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 \(x=\pm\dfrac{\sqrt{2}}{5}\)

a)  x + 30 % x = − 1 , 3

x 1 + 3 10 = − 13 10 13 10 x = − 13 10 x = − 1

b)  1 3 x + 2 5 x − 1 = 0

1 3 x + 2 5 x − 2 5 = 0 11 15 x = 2 5 x = 2 5 : 11 15 x = 6 11

c)  3 x − 1 2 − 5 x + 3 5 = − x + 1 5

3 x − 3 2 − 5 x − 3 = − x + 1 5 x = − 3 2 − 3 − 1 5 x = − 47 10

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

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

Bài giải:

a) 2 – 25x² = 0 => (√2)² – (5x)² = 0

=> (√2 – 5x)( √2 + 5x) = 0

Hoặc √2 – 5x = 0 => 5x = √2 => x = $\frac{\sqrt{2}}{5}$

Hoặc √2 + 5x = 0 => 5x = -√2 => x = - $\frac{\sqrt{2}}{5}$

b) x² - x + $\frac{1}{4}$ = 0 => x² – 2 . x . $\frac{1}{2}$ + ($\frac{1}{2}$)² = 0

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

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=\(\dfrac{1}{2}\)

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 ...

Bài 1: 

a) ĐKXĐ: \(x\ge2\)

Ta có: \(3\sqrt{x-2}+\sqrt{25x-50}=2^5\)

\(\Leftrightarrow3\sqrt{x-2}+5\sqrt{x-2}=32\)

\(\Leftrightarrow8\sqrt{x-2}=32\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

hay x=18(thỏa ĐK)

Vậy: S={18}

\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

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!!!!!!!!

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

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