x^4*4x^3*2+6x^2*2^2+4x*2^3+2^4+x^4+4x^3*8+6x^2*8^2+4x*8^3+8^4=272

2x^4+40x^3+408x^2+2080x+4112=272

Đến đây là bt ra x = -4

x = -4 nha 

\(\left(x+2\right)^4+\left(x+2\right)^4=272\)

\(\Rightarrow2\left[\left(x+2\right)^4\right]=272\)

\(\Rightarrow\left(x+2\right)^4=136\)

Vậy...............

Cho bạn kết quả phân tích thôi, tự phân tích nha:D

a) \(\Leftrightarrow2\left(x+4\right)\left(x+10\right)\left(x^2+14x+64\right)=0\)

b)\(\Leftrightarrow2\left(x-3\right)\left(x-4\right)\left(x^2-7x+26\right)=0\)

Dạng này thì em : \(\frac{6+8}{2}=7\). 

Đặt x  + 7 =t

=> Phương trình ban đầu trở thành: \(\left(t+1\right)^4+\left(t-1\right)^4=272\)

<=> \(\left(t^4+4t^3+6t^2+4t+1\right)+\left(t^4-4t^3+6t^2-4t+1\right)=272\)

<=> \(2t^4+12t^2+2=272\)

<=> \(t^4+6t^2-135=0\)

<=> \(t^4+6t^2+9=144\)

<=> \(\left(t^2+3\right)^2=12^2\)

<=> \(\orbr{\begin{cases}t^2+3=12\\t^2+3=-12\end{cases}}\Leftrightarrow\orbr{\begin{cases}t^2=9\left(tm\right)\\t^2=-15\left(l\right)\end{cases}}\Leftrightarrow t=\pm3\)

Với t = 3  có: x + 7 = 3 <=> x =-4

Với t = -3 có: x +7 =-3 <=> x = -10

b) pt  \(\left(5-x\right)^4+\left(2-x\right)^4=17\)<=> \(\left(x-5\right)^4+\left(x-2\right)^4=17\)

Tương tự: \(\frac{5+2}{2}=\frac{7}{2}\)

Đặt: \(x-\frac{7}{2}=t\)

pt trở thành: \(\left(t-\frac{3}{2}\right)^4+\left(t+\frac{3}{2}\right)^4=17\)

<=> .... 

Làm thử tiếp nha.

Chú ý công thức : \(\left(a\pm b\right)^4=a^4\pm4a^3b+6a^2b^2\pm4ab^3+b^4\)

OK!

\(\left(t-\frac{3}{2}\right)^4+\left(t+\frac{3}{2}\right)^4=17\)

<=> \(\left(t^4+4.t^3.\frac{3}{2}+6t^2.\frac{9}{4}+4t.\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4\right)\)

\(+\left(t^4-4.t^3.\frac{3}{2}+6t^2.\frac{9}{4}-4.t\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4\right)=17\)

<=> \(2t^4+27t^2-\frac{55}{8}=0\)

<=> \(t^4+\frac{27}{2}t^2-\frac{55}{16}=0\)

<=> \(\left(t^4+2.t^2.\frac{27}{4}+\frac{729}{16}\right)-\frac{729}{16}-\frac{55}{16}=0\)

<=> \(\left(t^2+\frac{27}{4}\right)^2=49\)

<=> \(t^2+\frac{27}{4}=\pm7\)

<=> \(\orbr{\begin{cases}t^2=\frac{1}{4}\\t^2=-\frac{55}{4}\left(l\right)\end{cases}}\Leftrightarrow t=\pm\frac{1}{2}\). Thay vào tìm x nhé.

Đặt \(x+7=a\)

\(pt\Leftrightarrow\left(a-1\right)^4+\left(a+1\right)^4=272\)

\(\Leftrightarrow a^4-4a^3+6a^2-4a+1+a^4+4a^3+6a^2+4a+1=272\)

\(\Leftrightarrow2a^4+12a^2+2=272\)

\(\Leftrightarrow2a^4+12a^2-270=0\)

\(\Leftrightarrow2\left(a^4+6a^2-135\right)=0\)

\(\Leftrightarrow a^4-3a^3+3a^3-9a^2+15a^2-45a+45a-135=0\)

\(\Leftrightarrow a^3\left(a-3\right)+3a^2\left(a-3\right)+15a\left(a-3\right)+45\left(a-3\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left(a^3+3a^2+15a+45\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left[a^2\left(a+3\right)+15\left(a+3\right)\right]=0\)

\(\Leftrightarrow\left(a-3\right)\left(a+3\right)\left(a^2+15\right)=0\)

Vì \(a^2+15>0\forall x\)

\(pt\Leftrightarrow\left(a-3\right)\left(a+3\right)=0\)

Thay \(a=x+7\)ta có pt :

\(\left(x+7-3\right)\left(x+7+3\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-10\end{cases}}\)

Vậy....

\(\left(x+1\right)^4+\left(x+3\right)^4=272\)

mk thấy đề sai thì phải,sửa nha.

\(\left(x+1\right)^4+\left(x+3\right)^4=256\)

\(\left(x+1\right)^4+\left(x+3\right)^4=4^4\)

TH1 : \(\left(x+1\right)+\left(x+3\right)=4\)

\(x+1+x+3=4\)

\(2x+4=4\Leftrightarrow2x=0\Leftrightarrow x=0\)

TH2 : \(\left(x+1\right)+\left(x+3\right)=-4\)

\(x+1+x+3=-4\)

\(2x+4=-4\Leftrightarrow2x=-8\Leftrightarrow x=-4\)

Lâu lâu chưa lạm dụng đến,chỉ nhớ bình phương chia 2 TH thôi,có j thông cảm ạ.

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

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

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

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

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

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