So sánh
\(a,2^{30}+3^{30}+4^{30}v\text{à}3^{20}+6^{20}+8^{20}\)
\(b,2^{30}+3^{30}+4^{30}v\text{à}3.24^{10}\)
\(c,2^0+2^1+2^2+...+2^{50}v\text{à}2^{51}\)
So sánh\(2^{30}+3^{30}+4^{30}v\text{à}3.24^{10}\)
So sánh :
\(a,2^{30}v\text{à}3^{20}\)
\(b,5^{300}v\text{à}3^{500}\)
\(c,2^{24}v\text{à}3^{16}\)
\(d,\left(0,3\right)^{40}v\text{à}\left(0,1\right)^{20}\)
\(\text{a, }2^{30}=8^{10}\)
\(\text{ }3^{20}=\left(3^2\right)^{10}=9^{10}\)
\(\text{Vậy }2^{30}< 3^{20}\)
\(\text{b, }5^{300}=\left(5^3\right)^{100}=125^{100}\)
\(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(\text{Vậy }5^{300}< 243^{100}\)
\(\text{c, }2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(\text{Vậy ...}\)
So sánh :
\(10^{30}v\text{à}2^{100}\)
\(5^{300}v\text{à}3^{453}\)
\(29^{12}v\text{à}18^{17}\)
103và 2100
Ta có:1030=(103)10=100010
2100=(210)10=102410
Vì 1000<1024 nên 1030<2100
5300 và 3453
Ta có:5300=(52)150=25150
3453=(33)151=27151=27.27150
Vì 25 < 27.27 nên 5300<3453
nhớ k ch mình nhé
Câu 1: Chứng minh:
\(31.82+125.48+21.43=125.67=1500\)
Câu 2: So sánh:
1,\(\sqrt{51}-\sqrt{5}v\text{à}\sqrt{20}-\sqrt{6}\)
2,\(\sqrt{2}+\sqrt{8}v\text{à}\sqrt{3}+3\)
3,\(\sqrt{37}-\sqrt{14}v\text{à}6-\sqrt{15}\)
4,\(\sqrt{5}+\sqrt{10}v\text{à}5,3\)
So sánh:
a) 5^300 và 3^500
b) (-16)^11 và (-32)^9
c) (2^2)^3 và 2^2^3
d) 2^30 + 2^30 + 4^30 và 3^20 + 6^20 + 8^20
e) 4^30 và 3×24^10
g) 2^0 + 2^1 + 2^2 + 2^3 +...+ 2^50 và 2^51
so sánh
\(10^{30}v\text{à}2^{106}\)
\(333^{444}v\text{à}444^{333}\)
\(M=\frac{\text{2 . 6 . 10 + 4 . 12 . 20 + 6 . 18 . 30 + ..... + 20 . 60 . 100}}{\text{1 . 2 . 3 + 2 . 4 . 6 + 3 . 6 . 9 + ..... + 10 . 20 . 30}}\)
Rút gọn biểu thức trên nha.
\(M=\frac{2.6.10+4.12.20+...+20.60.100}{1.2.3+2.4.6+...+10.20.30}=\frac{2.6.10.1^3+2.6.10.2^3+...+2.6.10.10^3}{1.2.3.1^3+1.2.3.2^3+...+1.2.3.10^3}\)
\(=\frac{2.6.10.\left(1^3+2^3+...+10^3\right)}{1.2.3.\left(1^3+2^3+...+10^3\right)}=\frac{2.6.10}{1.2.3}=20\)
vậy M=20
So sánh:
a)\(2^{24}v\text{à}3^{16}\)
b)\(2^{300}v\text{à}3^{200}\)
c)\(71^5v\text{à}7^{20}\)
a) Ta có \(\hept{\begin{cases}2^{24}=\left(2^6\right)^4=64^4\\3^{16}=\left(3^4\right)^4=81^4\end{cases}}\)
Mà \(64< 81\)
\(\Rightarrow64^4< 81^4\)
\(\Rightarrow2^{24}< 3^{16}\)
b) Ta có \(\hept{\begin{cases}2^{300}=\left(2^3\right)^{100}=8^{100}\\3^{200}=\left(3^2\right)^{100}=9^{100}\end{cases}}\)
Mà 8 < 9
\(\Rightarrow8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
c) Ta có \(7^{20}=\left(7^4\right)^5=2401^5\)
Ta có 71 < 2401
\(\Rightarrow71^5< 2401^5\)
\(\Rightarrow71^5< 7^{20}\)
!! K chắc câu c
@@ Học tốt
Chiyuki Fujito
a) \(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
Ta thấy 8<9\(\Rightarrow8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
b) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
Thấy \(8< 9\Rightarrow8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)
c) \(7^{20}=\left(7^4\right)^5=2401^5\)
Ta thấy \(71< 2401\Rightarrow71^5< 2401^5\Rightarrow71^5< 7^{20}\)
Rút gọn phân số sau:
\(M=\frac{\text{2 . 6 . 10 + 4 . 12 . 20 + 6 . 18 . 30 + ..... + 20 . 60 . 100 }}{\text{1 . 2 . 3 + 2 . 4 . 6 + 3 . 6 . 9 + ..... + 10 . 20 . 30}}\)
\(M=\frac{2.6.10+4.12.20+6.18.30+...+20.60.100}{1.2.3+2.4.6+3.6.9+...+10.20.30}\)
\(=\frac{2.6.10.\left(1+2+3+...+10\right)}{1.2.3.\left(1+2+3+...+10\right)}\)
\(=20\)