2/3 = a + 36/81
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KV
3 tháng 12 2023
a) (38 - 60) + (20 - 38)
= 38 - 60 + 20 - 38
= (38 - 38) + (-60 + 20)
= 0 - 40
= -40
b) 75 - (20 + 75)
= 75 - 20 - 75
= (75 - 75) - 20
= 0 - 20
= -20
c) 32 + (60 - 32)
= 32 + 60 - 32
= (32 - 32) + 60
= 0 + 60
= 60
d) (81 - 36) - (81 - 36)
= 81 - 36 - 81 + 36
= (81 - 81) + (-36 + 36)
= 0 + 0
= 0
e) (2 + 4 + 6 + 8) - (1 + 3 + 5 + 7)
= 2 + 4 + 6 + 8 - 1 - 3 - 5 - 7
= (2 - 1) + (4 - 3) + (6 - 5) + (8 - 7)
= 1 + 1 + 1 + 1
= 4
f) (1 + 3 + 5 + 7 + ... + 99) - (2 + 4 + 6 + 8 + ... + 100)
= 1 + 3 + 5 + 7 + ... + 99 - 2 - 4 - 6 - 8 - ... - 100
= (1 - 2) + (3 - 4) + (5 - 6) + (7 - 8) + ... + (99 - 100)
= -1 - 1 - 1 - 1 - ... - 1 (50 chữ số 1)
= -50
15 tháng 10 2023
a: \(3^7\cdot27^5\cdot81^3=3^7\cdot3^{15}\cdot3^{12}=3^{34}\)
b: \(36^5:18^5=\left(\dfrac{36}{18}\right)^5=2^5=32\)
c: \(24\cdot5^2+5^2\cdot5^3=24\cdot25+25\cdot125=25\cdot149=3725\)
d: \(\dfrac{125^4}{5^8}=\dfrac{5^{12}}{5^8}=5^4=625\)
VN
Võ Ngọc Phương
VIP
15 tháng 10 2023
a) \(3^7.27^5.81^3\)
\(=3^7.\left(3^3\right)^5.\left(3^4\right)^3\)
\(=3^7.3^{15}.3^{12}\)
\(=3^{34}\)
b) \(36^5:18^5\)
\(=\left(\dfrac{36}{18}\right)^5\)
\(=2^5\)
c) \(24.5^5+5^2.5^3\)
\(=24.5^5+5^5\)
\(=5^5.\left(24+1\right)\)
\(=5^5.25\)
\(=5^5.5^2=5^7\)
d) \(125^4:5^8\)
\(=\left(5^3\right)^4:5^8\)
\(=5^{12}:5^8\)
\(=5^4\)
8 tháng 8 2023
B={x\(\in\)N|x=3k; 1<=k<=4}
C={x\(\in\)N|x=4*a2; 1<=a<=5}
D={x\(\in\)N|x=9*a2;1<=a<=4}
E={x\(\in\)N|x=4k; 0<=x<=4}
G={x\(\in\)N|x=(-3)^k; 1<=k<=4}
15 tháng 9 2023
a) \(A=\left\{x\in N|0\le x\le4\right\}\)
b) \(B=\left\{x\in N|x=4k;0\le k\le4;k\in N\right\}\)
c) \(C=\left\{x\in Z|x=\left(-3\right)^k;1\le k\le4;k\in N\right\}\)
d) \(D=\left\{x\in N|x=k^2;k=3a;1\le a\le4;a\in N\right\}\)
7 tháng 1 2024
a: \(A=2^{\dfrac{1}{3}}\cdot2^{\dfrac{2}{3}}=2^{\dfrac{1}{3}+\dfrac{2}{3}}=2^{\dfrac{3}{3}}=2^1=2\)
b: \(B=36^{\dfrac{3}{2}}=\left(6^2\right)^{\dfrac{3}{2}}=6^{2\cdot\dfrac{3}{2}}=6^3=216\)
c: \(C=36^{\dfrac{3}{2}}\cdot\left(\dfrac{1}{6}\right)^2=\left(6^2\right)^{\dfrac{3}{2}}\cdot\dfrac{1}{6^2}=\dfrac{6^{2\cdot\dfrac{3}{2}}}{6^2}=\dfrac{6^3}{6^2}=6\)
d: \(D=\sqrt{81}\cdot\left(\dfrac{1}{3}\right)^2=9\cdot\dfrac{1}{3^2}=9\cdot\dfrac{1}{9}=1\)
e: \(E=\left(3+2\sqrt{2}\right)^{50}\cdot\left(3-2\sqrt{2}\right)^{50}\)
\(=\left[\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)\right]^{50}\)
\(=\left(9-8\right)^{50}=1^{50}=1\)
f: \(F=120^{\sqrt{5}+1}\cdot120^{3-\sqrt{5}}\)
\(=120^{\sqrt{5}+1+3-\sqrt{5}}=120^4\)
g: \(G=\left(3+2\sqrt{2}\right)^{2019}\cdot\left(3\sqrt{2}-4\right)^{2018}\)
\(=\left(3+2\sqrt{2}\right)^{2018}\cdot\left(3\sqrt{2}-4\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=\left[\left(3+2\sqrt{2}\right)\left(3\sqrt{2}-4\right)\right]^{2018}\left(3+2\sqrt{2}\right)\)
\(=\left(9\sqrt{2}-12+12-8\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=\left(\sqrt{2}\right)^{2018}\cdot\left(3+2\sqrt{2}\right)=2^{\dfrac{1}{2}\cdot2018}\cdot\left(3+2\sqrt{2}\right)\)
\(=2^{1009}\cdot\left(3+2\sqrt{2}\right)\)
2/3 = a + 36/81
=> a = 2/3-26/81
=> a=28/81