2\right)=0\)

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)

28 tháng 4 2018

a) 5 - 4x = 3x - 9

\(\Leftrightarrow5-4x-3x+9=0\)

\(\Leftrightarrow14-7x=0\)

\(\Leftrightarrow7x=14\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

b) \(\left(x-4\right)\left(3x+9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(S=\left\{-3;4\right\}\)

c) \(\dfrac{x}{x+4}+\dfrac{12}{x-4}=\dfrac{4x+48}{x\cdot x-16}\)(1)

ĐKXĐ: \(x\ne\pm4\)

\(\left(1\right)\Leftrightarrow\dfrac{x\left(x-4\right)+12\left(x+4\right)-4x-48}{\left(x+4\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2-4x+12x+48-4x-48=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-4\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

d) \(4-2x=7-x\)

\(\Leftrightarrow4-2x-7+x=0\)

\(\Leftrightarrow-x-3=0\)

\(\Leftrightarrow-x=3\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

e) \(\left(x+4\right) \left(8-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\8-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

f) \(\dfrac{x}{x+5}+\dfrac{11}{x-5}=\dfrac{x+55}{x\cdot x-25}\left(2\right)\)

ĐKXĐ: \(x\ne\pm5\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x-5\right)+11\left(x+5\right)-x-55}{\left(x+5\right)\left(x-5\right)}=0\)

\(\Leftrightarrow x^2-5x+11x+55-x-55=0\)

\(\Leftrightarrow x^2+5x=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-5\left(KTM\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

g) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)-3x-1-10-12x}{6}=0\)

\(\Leftrightarrow9x+6-3x-1-10-12x=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy \(S=\left\{-\dfrac{5}{6}\right\}\)

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

\(\Leftrightarrow2x-3+5x-4x-12=0\)

\(\Leftrightarrow3x-15=0\)

\(\Leftrightarrow x=5\)

Vậy \(S=\left\{5\right\}\)

i) \(3x-6+x=9-x\)

\(\Leftrightarrow3x-6+x-9+x=0\)

\(\Leftrightarrow5x-15=0\)

\(\Leftrightarrow x=3\)

Vậy \(S=\left\{3\right\}\)

k)\(2t-3+5t=4t+12\)

\(\Leftrightarrow2t-3+5t-4t-12=0\)

\(\Leftrightarrow3t-15=0\)

\(\Leftrightarrow t=5\)

Vậy \(S=\left\{5\right\}\)

28 tháng 4 2018

c.ơn bạn