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= ( 2^3 . 5 . 7 ) .( 5^2.7^3)/ (2^2 . 5^2 . 7^4)
= ( 2^3 . 5 . 7 ) .(1/7.4)
= ( 8 . 5 . 7) .1/ 28
= 280. 1/28
= 10
\(\dfrac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\dfrac{2^3.5^37^4}{2^2.5^27^4}=2.3=6\)
\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
\(=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
\(=\frac{2.5}{1}=10\)
Ta có: \(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
= \(\frac{2^3.5.5^2.7.7^3}{2.2.5.5.7^2.7^2}\)=\(\frac{2^2.5^2.7^4.2.5}{2^2.5^2.7^4}\)= 2.5 = 10
(32.5.79).(35.53):(33.52.75)2
=(3².5.79).(35.5³):(36.5⁴.710)
=(32.35:36).(5.53:54).(79.710)
=32+5-6. 51+3-4. 79+10
=31. 5 . 719
\(A=1+7+7^2+7^3+...+7^{200}\)
\(\Rightarrow7A=7+7^2+7^3+...+7^{201}\)
\(\Rightarrow7A-A=\left(7+7^2+...+7^{201}\right)-\left(1+7+7^2+...+7^{200}\right)\)
\(\Rightarrow6A=7^{201}-1\)
\(\Rightarrow A=\frac{7^{201}-1}{6}\)
\(B=5^1+5^3+5^5+...+5^{101}\)
\(\Rightarrow5^2B=5^3+5^5+5^7+...+5^{103}\)
\(\Rightarrow25B-B=\left(5^3+5^5+...+5^{103}\right)-\left(5+5^3+...+5^{101}\right)\)
\(\Rightarrow24B=5^{103}-5\)
\(\Rightarrow B=\frac{5^{103}-5}{24}\)
\(D=1+a+a^2+a^3+...+a^n\)
\(\Rightarrow aD=a+a^2+a^3+...+a^{n+1}\)
\(\Rightarrow aD-D=\left(a+a^2+...+a^{n+1}\right)-\left(1+a+a^2+...+a^n\right)\)
\(\Rightarrow\left(a-1\right)D=a^{n+1}-1\)
\(\Rightarrow D=\frac{a^{n+1}-1}{a-1}\)
(23.5.7)(52.7)
=(5.7)(23.5)
=35(8.5)
=35.40
=140
(2.5.72)2
=(10.49)2
=4902
=240100
(23x5x7)x(52x7)
=(8x5x7)x(25x7)
=280x175
=49000