\(\left[\left(\frac{4}{3}\right)^{-2}.\left(\frac{3}{4}\right)^3\right]:\left(-\frac{2}{3}\right)^{-3}\)

\(=\frac{9}{16}.\frac{27}{64}:\left(-\frac{27}{8}\right)\)

\(=-\frac{9}{128}\)

[(3/4)^-2.(3/4)³]:(-2/3)^-3

=[16/9.3/64]:9/4

=1/12:9/4

=1/27

     \(\frac{1}{99.97}-\frac{1}{97.95}-........-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.........+\frac{1}{5.3}+\frac{1}{3.1}\right)\)

\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.......+\frac{1}{5.3}+\frac{1}{3.1}\right).\frac{2}{2}\)

\(=-\left(-\frac{2}{99.97}+\frac{2}{97.95}+......+\frac{2}{5.3}+\frac{2}{3.1}\right).\frac{1}{2}\)

\(=-\left(-\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{95}+.....+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}-1\right).\frac{1}{2}\)

\(=\left(\frac{1}{99}-1\right).\frac{1}{2}\)

\(=-\frac{98}{99}.\frac{1}{2}\)

\(=-\frac{49}{99}\)

Vì S lớn hơn dãy số trên 2^3 lần(mỗi thừa số lớn hơn 2^3 lần)

Nên S = 3025x2^3=3025x8=24200

Vậy S = 24200

nha

biết tổng trên là 3025

\(\left[\left(0,15\right)^3\right]^0+\left[\left(\frac{1}{7}\right)^2\right]^{-1}.\left[\left(-2\right)^3\right]^2:2^5\)

\(=1+49.64:32\)

\(=1+98\)

\(=99\)

[ ( 0,15 )³ ]⁰ + [ ( 1/7 )² ]^-1 . [ ( -2 )³ ]² : 2⁵

= 1 + 49 . 64 : 32

= 1 + 98

= 99

Xét trường hợp:

\(3^{41}:5^{31}=\left(3^4:5^3\right)^{10}.\frac{3}{5}\)

Vậy \(3^4:5^3=\frac{81}{125}< 1\)

Mà \(\frac{3}{5}< 1\)

Vậy \(3^{41}:5^{31}< 1\)

Nên: \(5^{31}>3^{41}\)

3^41 lớn hơn

ta có:

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

=> \(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)và \(x-2y+3z=14\)

\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)

\(=\frac{x-1-2y-4+3z+9}{8}=\frac{14-1-4+9}{8}=\frac{18}{8}=\frac{9}{4}\)

+) \(\frac{x-1}{2}=\frac{9}{4}\)=>  \(x-1=\frac{9}{2}\) =>   \(x=\frac{11}{2}\)

+) \(\frac{2y-4}{6}=\frac{9}{4}\)=>  \(2y-4=\frac{27}{2}\)=> \(2y=\frac{35}{2}\)=> \(y=\frac{35}{4}\)

+) \(\frac{3z-9}{12}=\frac{9}{4}\)=> \(3z-9=27\)=> \(3z=36\)=> \(z=12\)

Vậy x=

       y=

       z=

\(x^2-4x=0\)

\(\Rightarrow x\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

\(x^2-5x-6=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)

x²-4x=0

=>x(x-4)=0

=>x-4=0:x

=>x-4=0

=>x=4

sai đề rồi bạn ơi