Đặt x + 4 = t thì pt trở thành :

\(\left(t+1\right)^4+\left(t-1\right)^4=16\)

\(\Leftrightarrow\left(t^4+4t^3+6t^2+4t+1\right)-\left(t^4-4t^3+6t^2-4t+1\right)=16\)

\(\Leftrightarrow8t^3+8t-16=0\)

\(\Leftrightarrow8\left[t^2\left(t-1\right)+t\left(t-1\right)+2\left(t-1\right)\right]=0\)

\(\Leftrightarrow\left(t-1\right)\left(t^2+t+2\right)=0\)

\(\Leftrightarrow t-1=0\) ( do \(t^2+t+2=\left(t+\frac{1}{2}\right)^2+\frac{7}{4}>0\forall t\))

\(\Leftrightarrow t=1\Leftrightarrow x=-3\) ( TM )

bổ sung:tìm điều kiện xác định của phương trình

ĐKXĐ : x khác cộng trừ 2

ĐKXĐ: ...

Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow a^2=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)

\(a^2+\frac{8}{3}=\frac{10}{3}a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)

\(x^2+\left(16-x\sqrt{3}\right)^2=4\left(12-x\right)^2\)

\(\Leftrightarrow x^2+256-32\sqrt{3}x+3x^2=4\left(144-24x+x^2\right)\)

\(\Leftrightarrow4x^2-32\sqrt{3}x+256=576-96x+4x^2\)

\(\Leftrightarrow4x^2-4x^2-32\sqrt{3}x+96x+256-576=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x-320=0\)

\(\Leftrightarrow\left(96-32\sqrt{3}\right)x=320\)

\(\Leftrightarrow x=\frac{320}{96-32\sqrt{3}}=\frac{15+5\sqrt{3}}{3}\)

\(x^5+y^5-\left(x+y\right)^5\)

\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)

\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)

\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)

\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)

|x-9|=2x+5

Xét 3 TH

TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)

TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)

TH3: x=9 =>0=23(L)

Vậy  x= 4/3

Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)

\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)

\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)

Ta có:

\(\dfrac{2}{x-3}+\dfrac{3}{x+3}=\dfrac{3x+5}{x^2-9}\)

\(\dfrac{2\left(x+3\right)+3\left(x-3\right)}{x^2-9}=\dfrac{3x+5}{x^2-9}\)

\(5x-4=3x+5\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\)

\(\Leftrightarrow\left(x^2-x-20\right)\left(x^2-x-6\right)+24=0\)

\(\Leftrightarrow\left(x^2-x-13-7\right)\left(x^2-x-13+7\right)+24=0\)

\(\Leftrightarrow\left(x^2-x-13\right)^2-7^2+24=0\)

\(\Leftrightarrow\left(x^2-x-13\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-13=5\\x^2-x-13=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-x-18=0\\x^2-x-8=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x\cdot\frac{1}{2}+\frac{1}{4}=18+\frac{1}{4}\\x^2-2x\cdot\frac{1}{2}+\frac{1}{4}=8+\frac{1}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{2}\right)^2=\frac{73}{4}\\\left(x-\frac{1}{2}\right)^2=\frac{33}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1+\sqrt{73}}{2}\\x=\frac{1-\sqrt{73}}{2}\\x=\frac{1+\sqrt{33}}{2}\\x=\frac{1-\sqrt{33}}{2}\end{matrix}\right.\) ( TM )