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a) \(2^x\)\(-\)\(64\)\(=\)\(2^6\)
\(2^x\)\(-\)\(2^6\)\(=\)\(2^6\)
\(2^x\) \(=\)\(2^6\)\(+\)\(2^6\)
\(2^x\) \(=\) \(64\)\(+\)\(64\)
\(2^x\) \(=\) \(128\)
\(\Rightarrow\) \(2^x\) \(=\) \(2^7\)
b) \(\left(7x-11\right)\)\(^3\)\(=\)\(2^5\)\(.\)\(5^2\)\(+\)\(200\)
\(\left(7x-11\right)\)\(^3\)\(=\)\(32\)\(.\)\(25\)\(+\)\(200\)
\(\left(7x-11\right)\)\(^3\)\(=\) \(1000\)
\(\left(7x-11\right)\)\(^3\)\(=\) \(10^3\)
\(\Rightarrow\)\(7x-11\)\(=\)\(10\)
\(7x\) \(=\)\(10+11\)
\(7x\) \(=\) \(21\)
\(x\) \(=\) \(21\)\(:\)\(7\)
\(x\) \(=\) \(3\)
\(2^x-64=2^6\)
\(2^x-64=64\)
\(2^x=64+64\)
\(2^x=128\)
\(\)Vì \(128=2^7\) \(\Rightarrow x=7\)
(7x-11)3 = 25.52+ 23.52
(7x-11)3 = 23.52( 22+1)
(7x-11)3 = 23.53=(2.5)3
7x-11=10
x=3
(7x-11)3=25.52+200
(7x-11)3=32.25+200
(7x-11)3=800+200
(7x-11)3=1000
(7x-11)3=103
=> 7x-11=10
7x=10+11
7x=21
x=21/7
x=3
\(\Rightarrow\left(7n-11\right)^3=32\times25+200\)
\(\Rightarrow\left(7n-11\right)^3=1000\)
\(\Rightarrow7n-11=10\)
\(\Rightarrow7n=10+11\)
\(\Rightarrow n=21:7=3\)
\(\left(7n-11\right)^3=2^5.5^2+200\)
\(\left(7n-11\right)^3=32.25+200\)
\(\left(7n-11\right)^3=800+200\)
\(\left(7n-11\right)^3=1000\)
=>TH1:(7n-11)3=103
=>7n-11=10
=>7n=10+11=21
=>n=21:7=3
TH2: (7n-11)3=-103
=>7n-11=-10
=>7n=-10+11
=>7n=1
=>n=1:7=1/7
Mà n thuộc N nên n=3
Kết luận n=3
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)
\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)
Đến đây bn tự tính nhé !!
Nhận thấy \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
=> \(\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\forall x\)
Dấu "=" xảy ra <=> \(2x+\frac{1}{3}=0\Rightarrow x=-\frac{1}{6}\)
Vậy Min A = -1 <=> X = -1/6
a, \(\left(2x+\frac{1}{3}\right)^{44}\ge0\forall x\)
\(\Rightarrow\left(2x+\frac{1}{3}\right)^{44}-1\ge-1\)
Dấu "=" xảy ra <=> 2x+1/3=0 <=> x= -1/6
\(\left(7x-11\right)^3=2^5.5^2+200\)
=> \(\left(7x-11\right)^3=32.25+200\)
=>\(\left(7x-11\right)^3=800+200\)
=>\(\left(7x-11\right)^3=1000\)
=>\(\left(7x-11\right)^3=10^3\)
=> \(7x-11=10\)
=>\(7x=21\)
=>\(x=3\)
Vậy x = 3
\(\left(7x-11\right)^3=2^5\cdot5^2+200\)
\(\left(7x-11\right)^3=800+200\)
\(\left(7x-11\right)^3=1000\)
\(\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(7x=10+11\)
\(7x=21\)
\(x=21\div7\)
\(x=3\)
(7x−11)3 = 25 . 52 + 200
(7x−11)3 = 32 . 25 + 200
(7x−11)3 = 1000
(7x−11)3 = 103
7x−11 = 10
7x = 10 + 11
7x = 21
x = 21 : 7
x = 3
giúp mik ik mik tic cho