mọi người giải cạng kẽ nha!

=-6,2-4,5=-10,7

Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a,=0,9:(4/5.1,25+1:1/3)
=0,9:(1+3)

=0,9:4

=0,225

b,=9,6:6-0,6

=1,6-0,6

=1

c,=7/2.11/4-7/2.5/4

=7/2.(11/4-5/4)

=7/2.3/2

=21/4

\(\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(-\frac{1}{3}-\frac{16}{15}\right)\)

\(=\frac{3}{5}:\left(-\frac{7}{30}\right)+\frac{3}{5}:\left(-\frac{7}{5}\right)\)

\(=\frac{3}{5}.\left(-\frac{30}{7}\right)+\frac{3}{5}.\left(-\frac{5}{7}\right)\)

\(=\frac{3}{5}\left(-\frac{30}{7}-\frac{5}{7}\right)\)

\(=\frac{3}{5}.\left(-\frac{35}{7}\right)\)

\(=-3\)

a,     (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))

=  (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\))  \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\)  - \(\dfrac{21}{60}\))

= - \(\dfrac{3}{80}\)  \(\times\) (- \(\dfrac{2}{3}\))

= \(\dfrac{1}{40}\) 

b, (-1)³ + (- \(\dfrac{2}{3}\))² : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)

=  -1³ +   \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)

= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1

= 0

3/15 chia cả mẫu cả tử cho 3 được 1/5+2/5=3/5 nhé

Mn giúp mh với đg cần gấp lắm ln🥺🥺

Mọi người giúp mình với

\(a,\dfrac{15^3}{5^4}\)

\(=\dfrac{\left(3\cdot5\right)^3}{5^4}\)

\(=\dfrac{3^3\cdot5^3}{5^4}\)

\(=\dfrac{3^3}{5}\)

\(=\dfrac{27}{5}\)

\(---\)

\(b,\dfrac{21^3}{7^4}\)

\(=\dfrac{\left(3\cdot7\right)^3}{7^4}\)

\(=\dfrac{3^3\cdot7^3}{7^4}\)

\(=\dfrac{3^3}{7}\)

\(=\dfrac{27}{7}\)

\(---\)

\(c,\dfrac{6^6}{3^8}\)

\(=\dfrac{\left(2\cdot3\right)^6}{3^8}\)

\(=\dfrac{2^6\cdot3^6}{3^8}\)

\(=\dfrac{2^6}{3^2}\)

\(=\dfrac{64}{9}\)

#\(Toru\)