1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

x²+10x+25-4x(x+5)=0

⇔(x+5)²-4x(x+5)=0

⇔(x+5)(x+5-4x)=0

⇔(x+5)(5-3x)=0

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

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

\(\Leftrightarrow x^2-10x+25-4x^2-20x=0\)

\(\Leftrightarrow-3x^2-30x+25=0\)

\(\Leftrightarrow3x^2+30x-25=0\)

\(\text{Δ}=30^2-4\cdot3\cdot\left(-25\right)=900+300=1200>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-30-20\sqrt{3}}{6}=\dfrac{-15-10\sqrt{3}}{3}\\x_2=\dfrac{-15+10\sqrt{3}}{3}\end{matrix}\right.\)

a) x² + 10x + 25 - 4x² - 20x = 0

<=> 3x² + 10x - 25 = 0

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

Vậy S = 

\(\Leftrightarrow x^2-10x+25-4x^2-20x=0\)

\(\Leftrightarrow-3x^2-30x+25=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-15+10\sqrt{3}}{3}\\x=\dfrac{-15-10\sqrt{3}}{3}\end{matrix}\right.\)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

a.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1-x^2}=t\Rightarrow0\le t\le1\)

\(x^2=1-t^2\Rightarrow x^4=t^4-2t^2+1\)

Pt trở thành:

\(729\left(t^4-2t^2+1\right)+8t=36\)

\(\Leftrightarrow729t^4-1458t^2+8t+693=0\)

\(\Leftrightarrow\left(9t^2+2t-9\right)\left(81t^2-18t-77\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9t^2+2t-9=0\\81t^2-18t-77=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{82}-1}{9}\\t=\dfrac{1+\sqrt{78}}{9}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt{1-t^2}=...\)

b.

ĐKXĐ: ...

\(-3\left(10+4x-x^2\right)-5\sqrt{10+4x-x^2}+42=0\)

Đặt \(\sqrt{10+4x-x^2}=t\ge0\)

\(\Rightarrow-3t^2-5t+42=0\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-\dfrac{14}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{10+4x-x^2}=3\)

\(\Leftrightarrow x^2-4x-1=0\)

\(\Leftrightarrow x=...\)

10.

\((x^2-2x-3)(x^2+10x+21)=25\)

\(\Leftrightarrow (x-3)(x+1)(x+3)(x+7)=25\)

\(\Leftrightarrow [(x-3)(x+7)][(x+1)(x+3)]=25\)

\(\Leftrightarrow (x^2+4x-21)(x^2+4x+3)=25\)

Đặt \(x^2+4x-21=a\) thì pt trở thành:

\(a(a+24)=25\)

\(\Leftrightarrow a^2+24a-25=0\)

\(\Leftrightarrow (a-1)(a+25)=0\Rightarrow \left[\begin{matrix} a=1\\ a=-25\end{matrix}\right.\)

Nếu \(a=x^2+4x-21=1\Leftrightarrow x^2+4x-22=0\)

\(\Leftrightarrow (x+2)^2=26\Rightarrow x+2=\pm \sqrt{26}\Rightarrow x=-2\pm \sqrt{26}\) (t/m)

Nếu \(a=x^2+4x-21=-25\Leftrightarrow x^2+4x+4=0\Leftrightarrow (x+2)^2=0\Rightarrow x=-2\) (t/m)

Vậy \(x\in \left\{-2\pm \sqrt{26}; -2\right\}\)

11.

\(x^4-4x^3+10x^2+37x-14=0\)

\(\Leftrightarrow (x^4-4x^3+4x^2)+6x^2+37x-14=0\)

\(\Leftrightarrow x^4+2x^3-(6x^3+12x^2)+(22x^2+44x)-(7x+14)=0\)

\(\Leftrightarrow x^3(x+2)-6x^2(x+2)+22x(x+2)-7(x+2)=0\)

\((x+2)(x^3-6x^2+22x-7)=0\)

\(\Rightarrow \left[\begin{matrix} x+2=0\\ x^3-6x^2+22x-7=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x^3-6x^2+22x-7=0(*)\end{matrix}\right.\)

Đối với pt $(*)$ (ta sử dụng pp Cardano)

\(\Leftrightarrow (x^3-6x^2+12x-8)+10x+1=0\)

\(\Leftrightarrow (x-2)^3+10(x-2)+21=0\)

Đặt \(x-2=a-\frac{10}{3a}\) thì PT trở thành:

\((a-\frac{10}{3a})^3+10(a-\frac{10}{3a})+21=0\)

\(\Leftrightarrow a^3-\frac{1000}{27a^3}+21=0\)

\(\Leftrightarrow 27a^6+576a^3-1000=0\). Đặt \(a^3=t\) thì:

\(27t^2+576t-1000=0\)

\(\Rightarrow 27(t^2+\frac{64}{3}t+\frac{32^2}{3^2})=4072\)

\(\Leftrightarrow 27(t+\frac{32}{3})^2=4072\Rightarrow t=\pm\sqrt{\frac{4072}{27}}-\frac{32}{3}\)

\(\Rightarrow a=\sqrt[3]{\pm \sqrt{\frac{4072}{27}}-\frac{32}{3}}\)

\(x=2+a-\frac{10}{3a}\) với giá trị $a$ như trên.

P/s: Bài này mình thấy có vẻ không phù hợp với lớp 8.