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

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

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

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

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

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

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

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 

=> \(\left(x-\frac{1}{2}\right)^2=0\)

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

ê, sorry nhá cái đoạn nì quên:

\(x=\pm\sqrt{13}\)

\(x=\pm\sqrt{\frac{2}{25}}\)

p/s: nguồn: Thăm Tuy Thăm Tuy sorry t quên >:

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

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

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

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 \(S=\left\{28\right\}\)

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

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

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

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

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

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

Đáp án : D

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 \(x=\pm\dfrac{\sqrt{2}}{5}\)

`1)`

$a\big)\Delta=7^2-5.4.1=29>0\to$ PT có 2 nghiệm pb

$b\big)$

Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{7}{5}\\x_1x_2=\dfrac{1}{5}\end{matrix}\right.\)

\(A=\left(x_1-\dfrac{7}{5}\right)x_1+\dfrac{1}{25x_2^2}+x_2^2\\ \Rightarrow A=\left(x_1-x_1-x_2\right)x_1+\left(\dfrac{1}{5}\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\\ \Rightarrow A=-x_1x_2+\left(x_1x_2\right)^2\cdot\dfrac{1}{x_2^2}+x_2^2\)

\(\Rightarrow A=-x_1x_2+x_1^2+x_2^2\\ \Rightarrow A=\left(x_1+x_2\right)^2-3x_1x_2\\ \Rightarrow A=\left(\dfrac{7}{5}\right)^2-3\cdot\dfrac{1}{5}=\dfrac{34}{25}\)

Đáp án D