a.\(x^2-25=8\left(5-x\right)\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)-8\left(5-x\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+8\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-13\end{matrix}\right.\)

b.\(\dfrac{x-2}{x+2}-\dfrac{2\left(x-11\right)}{x^2-4}=\dfrac{3}{x-2}\) ; \(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x-2\right)\left(x-2\right)-2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x-2\right)^2-2\left(x-11\right)=3\left(x+2\right)\)

\(\Leftrightarrow x^2-4x+4-2x+22=3x+6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

2:

\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)

\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)

B=(x1+x2)^2-2x1x2

=3^2-2*(-7)

=9+14=23

C=căn (x1+x2)^2-4x1x2

=căn 3^2-4*(-7)=căn 9+28=căn 27

D=(x1^2+x2^2)^2-2(x1x2)^2

=23^2-2*(-7)^2

=23^2-2*49=431

D=9x1x2+3(x1^2+x2^2)+x1x2

=10x1x2+3*23

=69+10*(-7)=-1

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-1\\x_1x_2=-2+\sqrt{2}\end{matrix}\right.\)

\(A=\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-1}{-2+\sqrt{2}}=\dfrac{2+\sqrt{2}}{2}\)

\(B=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=\left(-1\right)^2-2\left(-2+\sqrt{2}\right)=5-2\sqrt{2}\)

\(a,x^2-6x+5=0\\ \Rightarrow\left(x^2-5x\right)-\left(x-5\right)=0\\ \Rightarrow x\left(x-5\right)-\left(x-5\right)=0\\ \Rightarrow\left(x-1\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

\(b,2x^2+4x-8=0\\ \Rightarrow x^2+2x-4=0\\ \Rightarrow\left(x^2+2x+1\right)-5=0\\ \Rightarrow\left(x+1\right)^2-\sqrt{5^2}=0\\ \Rightarrow\left(x+1+\sqrt{5}\right)\left(x+1-\sqrt{5}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1-\sqrt{5}\\x=-1+\sqrt{5}\end{matrix}\right.\)

\(c,4y^2-4y+1=0\\ \Rightarrow\left(2y-1\right)^2=0\\ \Rightarrow2y-1=0\\ \Rightarrow y=\dfrac{1}{2}\)

\(d,5x^2-x+2=0\)

Ta có:\(\Delta=\left(-1\right)^2-4.5.2=1-40=-39\)

Vì \(\Delta< 0\Rightarrow\) pt vô nghiệm

Lời giải:
Các pt bậc nhất là: $a, c, d, g$

a) x(4x + 2) = 4x² - 14

⇔ 4x² + 2x = 4x² - 14

⇔ 4x² - 4x² + 2x = -14

⇔ 2x = -14

⇔ x = -7

Vậy tập nghiệm S = ......

b) (x² - 9)(2x - 1) = 0

⇔ x² - 9 = 0 hoặc 2x - 1 = 0

⇔ x² = 9 hoặc 2x = 1

⇔ x = 3 hoặc -3 hoặc x = \(\dfrac{1}{2}\)

Vậy .......

c) \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{x^2-4}\) 

⇔ \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{\left(x-2\right)\left(x+2\right)}\)

ĐKXĐ: x - 2 ≠ 0 và x + 2 ≠ 0

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

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

`<=>(2x-1,5)(2x-0,5)=0`

`<=>[(x=0,75)(x=0,25):}`

`b)x^2+9=6x`

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

`<=>x-3=0`

`<=>x=3`

`c)(x^2-4)-3x-6=0`

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

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

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

`<=>[(x=-2),(x=5):}`

a: =>(2x-1-0,5)(2x-1+0,5)=0

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

=>x=0,25 hoặc x=0,75

b: =>x^2-6x+9=0

=>(x-3)^2=0

=>x-3=0

=>x=3

c: =>(x-2)(x+2)-3(x+2)=0

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

=>x=5 hoặc x=-2

\(a,=>x^3-2x^2+4x+2x^2-4x+8-x^3+2x-15=0\)

\(< =>2x-7=0< =>x=\dfrac{7}{2}\)

b,\(=>x\left(x^2-25\right)-\left(x+2\right)\left(x^2-2x+4\right)-3=0\)

\(< =>x^3-25x-x^3+2x^2-4x-2x^2+4x-8-3=0\)

\(< =>-25x-11=0\)

\(< =>x=-0,44\)