Thực hiên phép tính :
\(\frac{25^3.5^5}{6.5^{10}}\)
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\(A=\frac{\left(5^2\right)^3\cdot5^5}{6\cdot5^{10}}\)
\(A=\frac{5^{2+3}\cdot5^5}{6\cdot5^{10}}\)
\(A=\frac{5^5\cdot5^5}{6\cdot5^{10}}\)
\(A=\frac{5^{5+5}}{6\cdot5^{10}}\)
\(A=\frac{5^{10}}{6\cdot5^{10}}\)
\(\Rightarrow A=\frac{1}{6}\)
\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^5.5^5}{6.5^{10}}=\frac{1}{6}.\)
\(\frac{5^3.5^8}{25^4}\)
\(=\frac{5^{11}}{25^4}\)
\(=\frac{48828125}{390625}\)
\(\frac{5^3.5^8}{25^4}=\frac{5^8.5^3}{\left(5^2\right)^4}\)
\(\Rightarrow\frac{5^8.5^3}{5^8}=5^3=125\)
Hk tốt
=2^12.3^4.(3-1)/2^12.3^5(3+1)-5^10.7^3.(1-7)/5^9.7^3.(1+2^3)
2/3.4-5.(-6)/9
=1/6-(-10/3)
1/6+10/3
7/2
a;47.69-31(-41)
=3243-(-1271)
=4514
b,(-14)(-25)+25(-11)
=350+(-275)
=75
c,(-127).57+(-43).127
=(-7239)+(-5461)
=-12700
e,(-25).68+(-34).(-250)
=(-1700)+8500
=6800
f,(-2)^3.(-3)^2
=(-8).9
=-72
Câu d, cậu ghi ko rõ đề nên tớ ko làm
NHỚ TICK NHA
\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
Phạm Hồng Anh k tớ nhé
a) \(\frac{6^{15}+6^{13}}{6^{12}}=\frac{6^{13}.\left(6^2+1\right)}{6^{12}}=6.37=222\)
b) \(\frac{2^6.5^7}{10^6}=\frac{2^6.5^6.5}{10^6}=\frac{10^6.5}{10^6}=5\)
a, Ta có: \(A=\frac{6^{15}+6^{13}}{6^{12}}=\frac{6^{12}\left(6^3+6\right)}{6^{12}}=6^3+6\) \(=222\)
b, Ta có: \(B=\frac{2^6.5^7}{10^6}=\frac{2^6.5^6.5}{10^6}=\frac{\left(2.5\right)^6.5}{10^6}\) \(=\frac{10^6.5}{10^6}=5\)
\(A=\frac{125^{100}}{5^{298}}\cdot\frac{2^{160}}{4^{80}}=>A=\frac{\left(5^3\right)^{100}}{5^{298}}\cdot\frac{2^{160}}{\left(2^2\right)^{80}}\)
\(=>A=\frac{5^{300}}{5^{298}}\cdot\frac{2^{160}}{2^{160}}=>A=5^2\cdot1=>A=25\)
\(A=\frac{125^{100}}{5^{298}}.\frac{2^{160}}{4^{80}}\)
\(=\frac{\left(5^3\right)^{100}}{5^{298}}.\frac{2^{160}}{\left(2^2\right)^{80}}\)
\(=\frac{5^{300}}{5^{298}}.\frac{2^{160}}{2^{160}}\)
\(=5^2.1=25\)
Vậy \(A=25\)
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}++\frac{1}{14.19}+......+\frac{1}{44.49}\)
\(A=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+.....+\frac{5}{44.49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)
Đặt \(B=\frac{1-3-5-7-.......47-49}{89}\)
\(B=\frac{1-\left(3+5+7+......+47+49\right)}{89}\)
Từ 3 -> 49 có: (49-3):2+1=24(số hạng)
=>\(3+5+7+....+47+49=\frac{\left(49+3\right).24}{2}=624\)
=>\(B=\frac{1-624}{89}=\frac{-623}{89}=-7\)
Vậy \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)
= \(\frac{5^6.5^5}{6.5^{10}}\) = \(\frac{5^{11}}{6.5^{10}}\)= \(\frac{5}{6}\)
\(\frac{25^3.5^5}{6.5^{10}}\)=\(\frac{5^{11}}{6.5^{10}}\)=\(\frac{5}{6}\)
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