1)x^4+x^2-6x+1=0>>>x^4+4x^2+4-3x^2-6x-3=0>>>(x^2+2)^2=3(x-1)^2.

>>Sau đó giải bt.

2)Đặt x^2-x+1=a;x+1=b thì:x^3+1=ab.

Pt:2a+5b^2+14ab=0(tự giải nha)

\(\left(x+4\right)\left(x+6\right)\left(x-2\right)\left(x-12\right)=25x^2\)

\(\Leftrightarrow\left(x+3\right)\left(x+8\right)\left(x^2-15x+24\right)=0\)

\(x^4-8x^3+21x^2-24x+9=0\)

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

\(\Leftrightarrow\left(x-\frac{5+\sqrt{13}}{2}\right)\left(x-\frac{5-\sqrt{13}}{2}\right)=0\) (vì \(x^2-3x+3=\left(x-\frac{3}{2}\right)^2+0,75>0\))

\(\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{13}}{2}\\x=\frac{5-\sqrt{13}}{2}\end{cases}}\)

bn cứ quy đồng lần lượt 2 hạng tử đầu tiên là đc thôi

mk giải phần a k ra

Đk:\(x\ge1;x\le-2\)

Đặt \(t=\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}\)

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

Pttt: \(t^2+4t=12\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)

TH1: \(t=2\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-1\right)\left(x+2\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x^2+x-6=0\end{matrix}\right.\)\(\Rightarrow x=2\) (thỏa mãn)

TH2:\(t=-6\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=-6\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1< 0\\\left(x-1\right)\left(x+2\right)=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x^2+x-38=0\end{matrix}\right.\)\(\Rightarrow x=\dfrac{-1-3\sqrt{17}}{2}\) (thỏa mãn)

Vậy...

10 - { [ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 } = 5

[ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 = 10 - 5 = 5

( x : 3 + 17 ) : 10 + 3 : 2⁴ = 5 x 10

( x : 3 + 17 ) : 10 + 48 = 50

( x : 3 + 17 ) : 10 = 50 - 48

( x : 3 + 17 ) : 10 = 2

x : 3 + 17 = 2 x 10

x : 3 + 17 = 20

x : 3 = 20 - 17 = 3

x = 3 x 3 = 9

a) [(2x+14) : 4 - 3] : 2 = 1

(2x+14) : 4 - 3 = 1/2

(2x+14) : 4  = 1/2 + 3

(2x+14) : 4  = 7/2

2x+14 = 7/2 . 1/4

2x = 7/8 - 1/4

2x = 5/8

x= 5/8.1/2

x= 5/16

Còn phần b nữa ak

a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)

<=> \(-\frac{4}{3}x=-\frac{59}{24}\)

<=> \(x=\frac{59}{32}\)

Vậy S = { 59/32}

b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)

<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)

<=> \(-x=-8\)

<=> x = 8 

Vậy S = { 8 }

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

bậc nhất môt ẩn đây ak