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\(Q=\dfrac{8^5.\left(-5\right)^8-\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(Q=\dfrac{2^{15}.5^8+2^5.2^9.5^9}{2^{16}.5^7+4^8.5^8}\)
\(Q=\dfrac{2^{14}.5^8\left(2+5\right)}{2^{16}.5^7+2^{16}.5^8}\)
\(Q=\dfrac{2^{14}.5^8.7}{2^{16}.5^7\left(1+5\right)}\)
\(Q=\dfrac{5.7}{2^2.6}=\dfrac{35}{24}\)
\(\frac{8^5.\left(-5\right)^8+\left(-2\right)^5.10^9}{2^{16}.5^7+20^8}\)
\(=\frac{\left(2^3\right)^5.5^8+\left(-2\right)^5.\left(2.5\right)^9}{2^{16}.5^7+\left(2^2.5\right)^8}\)
\(=\frac{2^{15}.5^8+\left(-2\right)^5.2^9.5^9}{2^{16}.5^7+2^{16}.5^8}\)
\(=\frac{2^{15}.5^8-2^{14}.5^9}{2^{16}.5^7\left(1+5\right)}\)
\(=\frac{2^{14}.5^8\left(2-5\right)}{2^{16}.5^7.\left(1+5\right)}\)
\(=\frac{2^{14}.5^8.\left(-3\right)}{2^{16}.5^7.6}\)
\(=\frac{-5}{8}\)
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: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)
b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)
c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)
d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)
\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)
\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)
\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)
\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)
\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)
\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)
a: TH1: x<2
Pt sẽ là 5-x+2-x=5x
=>5x=-2x+7
=>x=1(nhận)
TH2: 2<=x<5
Pt sẽ là 5x=x-2+5-x=3
=>x=3/5(loại)
TH3: x>=5
Pt sẽ là 5x=x-5+x-2=2x-7
=>3x=-7
=>x=-7/3(loại)
b: \(A=\dfrac{2^6\cdot5^2+2^{11}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)
\(=\dfrac{2^6\cdot5^2\left(1+2^5\cdot5^7\right)}{2^{16}\cdot5^7\left(1+5\right)}=\dfrac{1+2^5\cdot5^7}{2^{10}\cdot5^5\cdot6}\)
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.
\(\dfrac{8^5\cdot\left(-5\right)^8+\left(-2\right)^5\cdot10^9}{2^{16}\cdot5^7+20^8}\)
\(=\dfrac{2^{15}\cdot5^8-2^{14}\cdot5^9}{2^{16}\cdot5^7+2^{16}\cdot5^8}\)
\(=\dfrac{2^{14}\cdot5^8\left(2-5\right)}{2^{16}\cdot5^7\cdot\left(1+5\right)}\)
\(=\dfrac{5\cdot\left(-3\right)}{4\cdot6}=\dfrac{-15}{24}=\dfrac{-5}{8}\)