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Bài 2:
\(A=\frac{8^5(-5)^8+(-2)^5.10^9}{2^{16}.5^7+20^8}\) \(=\frac{(2^3)^5(-5)^8+(-2)^5.2^9.5^9}{2^{16}.5^7+(2^2.5)^8}\)
\(=\frac{2^{15}.5^8-2^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{14}.5^8(2-5)}{2^{16}.5^7(1+5)}\)
\(=\frac{5(-3)}{2^2.6}=\frac{-5}{8}\)
Bài 3:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\)
Thay vào:
\(\frac{5a+3b}{5a-3b}=\frac{5bt+3b}{5bt-3b}=\frac{b(5t+3)}{b(5t-3)}=\frac{5t+3}{5t-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5dt+3d}{5dt-3d}=\frac{d(5t+3)}{d(5t-3)}=\frac{5t+3}{5t-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\) (đpcm)
Bài 4:
Ta có:
\(A=3+3^2+3^3+3^4+...+3^{100}\)
\(=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(=3(1+3+3^2+3^3)+3^5(1+3+3^2+3^3)+...+3^{97}(1+3+3^2+3^3)\)
\(=3.40+3^5.40+....+3^{97}.40\)
\(=120(1+3^4+....+3^{96})\vdots 120\)
Ta có đpcm.
a) \(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.10^9}{2^{16}.5^7+2.10^8}\)
\(=\frac{5-2^4.10}{2}\)
\(=5-8.10\)
\(=5-80\)
\(=-75\)
a) \(\frac{15^5.10^5}{6^6.25^6}\)= (15.10)^5/(6.25)^6=150^5/150^6=1/150
\(^{\frac{\left(5^4-5^3\right)^3}{125^4}=\frac{\left[5^3\cdot\left(5-1\right)\right]^3}{\left(5^3\right)^4}=\frac{\left[5^3\cdot4\right]^3}{5^3\cdot4}=\frac{\left(5^3\right)^3\cdot4^3}{5^{12}}=\frac{5^9\cdot4^3}{5^9\cdot5^3}=\frac{4^3}{5^3}}\)
\(A=\frac{4^6.9^5+69.120}{8^4.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3^2.115}{2^{12}.3^{12}+\left(2.3\right)^{11}}=\frac{3^2.2^3\left(115+2^9.3^8\right)}{6^{11}\left(6+1\right)}=\frac{115+2^9.3^8}{6^8.3.7}\)
\(B=\frac{10^4+5.10^3+5^4}{25}=\frac{\left(10^2\right)^2+2.5^2.10^2+\left(5^2\right)^2}{25}=\frac{\left(10^2+5^2\right)^2}{25}=\frac{125^2}{25}=\frac{25.625}{25}=625\)
\(C=\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{4^{10}.2^{10}+4^{10}}{4^4.4^7+4^4.2^4}=\frac{4^{10}\left(2^{10}+1\right)}{4^4.2^4\left(2^{10}+1\right)}=\frac{4^6}{2^4}=256\)
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
a) A=6.10^7+5.10^5+4.10^3+2.10
A= 6.10000000+5.100000+4.1000+2.10
A= 60000000+500000+4000+20
A= 60504020
b) B=12.10^8+17.10^7+5.10^4+3
B= 12.100000000+17.10000000+5.10000+3
B=1200000000+170000000+50000+3
B=1370050003