\(A=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^2\cdot9^2}=\dfrac{2^{13}\cdot3^7}{2^{15}\cdot3^6}=\dfrac{3}{4}\)

\(C=27\cdot\left(-\dfrac{3}{2}\right)^{-5}\cdot\left(-\dfrac{2}{5}\right)^{-4}:\left(\dfrac{2}{125}\right)^{-1}\)

\(=27\cdot\dfrac{-32}{243}\cdot\dfrac{625}{16}\cdot\dfrac{2}{125}\)

\(=\dfrac{-32}{9}\cdot\dfrac{1}{8}\cdot5\)

\(=-\dfrac{20}{9}\)

c: Ta có: \(\left(\dfrac{1}{2}\right)^{2x+1}=\dfrac{1}{8}\)

\(\Leftrightarrow2x+1=3\)

\(\Leftrightarrow2x=2\)

hay x=1

d: Ta có: \(\left(-\dfrac{1}{3}\right)^{x+3}=\dfrac{1}{81}\)

\(\Leftrightarrow x+3=4\)

hay x=1

xong r còn j nữa

tổng của 3 số liên tiếp chia hết cho 6

đề ktr?

bạn có thể kết bạn với mình ko

a,(-23).(-3).4.(-7)

= 39.4.(-7)

= 156.(-7)

= 1092

b, |-35| +(-|15|)

= 35 + (-15)

= 20

c, 125 . (-25)+25 . 225

= -125 . 25+ 25. 225

= 25.(-125+225)

= 25 . 100

=2500

8 - 3n = 11 - (3n + 3 ) = 11 - 3(n+1)

Mà 3(n+1) chia hết n+1

=> 11 chia hết n+1

Với n+1 = -11 => n = -12

Với n+1 = -1 => n = -2

Với n+1 = 1 => n = 0

Với n+1 = 11 => n = 10

Vậy n thuộc {-12 ; -2 ; 0 ; 10}

c) Để A>-1 thì A+1>0

\(\Leftrightarrow\dfrac{1-x}{x+1}+1>0\)

\(\Leftrightarrow\dfrac{1-x+x+1}{x+1}>0\)

\(\Leftrightarrow\dfrac{2}{x+1}>0\)

mà 2>0

nên x+1>0

hay x>-1

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)

a) Ta có: \(A=\left(\dfrac{x+1}{x-1}-\dfrac{1-x}{x+1}+\dfrac{4x^2}{1-x^2}\right):\dfrac{2x^2-2}{x^2-2x+1}\)

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

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

\(=\dfrac{-2x^2+2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{-2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2\left(x+1\right)}\)

\(=\dfrac{-2\cdot\left(x-1\right)}{2\left(x+1\right)}\)

\(=\dfrac{1-x}{x+1}\)

ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Để A=2 thì \(\dfrac{1-x}{x+1}=2\)

\(\Leftrightarrow1-x=2\left(x+1\right)\)

\(\Leftrightarrow1-x-2x-2=0\)

\(\Leftrightarrow-3x=1\)

hay \(x=-\dfrac{1}{3}\left(thỏa\right)\)