a)\(\dfrac{27^4.4^3}{9^5.8^2}\)

=\(\dfrac{3^{12}.2^6}{3^{10}.2^6}\)

=3\(^2\)=9

b)\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}\)

=\(\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}\)

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

\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=\dfrac{3^{12}}{3^{10}}=3^2=9\)

_________

\(\dfrac{3^{29}.4^{16}}{27^9.8^{11}}=\dfrac{3^{29}.\left(2^2\right)^{16}}{\left(3^3\right)^9.\left(2^3\right)^{11}}=\dfrac{3^{29}.2^{32}}{3^{27}.2^{33}}=\dfrac{1}{3^2.2}=\dfrac{1}{9.2}=\dfrac{1}{18}\)

Mình sửa phần cuối câu 2:

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

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{3^{10}.2^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{3^9.2^8.\left(3.2^2-1.1.5\right)}{3^8.2^{10}.\left(3.1+2^2\right)}\)

\(=\frac{3^9.2^8.7}{3^8.2^{10}.7}\)

\(=\frac{3}{2^2}=\frac{3}{4}\)

Bài làm :

\(A=\frac{6^{10}-3^9.2^8.5}{27^3.4^5+16^3.9^4}\)

\(=\frac{\left(2.3\right)^{10}-3^9.2^8.5}{\left(3^3\right)^3.\left(2^2\right)^5+\left(2^4\right)^3.\left(3^2\right)^4}\)

\(=\frac{2^{10}.3^{10}-3^9.2^8.5}{3^9.2^{10}+2^{12}.3^8}\)

\(=\frac{2^8.3^9.\left(2^2.3-5\right)}{3^8.2^{10}.\left(3+2^2\right)}\)

\(=\frac{3.7}{2^2.7}\)

\(=\frac{3}{4}\)

Học tốt

Câu 1 : \(1,321338308x10^{-4}\)

Câu 2 : \(1316,572106\)

Câu 3 : \(1,641302619x10^{-13}\)

Ủng hộ nhé,tớ đang âm.

mấy câu này dễ

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)

\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)

.......

~ Chúc học tốt ~ 

Ai ngang qua xin để lại 1 L - I - K - E

\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)

\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)

\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{6}-\frac{1}{24360}\)

\(3A=\frac{1353}{8120}\)

\(A=\frac{1353}{8120}:3\)

\(A=\frac{451}{8120}\)

Ta có:3A=\(\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+.............+\frac{3}{27.28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)

\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)

\(3A=\frac{1353}{8120}\Rightarrow A=\frac{451}{8120}\)

\(1.\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)

\(2.\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2}{3}\)

Ý 1:

\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)

Ý 2:

\(\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{\left(2^3\right)^5.3^{15}}{2^{14}.\left(3^4\right)^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2^{14}.2.3^{15}}{2^{14}.3^{15}.3}=\dfrac{2}{3}\)

\(\frac{3^8.2^6+3^8.2^8}{3^9.2^6-3^8.2^8}\)

=\(\frac{3^8.2^6.\left(1+2^2\right)}{3^8.2^6\left(3-2^2\right)}\)

=\(\frac{5}{-1}\)

=\(-5\)

\(\frac{3^8.2^6+9^4.4^4}{3^9.4^3-3^8.2^8}=\frac{3^8.2^6+3^8.2^8}{3^9.2^6-3^8.2^8}=\frac{3^8.2^6\left(1+2^2\right)}{3^8.2^6\left(3-2^2\right)}\)

=\(\frac{3^8.2^6.5}{3^8.2^6.\left(-1\right)}=\frac{5}{-1}=\frac{-5}{1}=5\)

xin lỗi mk viết nhầm =-5 chứ ko phải =5

CÂU 1 = -59/111

 CÂU 2 = 11/63

cảm ơn kết quả thì mik b òi nhưng mik cần cách làm

b)=1/5.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/44-1/49).2-1-3-5-7-...-49/89

=1/5.(1/4-1/49).2-(1+3+5+7...+49)/89

=1/5.45/196.2-625/89

=9/196.-623/89

=9/196.-7

=9/28

h cho mình nha ! Chúc bạn học tốt

\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)

\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)

\(=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)

\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)

\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)

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