\(3,\\ A=1-8x^3+8x^3-8=-7\\ B=\left(3x-y\right)\left(9x^2+3xy+y^2\right)+x^3+y^3-27x^3\\ B=27x^3-y^3+x^3+y^3-27x^3=x^3\)

Bài 4: 

a: Ta có: \(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=17\)

\(\Leftrightarrow x^3-25x-x^3-8=17\)

\(\Leftrightarrow-25x=25\)

hay x=-1

b: Ta có: \(8\left(x-\dfrac{1}{2}\right)\left(x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)-4x\left(2x^2-x+1\right)+2=0\)

\(\Leftrightarrow8x^3-1-8x^3+4x^2-4x+2=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

\(\frac{\left(x+1\right)3}{111\cdot3}=\frac{3x+3}{333}\)

\(\frac{\left(y+2\right)2}{222\cdot2}=\frac{2y+4}{444}\)

Ta có: \(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{3x+3}{333}=\frac{2y+4}{444}=\frac{z+3}{333}=\frac{3x+3+2y+4+z+3}{333+444+333}=\frac{\left(3x+2y+z\right)+\left(3+4+3\right)}{1110}=\frac{989+10}{1110}=\frac{999}{1110}=\frac{9}{10}\)

\(\frac{3x+3}{333}=\frac{9}{10}\Rightarrow3x+3=\frac{2997}{10}\Rightarrow3x=\frac{2967}{10}\Rightarrow x=\frac{989}{10}=98,9\)

Tìm y và z tương tự nhé! Ko hiểu chỗ nào thì nói tớ!

thanks:)

999³³³ = ( 333³ )³³³ = 333⁹⁹⁹

=> 333⁹⁹⁹ = 333⁹⁹⁹

=> 999³³³ = 333⁹⁹⁹

sai rồi bạn cậu bé tiếm pro ơi

Cậu Bé Tiến Pro ====>>

\(333^{999}=\left(333^3\right)^{333}>\left(333.3\right)^{333}=999^{333}\)

\(\Rightarrow333^{999}>999^{333}\)

p/s: lần sau so sánh có máy tính thì bấm xem nếu = nhau thì = 0 ấy dễ bt mk làm sai hay đúng 

              TA CÓ : 333⁴⁴⁴= 333^4.111=(333⁴)¹¹¹=12296370321¹¹¹ (1)

                            444³³³=444^3.111=(444³)¹¹¹=87528384¹¹¹         (2)

                         TỪ (1) VÀ  (2) => 333⁴⁴⁴ > 444³³³

\(A=333^{444}=111^{444}.3^{4.111}=111^{444}.81^{111}\)

\(B=444^{333}=111^{333}.4^{3.111}=111^{333}.64^{111}\)

Ta thấy *)444>333 nên \(111^{444}>111^{333}\)(1)

             *)81>64 nên \(81^{111}>64^{111}\)(2)

Từ (1) và (2) suy ra \(111^{444}.81^{111}>111^{333}.64^{111}\)

Vậy A>B

ta co 

A =333^444=(333^4)¹¹¹va B =(444^3)¹¹¹

^{ta duoc 333^4>444^3}

^{suy ra A>B}

^{vay A > B}

b,\(A=\frac{4}{3x-6}-\frac{x}{x^2-4}\)

\(A=\frac{4}{3\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{4x+8}{3\left(x-2\right)\left(x+2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{x-8}{3\left(x-2\right)\left(x+2\right)}\)

c, Thay x = 1 vào A ta đc

\(\frac{1-8}{3\left(1-2\right)\left(1+2\right)}=\frac{7}{9}\)

a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x-6\ne0\\x^2-4\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne6\\x^2\ne4\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne2\\x\ne\pm2\end{cases}\Leftrightarrow}x\ne\pm2}\)

Vậy A xác định khi \(x\ne\pm2\)

b) \(A=\frac{4}{3x-6}-\frac{x}{x^2-4}\left(x\ne\pm2\right)\)

\(\Leftrightarrow A=\frac{4}{3\left(x-2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4x+8}{3\left(x+2\right)\left(x-2\right)}-\frac{3x}{3\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow A=\frac{4x+8-3x}{3\left(x-2\right)\left(x+2\right)}=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\)

Vậy \(A=\frac{x+8}{3\left(x-2\right)\left(x+2\right)}\left(x\ne\pm2\right)\)

c) Thay x=1 (tmđk) vào A ta có: \(A=\frac{1+8}{3\left(1-2\right)\left(1+2\right)}=\frac{9}{-9}=-1\)

Vậy \(A=-1\)khi x=1

ơ câu b là 8 mũ 39 nha. Xin lỗi các bạn