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\(a^3.a^9=a^{12};\left(a^5\right)^7=a^{35}\)
\(\left(a^6\right)^4.a^{12}=a^{24}.a^{12}=a^{36};5^6:5^4-3^3:3^2=25-3=22\)
\(4.5^2-2^5.3^0=100-32=68\)
ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)
mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)
\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)
\(\Rightarrow A< B\)
Câu hỏi của nguyen van nam - Toán lớp 6 - Học toán với OnlineMath
A=1+5+5^2+..+5^9/1+5+5^2+...+5^8
=1+5^9/1+5+5^2+...+5^8
B=1+3+3^2+..+3^9/1+3+3^2+..+3^8
=1+3^9/1+3+3^2+..+3^8
đặt A' =1+5+5^2+...+5^8
5A'=5+5^2+5^3+...+5^9
5A'-A'=5+5^2+5^3+...+5^9-5-1-5-5^2-...-5^8
4A'=5^9-1=>A'=(5^9-1):4
tương tự B'=(3^9-1):4
A=1+5^9/(5^9-1)/4=4.5^9/5^9-1
B=1+3^9/(3^9-1)/4=4.3^9/3^9-1
=> A<B
a) A = 82 = 64
B = 9 + 25 = 34
Vậy A > B
b) C = 82 = 64
D = 27 + 125 = 152
Vậy C < D
Bạn nào k mình mình sẽ k lại 3 cái nha ! Hứa đó
a/ A = (3 + 5)2 = 82 = 64
B = 32 + 52 = 9 + 25 = 34
vì 64 > 34
=> A > B
b/ C = (3 + 5)2 = 82 = 64
D = 33 + 53 = 27 + 125 = 152
64 < 152
=> C<D
1)
a)
\(\dfrac{-5}{11}\cdot\dfrac{4}{7}+\dfrac{-5}{11}\cdot\dfrac{3}{7}-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{8}{11}\\ =\dfrac{-5}{11}\cdot1-\dfrac{8}{11}\\ =\dfrac{-5}{11}-\dfrac{8}{11}\\ =\dfrac{-5}{11}+\dfrac{-8}{11}\\ =\dfrac{-13}{11}\)
b)
\(\left(\dfrac{2}{9}:\dfrac{5}{3}+\dfrac{1}{3}:\dfrac{5}{3}\right)^2-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =\left(\dfrac{2}{9}\cdot\dfrac{3}{5}+\dfrac{1}{3}\cdot\dfrac{3}{5}\right)^2-\left(\dfrac{-7}{24}\right)\\ =\left[\dfrac{3}{5}\cdot\left(\dfrac{2}{9}+\dfrac{1}{3}\right)\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{3}{5}\cdot\dfrac{5}{9}\right]^2+\dfrac{7}{24}\\ =\left[\dfrac{1}{3}\right]^2+\dfrac{7}{24}\\ =\dfrac{1}{9}+\dfrac{7}{24}\\ =\dfrac{29}{72}\)
c) \(14-\left|\dfrac{-3}{4}\right|-\left(\dfrac{1}{3}-\dfrac{5}{8}\right)\\ =14-\dfrac{3}{4}-\left(\dfrac{-7}{24}\right)\\ =14+\dfrac{-3}{4}+\dfrac{7}{24}\\ =13\dfrac{13}{24}\)
a: \(A=2\cdot9\cdot5=18\cdot5=90\)
Ư(90)={1;2;3;5;6;9;10;15;18;30;45;90}
b: \(A=4\cdot3\cdot5=12\cdot5=60\)
Ư(60)={1;2;3;4;5;6;10;12;15;20;30;60}
c: \(A=6\cdot25=150\)
Ư(150)={1;2;3;5;6;10;15;25;30;50;75;150}
d: \(A=8\cdot3=24\)
Ư(24)={1;2;3;4;6;8;12;24}
\(A=\left(9.25.32\right):\left(8.9\right)\\ =7200:72\\ =100\)
\(A=\left(3^2.5^2.2^5\right):\left(2^3.3^2\right)\\ =\left(9.25.32\right):\left(8.9\right)\\ =7200:72\\ =1000\)