đkxđ: x khác 0

\(\Leftrightarrow8.\left(x+\dfrac{1}{x}\right)\left(x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)+4\left(x^2+\dfrac{1}{x^2}\right)^2=x^2+8x+16\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(8.x+\dfrac{1}{x}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\right]+4\left(x^4+2+\dfrac{1}{x^2}\right)-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left[\left(\dfrac{8x^2+1}{x}-4x^2-\dfrac{4}{x^2}\right)\right]+4x^4+8+\dfrac{4}{x^2}-x^2-8x-16=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{x\left(8x^2+1\right)}{x^2}-\dfrac{4x^2.x^2}{x^2}-\dfrac{4}{x^2}\right)+......=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(\dfrac{8x^3+x-4x^4-4}{x^2}\right)+...=0\)

\(\Leftrightarrow\dfrac{x^2}{x}.-\dfrac{4x^4+8x^3+x-4}{x^2}+.....=0\)

\(\Leftrightarrow-\dfrac{4x^6+8x^5+x^3-4x^2}{x^3}+\dfrac{4x^4+8+4x^2}{1}-\dfrac{x^2-8x-16}{1}=0\)

\(\Leftrightarrow......+\dfrac{x^3.\left(4x^4+8+4x^2\right)}{x^3}-\dfrac{x^3\left(x^2-8x-16\right)}{x^3}=0\)

\(\Leftrightarrow-4x^6+8x^5+x^3-4x^2+4x^7+8x^3+4x^5-x^5+8x^4+16x^3=0\)

\(\Leftrightarrow4x^7-4x^6+12x^5+8x^4+25x^3-4x^2=0\)

=> x=0 ( loại , ko tm)

Vậy pt vô nghiệm

Phương pháp:

Đặt \(x+\dfrac{1}{x}=a\Rightarrow a^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow a^2-2=x^2+\dfrac{1}{x^2}\)

Thay vào pt

\(x\ne0:đặt:x+\dfrac{1}{x}=t\)

\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)

\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)

\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)

\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)

(bn xem lại đề)

`(x^2-x+1)^4+4x^4=5x^2(x^2-x+1)^2`

Đặt `a=(x^2-x+1)^2,b=x^2`

`pt<=>a^2+4b^2=5ab`

`<=>a^2-5ab+4b^2=0`

`<=>a^2-ab-4ab+4b^2=0`

`<=>a(a-b)-4b(a-b)=0`

`<=>(a-b)(a-4b)=0`

`<=>` $\left[ \begin{array}{l}a=b\\a=4b\end{array} \right.$

`+)a=b`

`<=>x^2=(x^2-x+1)^2`

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

`<=>(x-1)^2=0` do `x^2+1>0`

`<=>x=1`

`+)a=4b`

`<=>x^2=4(x^2-x+1)^2`

`<=>x^2=(2x^2-2x+1)^2`

`<=>(2x^2-x+1)(2x^2-3x+1)=0`

`+)2x^2-x+1=0`

`<=>x^2-1/2x+1/2=0`

`<=>(x-1/4)^2+7/16=0` vô lý

`+)2x^2-3x+1=0`

`<=>2x^2-2x-x+1=0`

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

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

`<=>` $\left[ \begin{array}{l}x=1\\x=\dfrac{1}{2}\end{array} \right.$

Vậy `S={1,1/2}`

a) \(x^4-x^2+\dfrac{1}{4}-\dfrac{225}{4}=0\\ \left(x^2-\dfrac{1}{2}\right)^2-\dfrac{15}{2}^2=0\\ \left(x+7\right)\left(x-8\right)=0\\ \left[{}\begin{matrix}x=8\\x=-7\end{matrix}\right.\)

Vậy x = 8 hoặc x = -7

a: Ta có: \(x^4-x^2-56=0\)

\(\Leftrightarrow x^4-8x^2+7x^2-56=0\)

\(\Leftrightarrow\left(x^2-8\right)\left(x^2+7\right)=0\)

\(\Leftrightarrow x^2-8=0\)

hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)

TH1: \(x\ge2\)

\(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=4\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=4\)

\(\Leftrightarrow x^4-5x^2=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\sqrt{5}\left(loại\right)\\x=\sqrt{5}\end{matrix}\right.\)

TH2: \(x< 2\)

\(-\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)=4\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=-4\)

\(\Leftrightarrow x^4-5x^2+8=0\)

\(\Leftrightarrow\left(x^2-\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0\) (vô nghiệm)

Vậy \(x=\sqrt{5}\)

