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Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=K\)
\(\Rightarrow a=cK;b=dK\)
Khi đó: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(cK\right)^2+c^2}{\left(dK\right)^2+d^2}=\frac{c^2.K^2+c^2}{d^2.K^2+d^2}=\frac{c^2\left(K^2+1\right)}{d^2\left(K^2+1\right)}=\frac{c^2}{d^2}=\frac{ac}{bd}\)(Do \(\frac{a}{b}=\frac{c}{d}\))
Vậy: \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
\(\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}< 1\)
\(A=\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}\)
A=\(\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
A=\(1-\frac{1}{10^2}\)
A=\(1-\frac{1}{100}\)
A=\(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^22^2}+\frac{5}{2^23^2}+\frac{7}{3^24^2}+....+\frac{19}{9^210^2}< 1\)
\(a.\)
\(8^7-2^{18}\)
\(=\left(2^3\right)^7-2^{18}\)
\(=2^{21}-2^{18}\)
\(=2^{18}.2^3-2^{18}\)
\(=2^{18}\left(2^3-1\right)\)
\(=2^{18}.7\)
\(=2^{17}.7.2⋮14\)
Vậy \(8^7-2^{18}⋮14\)
\(b.\)
\(5^5-5^4+5^3\)
\(=5^3\left(5^2-5+1\right)\)
\(=5^3.21\)
\(=5^3.7.3⋮7\)
Vậy \(5^5-5^4+5^3⋮7\)
\(c.\)
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11⋮11\)
Vậy \(7^6+7^5-7^4⋮11\)
a)Ta có:\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)
=>\(7^6+7^5-7^4⋮55\)
b)\(A=1+5+5^2+...+5^{50}\)
\(5A=5\left(1+5+5^2+...+5^{50}\right)=5+5^2+5^3+...+5^{51}\)
\(5A-A=5+5^2+5^3+...+5^{51}-\left(1+5+5^2+...+5^{50}\right)\)
\(4A=5^{51}-1\)
\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)
a) \(7^6+7^5+7^4=7^4\left(7^2+7+1\right)\)
= \(7^4.55\)
Vậy: \(7^6+7^5+7^4\) chia hết cho 55.
b) A= \(1+5+5^2+5^3+5^4+.....+5^{50}\)
5A= 5+\(5^2+5^3+5^4+5^{51}\)
5A-A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-\left(1+5^2+5^3+5^4+......+5^{51}\right)\)
4A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-1-5-5^2-5^3-5^4-.......-5^{50}\)
= \(5^{51}-1\)
Vậy A= \(\left(5^{51}-1\right):4\)
Tick mk nha!
a)Ta có: \(VT=7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
Vậy \(7^6+7^5-7^4⋮55\)
b)\(A=1+5+...+5^{100}\)
\(5A=5\left(1+5+...+5^{100}\right)\)
\(5A=5+5^2+...+5^{101}\)
\(5A-A=\left(5+5^2+...+5^{101}\right)-\left(1+5+...+5^{100}\right)\)
\(4A=5^{101}-1\Rightarrow A=\frac{5^{101}-1}{4}\)
a) \(\frac{1}{7}+\frac{6}{7}:\frac{3}{7}\)
\(=\frac{1}{7}+\frac{6}{7}.\frac{7}{3}\) (nhân nghịch đảo)
\(=\frac{1}{7}+2\)
\(=\frac{15}{7}\)
b) \(\frac{4}{5}-\frac{1}{5}.\left(-3\right)\)
\(=\frac{4}{5}-\left(-\frac{3}{5}\right)\)
\(=\frac{7}{5}\)
c) \(\frac{3}{7}+\left(\frac{-5}{2}\right)-\left(-\frac{3}{5}\right)\)
\(=\frac{3}{7}-\left(-\frac{5}{2}\right)+\frac{3}{5}\)
\(=\frac{30}{70}+\frac{175}{70}+\frac{42}{70}\)
\(=\frac{30+175+42}{70}\)
\(=\frac{247}{70}\)
d) viết lại đề hộ mình nhé
a, \(7^6+7^5-7^4=7^4\left[7^2+4-1\right]=7^4\cdot55⋮55\)
b, \(A=1+5+5^2+5^3+...+5^{50}\)
\(\Rightarrow5A=5+5^2+5^3+5^4+...+5^{51}\)
\(\Rightarrow5A-A=\left[5+5^2+5^3+5^4+...+5^{51}\right]-\left[1+5+5^2+5^3+...+5^{50}\right]\)
\(\Rightarrow4A=5^{51}-1\Leftrightarrow A=\frac{5^{51}-1}{4}\)