a, \(-\dfrac{315}{540}\) = \(\dfrac{-315:45}{540:45}\) = \(\dfrac{-7}{12}\)        b,   \(\dfrac{25.13}{26.35}\) = \(\dfrac{25.13:5:13}{26.35:13:5}\) = \(\dfrac{5}{14}\)

c,     \(\dfrac{6.9-2.17}{63.3-119}\) = \(\dfrac{2.3.9-2.17}{7.9.3-7.17}\) = \(\dfrac{2.(27-17)}{7.(7-17)}\) = \(\dfrac{2}{7}\)

d, \(\dfrac{3.13-13.18}{15.40-80}\) = \(\dfrac{3.13(1-6)}{40.(15-2)}\) = \(\dfrac{-3.13.5}{40.13}\) = \(\dfrac{-15}{40}\) = \(\dfrac{-15:5}{40:5}\) = \(-\dfrac{3}{8}\)

A=9×11×(11² +1)×(11⁴ +1)×(11⁸ +1)

= 9 . 11 (  11⁶ + 11² + 11⁴ + 1 )  ( 11⁸ + 1)

= 9 . 11 ( 11¹⁴+ 11⁶+ 11¹⁰+ 11² 11¹²+11⁴+ 1)

Hơi tắt , ý kiến riêng , bạn nên nhờ thầy cô để có kết quả chuẩn xác hơn nha

Hok tốt

A= 99/120 x(11^2 -1)x(11^2 +1)x(11^4 +1)(11^8 +1) (11^2 -1 =120)

=99/120 x(11^4 -1)(11^4 +1)(11^8 +1)

=99/120 x(11^8 -1)(11^8 +1)

=99/120 x(11^16 -1)

Không nên tính kết quả cụ thể. Chúc bạn học tốt

để rút gọn đc thì 63 chia hết cho x+1

=>x+1 thuộc Ư(63)

mà Ư(63)=1,3,7,9,2163

=>x+1=1,3,7,9,21,63

x=0,2,6,8,20,62

bài 1: 

2(x^2-9).4(x^2-1)

=(2x^2-18)(4x^2-4)

=8x^4-8x^2-72x^2+72

=8x^4-80x^2+72

\(Bai1:2\left(x-3\right)\left(x+3\right)+4\left(x-1\right)\left(x+1\right)\)

\(=2\left(x^2-9\right)+4\left(x^2-1\right)\)

\(=2x^2-18+4x^2-4\)

\(=6x^2-22\)

\(Bai2:-\left(6x-1\right)\left(3-2x\right)+\left(3x-2\right)\left(4x-3\right)=17\)

\(\Leftrightarrow-\left(18x-12x^2-3+2x\right)+12x^2-9x-8x+6=17\)

\(\Leftrightarrow-18x+12x^2+3-2x+12x^2-9x-8x+6=17\)

\(\Leftrightarrow24x^2-37x+9-17=0\)

\(\Leftrightarrow24x^2-37x-8=0\)

Đề sai??

a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)

\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)

\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)

\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)

\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)

bạn xem lại đề đc k

hạng tử cuối là \(2^{2019}\)

đúng ko bạn

bạn lấy 2p-p thì ra nha

\(2;A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(\frac{1-x}{x+2}\right)\)

\(ĐKXĐ:\hept{\begin{cases}x^2-4\ne0\\1-x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne\pm2\\x\ne1\end{cases}}\)

\(a,A=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right).\frac{x+2}{1-x}\)

\(A=\left(\frac{x+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\right).\frac{x+2}{1-x}\)

\(A=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1-x}=\frac{-6}{\left(x-2\right)\left(1-x\right)}\)

b, Khi x = -4

\(A=\frac{-6}{\left(-4-2\right)\left(1+4\right)}=\frac{-6}{-6.5}=\frac{1}{5}\)

cảm ơn bạn

d: \(\left(x-2\right)\left(x^2+2x+4\right)\left(x+2\right)\left(x^2-2x+4\right)\)

\(=\left(x^3-8\right)\left(x^3+8\right)\)

\(=x^6-64\)