\(\left(x-2\right)\left(x-1\right)\left(x-4\right)\left(x-8\right)=4x^2\)

\(\Leftrightarrow[\left(x-2\right)\left(x-4\right)][\left(x-1\right)\left(x-8\right)]=4x^2\)

\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-9x+8\right)=4x^2\)

thấy \(x=0;2\) không phải nghiệm của phương trình nên ta chia hai vế của pt cho \(x^2\) ta được \(:\)

\(\Leftrightarrow\left(x+\dfrac{8}{x}-9\right)\left(x+\dfrac{8}{x}-6\right)=4\)

\(Đặt:\) \(x+\dfrac{8}{x}=a\) thì pt trở thành \(:\)

\(\left(a-6\right)\left(a-9\right)=4\)

\(\Leftrightarrow a^2-15a+50=0\)

\(\Leftrightarrow\left(a-5\right)\left(a-10\right)=0\Leftrightarrow\left\{{}\begin{matrix}a=5\\a=10\end{matrix}\right.\)

\(Với\) \(a=5\) thì \(x+\dfrac{8}{x}=5\Leftrightarrow x^2-5x+8=0\left(vônghiem\right)\)

\(Với\) \(a=10\) thì \(x+\dfrac{8}{x}=10\Leftrightarrow x^2-10x+8=0\Leftrightarrow\left\{{}\begin{matrix}x=5-căn17\\x=5+căn17\end{matrix}\right.\)

\(Vậy...\)

căn bậc 2 của \(17\) đấy á

Đặt \(\left(x^2-x+1\right)^2=a;x^2=b\left(a,b\ge0\right)\)

\(PT\Leftrightarrow a^2-10ab+9b^2=0\\ \Leftrightarrow a^2-9ab-ab+9b^2=0\\ \Leftrightarrow\left(a-b\right)\left(a-9b\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=b\\a=9b\end{matrix}\right.\\ \forall a=b\Leftrightarrow\left(x^2-x+1\right)^2-x^2=0\\ \Leftrightarrow\left(x^2-2x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow x=1\\ \forall a=9b\Leftrightarrow\left(x^2-x+1\right)^2-9x^2=0\\ \Leftrightarrow\left(x^2-4x+1\right)\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x^2+x\right)^2-5\left(x^2+x\right)-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$

$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$

$\Leftrightarrow x-2=0$ hoặc $4-x=0$

$\Leftrightarrow x=2$ hoặc $x=4$ (tm)

Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$

$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$

Với $4x^3-3x^2+6x-4=0(*)$

Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$

Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:

$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$

Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)

Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)

Nãy mình tìm được một cách giải tương tự cho câu 2.

PT \(\Leftrightarrow\left(x-1\right)\left(4x^3-3x^2+6x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x^3-3x^2+6x-4=0\left(1\right)\end{matrix}\right.\)

Vậy pt có 1 nghiệm bằng 1.

\(\left(1\right)\Rightarrow8x^3-6x^2+12x-8=0\)

\(\Leftrightarrow7x^3+x^3-6x^2+12x-8=0\)

\(\Leftrightarrow\left(x-2\right)^3=-7x^3\)

\(\Leftrightarrow x-2=-\sqrt[3]{7}x\)

\(\Leftrightarrow x=\dfrac{2}{1+\sqrt[3]{7}}\)

Vậy pt có nghiệm \(S=\left\{1;\dfrac{2}{1+\sqrt[3]{7}}\right\}\)

Lưu ý: Nghiệm của người kia hoàn toàn tương đồng với nghiệm của mình (\(\dfrac{2}{1+\sqrt[3]{7}}=\dfrac{1}{4}\left(1-\sqrt[3]{7}+\sqrt[3]{49}\right)\))

bla ta da dech hiu

x^2+4x+4 +x^4+16x^3+96x^2+256x+256= -x^3-9x^2-28x-28

(x^2+4x+4)+  ( x^4 + 16x^3 + 96x^2 + 256x+ 256) + (x^3+9x^2+28x+28)=0

x^4+ 17 x^3 + 106x^2 + 288x + 288=0

x^4+ 3x^3+ 14x^3+42x^2+ 64x^2+192x+96x+288=0

(x+3)(x^3+14x^2+64x+96)=0

(x+3)(x^3+6x^2+8x^2+48x+16x+96)=0

(x+3)(x+6)(x^2+8x+16)=0

(x+3)(x+6)(x+4)^2=0

Vậy x=-3 hay x=-6 hay x=-4

cảm ơn bạn Anime lover :DD nhé