Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

a) (a mũ m)ⁿ = a mũ m.n

=> (a mũ m)ⁿ = (a^m)ⁿ = a^m.n

a mũ m.n = a^m.n

Vậy (a^m)ⁿ = a^m.n .

b) (a.b)mũ n = a mũ n . b mũ n

=> (a.b)mũ n = (a.b)ⁿ = aⁿ . bⁿ

a mũ n . b mũ n = aⁿ . bⁿ

Vậy (a.b)ⁿ = aⁿ .bⁿ .

Bài 1.

a) \(12^3.3^3=\left(12.3\right)^3=36^3.\)

b) \(2^5.8^4=2^5.\left(2^3\right)^4=2^5.2^{12}=2^{17}.\)

c) \(3^8.9^0.27^2=3^8.1.\left(3^3\right)^2=3^8.3^6=3^{14}.\)

d) \(2^4.5^4=\left(2.5\right)^4=10^4.\)

e) \(2^4.4^3=2^4.\left(2^2\right)^3=2^4.2^6=2^{10}.\)

Bài 2.

a) \(5^x=259\)

Vì 5 khi nâng lên luỹ thừa bậc mấy thì chữ số tận cùng của kết quả luôn bằng 5.

Mà 259 có tận cùng là 9

\(\Rightarrow5^x=259\) (vô lý)

\(\Rightarrow\) Phương trình vô nghiệm.

b) \(\left(7x-11\right)^3=2^5.5^2+260\)

\(\Leftrightarrow\left(7x-11\right)^3=800+260\)

\(\Leftrightarrow\left(7x-11\right)^3=1060\)

\(\Leftrightarrow7x-11=\sqrt[3]{1060}\)

\(\Leftrightarrow7x=\sqrt[3]{1060}+11\)

\(\Leftrightarrow x=\dfrac{\sqrt[3]{1060}+11}{7}\).

a)12 mũ 3 nhân 3 mũ 3 bằng=36 mũ 3

giúp mk đi mk vội lắm mai mk kiểm tra rồi các bạn ơi

a: \(3C=3+3^2+...+3^{n+1}\)

=>\(2C=3^{n+1}-1\)

hay \(C=\dfrac{3^{n+1}-1}{2}\)

b: \(m\cdot S=m+m^2+m^3+...+m^{n+1}\)

=>\(S\left(m-1\right)=m^{n+1}-1\)

hay \(S=\dfrac{m^{n+1}-1}{m-1}\)

ta co \(3^3=27\) > 25

theo de bai, ta co 25 < \(3^n=3^n\) > \(3^2\left(1\right)\)

ta co \(3^5=\) 243 < 250 < \(3^6\)

theo de bai ra ta co \(3^n\) < 250 \(\Rightarrow3^n\) < \(3^6\left(2\right)\)

tu \(\left(1\right)va\left(2\right)\),suy ra 25 < \(3^3,3^4,3^5\)< 250

\(\Rightarrow n\in\left\{3,4,5\right\}\)

vay \(n\in\left\{3,4,5\right\}\)

Làm ơn viết cái đề rõ hơn dc ko vậy?

-_- Viết ra đi cậu. Khó nhìn chết được.

\(2014a^2+b^2+c^2\) / \(a^2\) = \(a^2+2014b^2+c^2\) /b\(^2\) = \(a^2+b^2+2014c^2\) /c\(^2\)

P = \(2015a^2+b^2\) /c\(^2\) + \(2015b^2\) +\(c^2\) / a\(^2\) + 2015\(c^2+a^2\)/b\(^2\)