1. So Sánh
a, \(25^{15}\)và \(8^{10}.3^{30}\)
b, \(\frac{4^{15}}{7^{30}}\) và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
So sánh
a,\(25^{15}\)và \(8^{10}.3^{30}\)
b\(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
a) Ta có: \(25^{15}=\left(5^2\right)^{15}=5^{30}\)
\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}\)\(=2^{30}.3^{30}=6^{30}\)
Vì \(5^{30}< 6^{30}\)nên \(25^{15}< 8^{10}.3^{30}\)
b) Ta có: \(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Vì \(2^{30}< 3^{30}\)nên \(\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\)hay \(\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
_Học tốt_
so sánh: \(\frac{4^{15}}{7^{10}}\) và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
Ta có : \(\frac{4^{15}}{7^{10}}=\frac{\left(2^2\right)^{15}}{7^{10}}=\frac{2^{30}}{7^{10}}\)
\(\frac{8^{10}.3^{30}}{7^{30}.1^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}}=\frac{2^{30}.3^{30}}{7^{30}}=\frac{\left(2.3\right)^{30}}{7^{30}}=\frac{6^{30}}{7^{30}}\)
Mà : \(\frac{2^{30}}{7^{10}}=\frac{\left(2^3\right)^{10}}{7^{10}}=\frac{8^{10}}{7^{10}}\)
\(\frac{6^{30}}{7^{30}}=\frac{\left(6^3\right)^{10}}{\left(7^3\right)^{10}}=\frac{216^{10}}{343^{10}}\)
Vì : \(\frac{8}{7}>\frac{216}{343}\Rightarrow\frac{8^{10}}{7^{10}}>\frac{216^{10}}{343^{10}}\)
\(\Rightarrow\frac{4^{15}}{7^{10}}>\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
giúp mình vs
cho n là số tự nhiên
a, (n+ 10) (n+ 15) chia hết cho 2
b, n (n+ 1) (n+2) chia hết cho 2 và 3
c, n (n+ 1) (2n+1) chia hết cho 2 và 3
So sánh:
\(\frac{4^{15}}{7^{30}}VS\frac{8^{10}×3^{30}}{7^{30}×4^{15}}\)
So sánh: \(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}\)
Ai nhanh nhất, đúng nhất mk tick cho.
Ta có:
\(VT:\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\)
\(VP:\frac{8^{10}\cdot3^{30}}{7^{30}.4^{15}}=\frac{\left(2^3\right)^{10}.3^{30}}{7^{30}.\left(2^2\right)^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{3^{30}}{7^{30}}\)
Ta thấy :\(\frac{2^{30}}{7^{30}}vs\frac{3^{30}}{7^{30}}\)có:
\(\orbr{\begin{cases}2^{30}< 3^{30}\\7^{30}=7^{30}\end{cases}\Rightarrow\frac{2^{30}}{7^{30}}< \frac{3^{30}}{7^{30}}\Leftrightarrow\frac{4^{15}}{7^{30}}< \frac{8^{10}.3^{30}}{7^{30}.4^{15}}}\)
Chúc bn hok tốt
So sánh \(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}\cdot3^{30}}{7^{30}\cdot4^{15}}\)
\(\frac{4^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}\) và \(\frac{8^{10}.3^{30}}{7^{30}.4^{15}}=\frac{2^{30}.3^{30}}{7^{30}.2^{30}}=\frac{2^{30}}{7^{30}}\)Vậy hai vế bằng nhau
so sánh \(\frac{4^{15}}{7^{30}}\) và \(\frac{8^{10}.3^{10}}{7^{30}.4^{15}}\)
Ta có: \(\frac{4^{15}}{7^{30}}\)=\(\frac{\left(2^2\right)^{15}}{7^{30}}\)=\(\frac{2^{30}}{7^{30}}\)=\(\frac{\left(2^3\right)^{10}}{7^{30}}\)=\(\frac{8^{10}}{7^{30}}\)
\(\frac{8^{10}.3^{10}}{7^{30}.4^{15}}\)=\(\frac{\left(2^3\right)^{10}.3^{10}}{7^{30}.\left(2^2\right)^{15}}\)=\(\frac{2^{30}.3^{10}}{7^{30}.2^{30}}\)=\(\frac{3^{10}}{7^{30}}\)
Vì 810>310 \(\Rightarrow\)\(\frac{8^{10}}{7^{30}}\)>\(\frac{3^{10}}{7^{30}}\)
Hay \(\frac{4^{15}}{7^{30}}\)>\(\frac{8^{10}.3^{10}}{7^{30}.4^{15}}\)
So sánh:
\(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}\times3^{30}}{7^{30}\times4^{15}}\)
So sánh:
\(\frac{4^{15}}{7^{30}}\)và \(\frac{8^{10}\times3^{30}}{7^{30}\times4^{15}}\)
Ta có:
\(\frac{4^{15}}{7^{30}}=\frac{\left(2^2\right)^{15}}{7^{30}}=\frac{2^{30}}{7^{30}}=\left(\frac{2}{7}\right)^{30}\)
\(\frac{8^{10}x3^{30}}{7^{30}x4^{15}}=\frac{\left(2^3\right)^{10}x3^{30}}{7^{30}x\left(2^2\right)^{15}}=\frac{2^{30}x3^{30}}{7^{30}x2^{30}}=\frac{3^{30}}{7^{30}}=\left(\frac{3}{7}\right)^{30}\)
Nhận thấy, 2 số đều có cùng số mũ mà \(\frac{3}{7}>\frac{2}{7}\)
=> \(\frac{8^{10}x3^{30}}{7^{30}x4^{15}}>\frac{4^{15}}{7^{30}}\)
Bài 1: So sánh
a) \(-2^{30}\) và \(-3^{30}\)
b) \(35^5\) và \(6^{10}\)
Bài 2: Tính giá trị biểu thức
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
b) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)
\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)
Bài 2:
a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)
\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)
\(=-\dfrac{3}{5}\)
b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)
\(\Leftrightarrow8x-1=5\)
\(\Leftrightarrow8x=6\)
hay \(x=\dfrac{3}{4}\)
Bài 1:
a: \(-2^{30}=-8^{10}\)
\(-3^{30}=-27^{10}\)
mà 8<27
nên \(-2^{30}>-3^{30}\)
b: \(35^5=35^5\)
\(6^{10}=36^5\)
mà 35<36
nên \(35^5< 6^{10}\)