2^x+3.4^2=64
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LD
1
17 tháng 9 2015
5^2x-1=5^3.5^4
5^2x-1=5^7
=>2x-1=7
còn lại bn lm tiếp
Câu b hình như bn vik sai đề bài , đề bài đúng phải thế này :
7^x-3.4=296
7^x-12=296
7^x =296+12
7^x = 308
7^x = 7^3
=>x=3
c) 2^x.2^x+3=64^2:2^5
(2^x)^2 +3 = (2^6)^2 : 2^5
(2^x)^2 +3 = 2^12 : 2^5
(2^x.2) +3= 2^7
M
2
M
1
DP
31 tháng 7 2017
\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{x\left(x+1\right)}=\frac{64}{13}\)
\(\Leftrightarrow5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{64}{13}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{13}\div5\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{64}{65}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{65}\)
\(\Rightarrow x+1=65\Rightarrow x=65-1=64\)
\(\text{Vậy }x=64\)
4 tháng 10 2020
a) Ta có: \(\left(2^2\right)^3\cdot4^5\)
\(=2^6\cdot2^{10}\)
\(=2^{16}=65536\)
b) Ta có: \(\left[\left(-4\right)^2\right]^2\cdot6\)
\(=16^2\cdot6\)
\(=256\cdot6=1536\)
c) Ta có: \(\frac{16}{25}\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^2\cdot\left(\frac{4}{5}\right)^3\)
\(=\left(\frac{4}{5}\right)^5\)
\(=\frac{1024}{3125}\)
d) Ta có: \(\left(\frac{121}{64}\right)^2\cdot\left(-\frac{64}{11}\right)^2\)
\(=\frac{121^2}{64^2}\cdot\frac{64^2}{11^2}\)
\(=11^2=121\)
e) Ta có: \(\left[\left(-3\right)^3\right]^3\cdot271:125\)
\(=\left(-27\right)^3\cdot\frac{271}{125}\)
\(=\frac{-5334093}{125}\)
6 tháng 7 2016
Ta có: \(\frac{1}{2}+\frac{1}{3}< 2\cdot\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< 4\cdot\frac{1}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+...+\frac{1}{15}< 8\cdot\frac{1}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+...+\frac{1}{31}< 16\cdot\frac{1}{16}=1\)
\(\frac{1}{32}+\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}< 32\cdot\frac{1}{32}=1\)
Cộng từng vế của các BĐT trên ta có:
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 5\)
\(\Leftrightarrow64+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 69\)
\(\Leftrightarrow1+\frac{1}{1}+1+\frac{1}{2}+1+\frac{1}{3}+...+1+\frac{1}{63}< 69\)
\(\Leftrightarrow\frac{2}{1}+\frac{3}{2}+\frac{4}{3}+...+\frac{64}{63}< 69\)
\(\Leftrightarrow\frac{2^2}{1\cdot2}+\frac{3^2}{2\cdot3}+\frac{4^2}{3\cdot4}+...+\frac{64^2}{63\cdot64}< 69\)đpcm
Cho Linh xin 2 k nào :D
CT
0
8 tháng 9 2015
a) 5x - x = 64 \(\Rightarrow\) 4x = 64 \(\Rightarrow\) x = 16
b) \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
c) \(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
d) \(C=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\left(1-\frac{1}{99}\right)\)
\(=\frac{1}{2}\cdot\frac{98}{99}\)
\(=\frac{49}{99}\)
TG
1
X
9 tháng 2 2021
ĐKXĐ: \(x\ne0;x\ne-1\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)
\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)
\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)
\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)
\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)
Vậy x = 2009
ta có :
\(2^x+3.4^2=64\Leftrightarrow2^x+3.16=64\)
\(\Leftrightarrow2^x+48=64\Leftrightarrow2^x=64-48=16=2^4\)
vậy x =4