a) \(\frac{2}{5}+\left(\frac{3}{11}+\frac{-2}{5}\right)\)

\(=\frac{2}{5}+\frac{3}{11}+\frac{-2}{5}\)

\(=\left(\frac{2}{5}+\frac{-2}{5}\right)+\frac{3}{11}\)

\(=0+\frac{3}{11}\)

\(=\frac{3}{11}\)

b) \(\frac{-5}{7}\cdot\frac{4}{13}+\frac{-5}{7}\cdot\frac{9}{13}+\frac{-2}{7}\)

\(=\frac{-5}{7}\left(\frac{4}{13}+\frac{9}{13}\right)+\frac{-2}{7}\)

\(=\frac{-5}{7}\cdot1+\frac{-2}{7}\)

\(=\frac{-5}{7}+\frac{-2}{7}\)

\(=-1\)

=))

a) 2/5 + (3/11 -2/5 ) = 2/5 - 2/5 + 3/11 = 3/11 

b) -5/7*4/13+(-5/7 ) *9/13 + (-2/7) =-5/7( 4/13 + 9/13 ) -2/7 = -1

a ) (2,8x - 32 ) : 2/3 = -90 

<=> 2,8x -32 = -60

<=> 2,8x = -28

<=> x = -10 

b ) 4/5 + 5/7 :x =1/6

<=> 5/7:x = -19/30

<=> x = -150/133

a)\(\left(2,8x-32\right):\frac{2}{3}=-90\)

\(2,8x-32=-90\cdot\frac{2}{3}\)

\(2,8x-32=-60\)

\(2,8x=-60+32\)

\(2,8x=-28\)

\(x=-28:2,8\)

\(x=-10\)

Vậy x = -10

b) \(\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)

\(\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}\)

\(\frac{5}{7}:x=\frac{-19}{30}\)

\(x=\frac{5}{7}:\frac{-19}{30}\)

\(x=\frac{-150}{133}\)

Vậy x = -150/133

=))

a) \(\frac{-5}{8}\cdot\left(\frac{4}{9}+\frac{-7}{12}\right)\)

\(=\frac{-5}{8}\cdot\frac{-5}{36}\)

\(=\frac{25}{288}\)

b) \(\left(-3,2\right)\cdot\frac{-15}{64}+\left(0,8-2\frac{4}{5}\right):3\frac{1}{2}\)

\(=\frac{3}{4}+\left(-2\right):\frac{7}{2}\)

\(=\frac{3}{4}+\frac{-4}{7}\)

\(=\frac{5}{28}\)

=))

\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}\)

\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)

\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}\)

\(=\frac{n}{2\left(3n+2\right)}\)

1) 27⁴ x 81¹⁰

= (3³)⁴ x (3⁴)¹⁰

= 3^{3 x 4} x 3^{4 x 10} 

= 3¹² x 3⁴⁰

= 3⁵²

2) a) n¹⁶ = n³

=> n¹⁶ - n³ = 0 

=> n³ x (n¹³ - 1) = 0

=> \(\orbr{\begin{cases}n^3=0\\n^{13}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}n^3=0^3\\n^{13}=1^3\end{cases}\Rightarrow}\orbr{\begin{cases}n=0\\n=1\end{cases}}}\)

Vậy n = 0 hoặc n = 1

\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)

\(\Leftrightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)

\(\Leftrightarrow10x=100\Leftrightarrow x=100:10=10\)

\(\left(2x-3\right)^{10x}=\left(3-2x\right)^{100}\)

\(\Rightarrow\left(2x-3\right)^{10x}=\left(2x-3\right)^{100}\)

\(\Rightarrow10x\Leftrightarrow x=100:10=10\)

~Study well~ :)

Ta có : 4ⁿ : 2^{n + 3} = 64

=> (2²)ⁿ : 2^{n + 3} = 64

=> 2²ⁿ : 2^{n + 3} = 64

=> 2^{2n - (n + 3)} = 64

=> 2^{2n - n - 3} = 2⁷

=> 2^{n - 3} = 2⁷

=> n - 3 = 7

=> n      = 7 + 3

=> n      = 10

Vậy n = 10 

Theo bài ra ta có : \(\frac{23+n}{40+n}=\frac{3}{4}\)

\(\Rightarrow4.\left(23+n\right)=3.\left(40+n\right)\)

\(\Rightarrow92+4n=120+3n\)

\(\Rightarrow4n-3n=120-92\)

\(\Rightarrow n=28\)

Vậy số n cần tìm là 28

=>( 23+a)/(40+a)=3/4

=>4*(23+a)=3*(40+a)

=> 92+4a=120+3a

=> a=28

=.số N đó là 28