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25 tháng 10 2020

Ta có x(x + 1) - x(x - 3) = 0

=> x2 + x - x2 + 3x = 0

=> 4x = 0

=> x = 0

Vậy x = 0

25 tháng 10 2020

x( x + 1 ) - x( x - 3 ) = 0

⇔ x2 + x - x2 + 3x = 0

⇔ 4x = 0

⇔ x = 0

26 tháng 10 2017

Trần văn ổi ()

26 tháng 10 2017

đù khó thế

25 tháng 6 2016

a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)

\(S=\left\{1;\frac{1}{3}\right\}\)

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

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)

\(S=\left\{2;-3\right\}\)

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

=>x^2-3x-4+x^2+3+5x-5=0

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

10 tháng 8 2017

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow-\left(2x+5\right)=0\)

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

10 tháng 8 2017

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow-\left(2x+5\right)=0\)

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

10 tháng 8 2017

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

\(\Leftrightarrow x^3-9x^2+27x-27+x^3+9x^2+27x+27=0\)\(\Leftrightarrow2x^3+54x^2=0\)

\(\Leftrightarrow x^2\left(2x+54\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\2x+54=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-27\end{matrix}\right.\)

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

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=0\)\(\Leftrightarrow6x^2+2=0\)

\(\Leftrightarrow6x^2=-2\)

\(\Leftrightarrow x^2=-3\) ( vô lí)

Vậy pt vô nghiệm

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

\(d,4x^2+4x+1=0\)

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

\(\Rightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\Rightarrow x=-\dfrac{1}{2}\)

\(e,\left(x+2\right)^2-\left(x+3\right)^2=0\)

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

\(\Leftrightarrow-\left(2x+5\right)=0\)

\(\Leftrightarrow-2x-5=0\)

\(\Leftrightarrow-2x=5\Rightarrow x=-\dfrac{5}{2}\)

Học tốt nha you <3

7 tháng 5 2021

a,<=>x-x=3+5

<=>x=8

s:={8}

 

7 tháng 5 2021

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

Vậy tập nghiệm \(S=\left\{-2;3\right\}\)

2) \(\left(2x+3\right)\left(-x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=7\end{matrix}\right.\)

Vậy...

3) \(\left(x-1\right)\left(x+5\right)\left(-3x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\dfrac{8}{3}\end{matrix}\right.\)

Vậy...

4) \(x\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy...

19 tháng 4 2020

Giúp luôn Đức Hải Nguyễn câu e:

e, (x - 1)2 + 2(x - 1)(x + 2) + (x + 2)2 = 0

\(\Leftrightarrow\) (x - 1 + x + 2)2 = 0

\(\Leftrightarrow\) (2x + 1)2 = 0

\(\Leftrightarrow\) 2x + 1 = 0

\(\Leftrightarrow\) x = \(\frac{-1}{2}\)

Vậy S = {\(\frac{-1}{2}\)}

Chúc bn học tốt!!

19 tháng 4 2020

a) (x - 3)(5 - 2x) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\5-2x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\frac{5}{2}\end{matrix}\right.\)

b) (x + 5)(x - 1) - 2x(x - 1) = 0

<=> (x - 1)(x + 5 - 2x) = 0

<=> (x - 1)(5 - x) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\5-x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

c) 5(x + 3)(x - 2) - 3(x + 5)(x - 2) = 0

<=> (x - 2)[5(x + 3) - 3(x + 5)] = 0

<=> (x - 2)(5x + 3 - 3x - 15) = 0

<=> (x - 2)(2x - 12) = 0

<=> \(\left[{}\begin{matrix}x-2=0\\2x-12=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

d) (x - 6)(x + 1) - 2(x + 1) = 0

<=> (x + 1)(x - 6 - 2) = 0

<=> (x + 1)(x - 8) = 0

<=> \(\left[{}\begin{matrix}x+1=0\\x-8=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-1\\x=8\end{matrix}\right.\)

Câu e thì để mình nghĩ đã :)

#Học tốt!