ai giỏi toán giup vs

(x+6)⁴=4096

(x+6)⁴=8⁴

==> x+6=8 hoặc x+6=—8

==> x=8–6 hoặc x=—8–6

==> x= 2 hoặc x=—14

2^x—3=128

2^x—3=2⁷

==> x—3=7

x=7+3

x=10

Ss: 2²⁰¹⁸ và 16⁹⁰⁰

Ta có 16⁹⁰⁰=(2⁴)⁹⁰⁰=2³⁶⁰⁰

Vì 2²⁰¹⁸<2³⁶⁰⁰

Nên 2²⁰¹⁸<2³⁶⁰⁰

17²⁰ và 4⁴⁰

Ta có 17²⁰=(17²)¹⁰=289¹⁰

4⁴⁰=(4⁴)¹⁰=256¹⁰

vì 289¹⁰>256¹⁰

Nên 17²⁰ > 4⁴⁰

Cách khác

Ta có 4⁴⁰=4^2.20=(4²)²⁰=16²⁰

Vì 16²⁰< 17²⁰

Nên 17²⁰> 4⁴⁰

333⁴⁴⁴ và 444³³³

Ta có 333⁴⁴⁴=(3.111)^4.111=3^4.111.111^4.111=81¹¹¹.111^4.111

444³³³=(4.111)^3.111=4^3.111.111^3.111=64¹¹¹.111^3.111

Vì 81¹¹¹.111^4.111> 64¹¹¹.111^3.111

Nên 333⁴⁴⁴> 444³³³

a) ta có: 7^10 < 7^14 = (7^2)^7 = 49^7 < 50^7

=> 7^10 < 50^7

b) ta có: 5^30 = (5^3)^10 = 125^10 > 124^10

=> 5^30 > 124^10

c) ta có: 9^21 = (9^3)^7=729^7

phần d thì mk ko bk, xl bn nha

a,<

b,>

c,=

^_^

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

Trả lời

4¹⁰⁰=2²⁰⁰

2²⁰²

Vậy 2²⁰⁰ < 2²⁰² hay 4¹⁰⁰ < 2²⁰²

3⁰ và 5⁸

3⁰ < 5⁸

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d, 

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

a, \(4^{100}=\left(2^2\right)^{100}=2^{200}< 2^{202}\)

\(\Rightarrow\text{ }4^{100}< 2^{202}\)

b, \(3^0=1< 5^8\)

\(3^0< 5^8\)

c, \(\left(0,6\right)^0=1\)

\(\left(-0,9\right)^6=\left(0,9\right)^6\)

\(\Rightarrow\text{ }\left(0,6\right)^0< \left(-0,9\right)^6\)

d,

e, \(8^{12}=\left(2^3\right)^{12}=2^{36}=2^{16}\cdot2^{20}=2^{16}\cdot\left(2^4\right)^5=2^{16}\cdot16^5\)

\(12^8=\left(2^2\cdot3\right)^8=2^{16}\cdot3^8=2^{16}\cdot\left(3^2\right)^4=2^{16}\cdot9^4\)

Vì \(2^{16}\cdot16^5>2^{16}\cdot9^4\text{ }\Rightarrow\text{ }8^{12}>12^8\)

Bạn có thể ghi cho tiết đề bài và bạn muốn làm gì cho bài đó được không?

\(2^{x-3}=128\Leftrightarrow2^{x-3}=2^7\Rightarrow x-3=7\Leftrightarrow x=10\)

\(\left(x+6\right)^4=4096\Leftrightarrow\left(x+6\right)^4=2^{12}=\left(2^3\right)^4=8^4\Rightarrow x+6=8\Leftrightarrow x=2\)

\(2^{2018}=\left(2^4\right)^{504}.2^2==16^{504}.4< 16^{900}\\ \)

\(17^{20}>16^{20}=\left(4^2\right)^{20}=4^{40}\)

\(3^{444}=\left(3^4\right)^{111}=81^{111}>64^{111}=\left(4^3\right)^{111}=4^{333}\\ \)

a/

\(2^{1050}=\left(2^2\right)^{525}=4^{525}< 5^{525}< 5^{540}\)

b/

\(2^{161}>2^{160}=\left(2^4\right)^{40}=16^{40}>13^{40}\)

c/

\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}>2^{55}=\left(2^5\right)^{11}=32^{11}>31^{11}\)

#)Giải :

\(A=\frac{20^{18}+1}{20^{19}+1}\)và \(B=\frac{20^{17}+1}{20^{18}+1}\)

\(A=\frac{20^{18}+1}{20^{18+1}+1}\)và \(B=\frac{20^{17}+1}{20^{17+1}+1}\)

\(A=\frac{1}{20+1}\)và \(B=\frac{1}{20+1}\)

\(A=\frac{1}{21}\)và \(B=\frac{1}{21}\)

\(\Rightarrow A=B\)

       #~Will~be~Pens~#

A>20¹⁸ +1+19/20¹⁹ +1+19

A>20¹⁸+20/20¹⁹+20

A>20(20¹⁷+1)/20(20¹⁸+1)

A>20¹⁷+1/20¹⁸+1

=>A>B

Chúc bạn học tốt

\(A=\frac{20^{18}+1}{20^{19}+1}\)

\(A< \frac{20^{18}+1+19}{20^{19}+1+19}=\frac{20^{18}+20}{20^{19}+20}=\frac{20\left(20^{17}+1\right)}{20\left(20^{18}+1\right)}=B\)

\(\Rightarrow A< B\)

                                     #Louis