tớ cũng đang học cái đấy nè

     225^10

a = _________  = 3^10

      75^10

       2^15 * 3^8             2^15 * 3^8

c=  _____________= --------------------  =2^3 * 3^5

      2^3 * 3^3 * 2^9       2^12 * 3^3

tớ có 1 bài giống y cậu nè

mình làm bài này rồi nè

a.

\(\frac{45^{10}\times5^{20}}{75^{15}}=\frac{\left(3^2\times5\right)^{10}\times5^{20}}{\left(3\times5^2\right)^{15}}=\frac{3^{20}\times5^{10}\times5^{20}}{3^{15}\times5^{30}}=3^5=243\)

b.

\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,8\right)^5}{\left(0,4\right)^5}\times\frac{1}{\left(0,4\right)}=\left(\frac{0,8}{0,4}\right)^5\times\frac{1}{\frac{4}{10}}=2^5\times\frac{5}{2}=2^4\times5=16\times5=80\)

c.

\(\frac{2^{15}\times9^4}{6^6\times8^3}=\frac{2^{15}\times\left(3^2\right)^4}{\left(2\times3\right)^6\times\left(2^3\right)^3}=\frac{2^{15}\times3^8}{2^6\times3^6\times2^9}=3^2=9\)

Chúc bạn học tốt ^^

b)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(2.0,4\right)^5}{\left(0,4\right)^6}=\frac{2^5.\left(0,4\right)^5}{\left(0,4\right)^6}=\frac{2^5}{0,4}=\frac{2^5}{\frac{2}{5}}=\frac{2^4}{5}=\frac{16}{5}\)

c)\(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^{12}}{3^6.2^6.2^9}=3^6\)

a)\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{15^{10}.3^{10}.5^{20}}{5^{15}.15^{15}}=\frac{3^{10}.5^5}{15^5}=\frac{3^{10}.5^5}{5^5.3^5}=3^5\)

Thanks bn!!

\(\frac{45^{10}\cdot5^{20}}{75^{15}}=\frac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(5^2\cdot3\right)^{15}}\)  

=\(\frac{3^{20}\cdot5^{10}\cdot5^{20}}{5^{30}\cdot3^{15}}=\frac{3^{20}\cdot5^{30}}{5^{30}\cdot3^{15}}=3^5=243\)  

k cho ah nha

\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)

\(\Leftrightarrow B=\frac{2}{15}+\frac{-1}{9}\)

\(\Leftrightarrow B=\frac{12}{90}+\frac{-10}{90}\)

\(\Leftrightarrow B=\frac{2}{90}\)

\(\Leftrightarrow B=\frac{1}{45}\)

\(B=\frac{2}{15}=\left(\frac{5}{9}+\frac{-6}{9}\right)\)

\(B=\frac{2}{15}+\frac{-1}{9}\)

\(B=\frac{18}{135}+\frac{-15}{135}\)

\(B=\frac{3}{135}\)

\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)

\(B=\frac{2}{15}+-\frac{1}{9}\)

\(B=\frac{1}{45}\)

\(Linh\)\(Yukiko\)

neu de bai bai 1 la tinh x+y thi mik lam cho

đăng từng này thì ai làm cho 

We have \(P=\frac{x^4+2x^2+2}{x^2+1}\)

\(\Rightarrow P=\frac{x^4+2x^2+1+1}{x^2+1}\)

\(=\frac{\left(x^2+1\right)^2+1}{x^2+1}\)

\(=\left(x^2+1\right)+\frac{1}{x^2+1}\)

\(\ge2\sqrt{\frac{x^2+1}{x^2+1}}=2\)

(Dấu "="\(\Leftrightarrow x=0\))

Vậy \(P_{min}=2\Leftrightarrow x=0\)

a/ (-3,2).\(\frac{-15}{64}\)+(0,8-2\(\frac{4}{5}\)):1\(\frac{23}{24}\)

=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(\(\frac{4}{5}\)-\(\frac{14}{5}\)):\(\frac{47}{24}\)

=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(-2):\(\frac{47}{24}\)

= \(\frac{3}{4}\)+\(\frac{-48}{47}\)

=\(\frac{-51}{188}\)

b/ 1\(\frac{13}{15}\).3.(0,5)\(^2\).3+(\(\frac{8}{15}\)-1\(\frac{19}{60}\)):1\(\frac{23}{24}\)

= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{8}{15}\)-\(\frac{79}{60}\)):\(\frac{47}{24}\)

= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

= \(\frac{28}{5}\).\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

= \(\frac{7}{5}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

= \(\frac{21}{5}\)+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)

= \(\frac{21}{5}\)+(\(\frac{-2}{5}\))

= \(\frac{19}{5}\)

mk làm hơi dài dòng chút 

CHÚC BẠN HỌC TỐT

=\(\left[\dfrac{\left(0,4.2\right)^5}{\left(0,4\right)^6}+\dfrac{2^9.2^6.3^8}{\left(3.2\right)^6.2^9}\right]=\left[\dfrac{\left(0,4\right)^5.2^5}{\left(0,4\right)^6}+\dfrac{2^6.3^8}{3^6.2^6}\right]\)

=\(\left[\dfrac{2^5}{0,4}+3^2\right]\)

=\(\left[80+9\right]=89\)

\(\left[\dfrac{\left(2.0,4\right)^5}{0,4,0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^6.2^9}\right]\div\dfrac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

\(=\left[\dfrac{2^5.0.4^5}{0,4.0,4^5}+\dfrac{2^{15}.3^8}{3^6.2^{15}}\right]\div3^5\)

\(=\left[\dfrac{2^5}{0,4}+3^2\right]\div243\)

\(=80+\left(3^5\div3^2\right)\)

\(=80+3^3\)

\(=80+27\)

\(=107\)