\(\frac{2^{10}.3^8-2^8.3^9.4}{2^{10}.3^6.9+2^{10}.3^9}=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}\left(3^9-3^9\right)}=\frac{3^8-3^9}{3^9-3^9}=\frac{3^8\left(1-3\right)}{3^9\left(1-1\right)}=\frac{3^8.\left(-2\right)}{3^9.0}=0\)

Bằng 0 nha bn

phải bằng- \(\frac{1}{2}\)

a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)

b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)

c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)

d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)

`a, -21/15 . (-10)/14 = 210/210=1`

`b, (-2/3)^2= 4/9`

`c, (-3)/4 . (-4)/5 . 16/9= 192/180=16/15`

`d, (-8)/3 . 5/6= -40/18=-20/9`

`e, 16/30 . 5/12= 8/15 . 5/12=40/180=2/9`

`f, 13/30 . (-1)/5= -13/150`

`g, 2/21 . 3/28= 6/588= 1/98`

`h, (-3/4)^3= -27/64`

a) \(\dfrac{2}{5}-\dfrac{3}{15}\)

\(=\dfrac{2}{5}-\dfrac{3:3}{15:3}\)

\(=\dfrac{2}{5}-\dfrac{1}{5}\)

\(=\dfrac{1}{5}\)

b) \(\dfrac{9}{27}-\dfrac{2}{9}\)

\(=\dfrac{9:3}{27:3}-\dfrac{2}{9}\)

\(=\dfrac{3}{9}-\dfrac{2}{9}\)

\(=\dfrac{1}{9}\)

c) \(\dfrac{18}{24}-\dfrac{4}{8}\)

\(=\dfrac{18:6}{24:6}-\dfrac{4:2}{8:2}\)

\(=\dfrac{3}{4}-\dfrac{2}{4}\)

\(=\dfrac{1}{4}\)

d) \(\dfrac{6}{16}-\dfrac{10}{64}\)

\(=\dfrac{6\times2}{16\times2}-\dfrac{10:2}{64:2}\)

\(=\dfrac{12}{32}-\dfrac{5}{32}\)

\(=\dfrac{7}{32}\)

=>\(D=7\left(\dfrac{5}{42\cdot37}+\dfrac{1}{42\cdot43}+\dfrac{6}{43\cdot49}+\dfrac{10}{49\cdot59}\right)\)

\(=7\left(\dfrac{1}{37}-\dfrac{1}{42}+\dfrac{1}{42}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{59}\right)\)

=7(1/37-1/59)

=7*22/2183

\(E=5\left(\dfrac{8}{37\cdot45}+\dfrac{2}{45\cdot47}+\dfrac{3}{47\cdot50}+\dfrac{9}{50\cdot59}\right)\)

\(=5\left(\dfrac{1}{37}-\dfrac{1}{45}+\dfrac{1}{45}-\dfrac{1}{47}+...+\dfrac{1}{50}-\dfrac{1}{59}\right)\)

=5(1/37-1/59)

=>D/E=7/5

nhớ cho cả hướng dẫn giải

a: \(A=\dfrac{16^5\cdot15^5}{2^{10}\cdot3^5\cdot5^4}=\dfrac{2^{20}\cdot3^5\cdot5^5}{2^{10}\cdot3^5\cdot5^4}=2^{10}\cdot5=5120\)

b: \(B=\dfrac{2^{15}\cdot3+2^{19}\cdot10}{2^{12}\cdot26}=\dfrac{2^{15}\left(3+2^4\cdot10\right)}{2^{13}\cdot13}=2^2\cdot\dfrac{163}{13}=\dfrac{652}{13}\)

Tính bình thường dc ko bn

\(C=\dfrac{2^6\cdot3^{10}}{3^9\cdot2^6}=3\\ D=\dfrac{3^{24}\cdot3^{10}}{3^{21}\cdot3^{11}}=\dfrac{3^{34}}{3^{32}}=3^2=9\\ F=\dfrac{2^{45}\cdot5^{14}}{5^{15}\cdot2^{47}}=\dfrac{1}{2^2\cdot5}=\dfrac{1}{20}\\ G=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{1\cdot5}{2}=\dfrac{5}{2}\)

C=3

D=9

F=1/20

G=5/2

Em ko giải chi tiết vì nó lâu

Mong thông cảm!

C= \(\dfrac{8^2.9^5}{3^9.4^3}\)

   = \(\dfrac{\left(2^3\right)^2.\left(3^2\right)^5}{3^9.\left(2^2\right)^3}\)

   = \(\dfrac{2^6.3^{10}}{3^9.2^6}\)

   = \(\dfrac{3^1}{1}\)

   = \(\dfrac{3}{1}\)

a.

\(\sqrt[3]{125}.\sqrt[3]{\frac{16}{10}}.\sqrt[3]{-0,5}=\sqrt[3]{125.\frac{16}{10}.(-0,5)}=\sqrt[3]{-100}\)

b.

\(=1+\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2})^3-1}=1+\sqrt[3]{2}-1=\sqrt[3]{2}\)

c.

\(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+\sqrt[3]{(\sqrt{3}+1)^3}=\sqrt{3}+\sqrt{3}+1=2\sqrt{3}+1\)

d.

\(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt[3]{(\sqrt{3}+1)^3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}+1}=\sqrt{3}+1\)

e.

Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)

Khi đó:

$a^3+b^3=4$

$ab=\frac{2}{3}$

$E^3=(a+b)^3=a^3+b^3+3ab(a+b)$
$E^3=4+2E$

$E^3-2E-4=0$
$E^2(E-2)+2E(E-2)+2(E-2)=0$

$(E-2)(E^2+2E+2)=0$

Dễ thấy $E^2+2E+2>0$ nên $E-2=0$

$\Leftrightarrow E=2$