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11 tháng 12 2018

c, \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=>x-\(\dfrac{1}{2}\)=0
<=> x=\(\dfrac{1}{2}\)

a: =>x(x^2-13)=0

=>\(x\in\left\{0;\sqrt{13};-\sqrt{13}\right\}\)

b: =>25x^2=2

=>x^2=2/25

hay \(x=\pm\dfrac{\sqrt{2}}{5}\)

11 tháng 12 2018

\(a,x^3-13x=0\)

\(x.\left(x^2-13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)

\(b,2-25x^2=0\)

\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)

11 tháng 12 2018

a, x 3 - 13 x = 0

=> x ( x 2 - 13 ) = 0

=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)

b, 2 - 25 x 2 = 0

=> 25 x 2 = 2

=> x 2 = 0,08

=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)

x, x 2 - x + \(\frac{1}{4}\)= 0 

=> \(\left(x-\frac{1}{2}\right)^2=0\)

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

18 tháng 9 2017

a)(x+1)(x2+2x)=(x+1)x(x+2)=0

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

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

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

c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)

=0

\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)

d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)

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

20 tháng 4 2017

Bài giải:

a) 2 – 25x2 = 0 => (√2)2 – (5x)2 = 0

=> (√2 – 5x)( √2 + 5x) = 0

Hoặc √2 – 5x = 0 => 5x = √2 => x = 25

Hoặc √2 + 5x = 0 => 5x = -√2 => x = - 25

b) x2 - x + 14 = 0 => x2 – 2 . x . 12 + (12)2 = 0

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

23 tháng 9 2017

a) 2-25x2=0

<=>-25x2=-2

<=>25x2=2

<=>x2=\(\dfrac{2}{25}\)

<=>x=\(\sqrt{\dfrac{2}{25}}\)

b)x2-x +\(\dfrac{1}{4}\) =0

<=>(x - \(\dfrac{1}{2}\))2 = 0

<=> x-\(\dfrac{1}{2}\) =0

<=>x=\(\dfrac{1}{2}\)

18 tháng 2 2021

Bài 2 :

a, Ta có : \(x^2-5x+4< 0\)

\(\Leftrightarrow x^2-x-4x+4< 0\)

\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)< 0\)

\(\Leftrightarrow\left(x-4\right)\left(x-1\right)< 0\)

Vậy ...

b, Ta có : \(\dfrac{x-3}{x+1}< 1\)

\(\Leftrightarrow\dfrac{x-3}{x+1}-\dfrac{x+1}{x+1}< 0\)

\(\Leftrightarrow\dfrac{x-3-x-1}{x+1}=\dfrac{-4}{x+1}< 0\)

Thấy - 4 < 0

Nên để \(-\dfrac{4}{x+1}< 0\) <=> x + 1 > 0 ( TH A, B trái dấu )

Vậy ...

18 tháng 2 2021

thanks bạn nhiều lắm,bạn biết làm bài 1 khum ạ

10 tháng 8 2018

a) \(x^4+x^2-2=0\)
\(\Leftrightarrow x^4+2x^2-x^2-2=0\)
\(\Leftrightarrow x^2\left(x^2+2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+2=0\) hoặc \(x+1=0\) hoặc \(x-1=0\)
. \(x^2+2=0\Leftrightarrow x^2=-2\) (vô nghiệm)
.. \(x+1=0\Leftrightarrow x=-1\)
... \(x-1=0\Leftrightarrow x=1\)
Vậy \(S=\left\{\pm1\right\}\)

b) \(x^4-13x^2+36=0\)
\(\Leftrightarrow x^4-9x^2-4x^2+36=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-4\left(x^2-9\right)=0 \)
\(\Leftrightarrow\left(x^2-9\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow x+3=0\) hoặc \(x-3=0\) hoặc \(x+2=0\) hoặc \(x-2=0\)
. \(x+3=0\Leftrightarrow x=-3\)
.. \(x-3=0\Leftrightarrow x=3\)
... \(x+2=0\Leftrightarrow x=-2\)
.... \(x-2=0\Leftrightarrow x=2\)

Vậy \(S=\left\{\pm3;\pm2\right\}\)
Câu C bạn ghi ko rõ lém!!!!!!!!

15 tháng 7 2018

a) \(4.\left(x-1\right)^2-9=0\)

\(\Rightarrow4.\left(x-1\right)^2=9\)

\(\Rightarrow\left(x-1\right)^2=9:4=\dfrac{9}{4}=\left(\pm\dfrac{3}{2}\right)^2\)

\(\Rightarrow x-1=\pm\dfrac{3}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

vậy\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

b) \(\dfrac{1}{4}-9.\left(x-1\right)^2=0\)

\(\Rightarrow9.\left(x-1\right)^2=\dfrac{1}{4}\)

\(\Rightarrow\left(x-1^2\right)=\dfrac{1}{36}=(\pm\dfrac{1}{6})^2\)

\(\Rightarrow x-1=\pm\dfrac{1}{6}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{6}\\x-1=\dfrac{-1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)

vậy \(\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)

15 tháng 7 2018

e) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\)

\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}=\left(\pm\dfrac{1}{4}\right)^2\)

\(\Rightarrow2x+\dfrac{3}{4}=\pm\dfrac{1}{4}\)

\(\Rightarrow\)\(\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{4}\\2x+\dfrac{3}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

24 tháng 8

a)

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9=3^2\)

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

\(\Rightarrow\left(5x+1\right)^2=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

 

 

30 tháng 12 2016

b)   ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0

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

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

<=> x( 2x - 3 ) = 0

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

Vậy .....

30 tháng 12 2016

a, 25x^2 - 1 - (5x -1)(x+2)=0

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

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

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

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

21 tháng 7 2018

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

<=> \(x\left(x^2-25\right)-\left(x^3+2x^2+4x-2x^2-4x-8\right)=3\)

<=> \(x^3-25x-x^3-2x^2-4x+2x^2+4x+8=3\)

<=> \(-25x+8=3\)

<=> \(-25x=-5\)

<=> \(x=\frac{1}{5}\)

\(25x^2-2=0\)

<=> \(\left(5x\right)^2=2\)

<=> \(\hept{\begin{cases}5x=\sqrt{2}\\5x=-\sqrt{2}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)

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

<=> \(\hept{\begin{cases}x+2=2x-1\\x+2=-2x+1\end{cases}}\)

<=> \(\hept{\begin{cases}-x=-3\\3x=-1\end{cases}}\)

<=> \(\hept{\begin{cases}x=3\\x=\frac{-1}{3}\end{cases}}\)

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

<=> \(\left(x+2\right)^2-\left(x^2-4\right)=0\)

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

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

<=> \(\left(x+2\right).4=0\)

<=> \(x+2=0\)

<=> \(x=-2\)

câu còn lại tương tự nha