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a)
\(4\left(x-1\right)-\left(3x+2\right)=16\)
=> 4x - 4 - 3x - 2 = 16
=> x - 6 = 16
=> x = 22
Vậy .......
b)
\(7\left(x-5\right)+5\left(15-x\right)=130\)
=> 7x - 35 + 75 - 5x = 130
=> 2x + 40 = 130
=> 2x = 90
=> x = 45
Vậy........
c)
\(30\left(x-2\right)-6\left(x-5\right)-22x=100\)
=> 30x - 60 - 6x + 30 - 22x = 100
=> 2x - 30 = 100
=> 2x = 130
=> x = 65
Vậy.....
d) \(x-17+x=x-23\)
=> 2x - 17 = x - 23
=> x = -6
Vậy .......
Đặt \(A=2^{2x}+2^{2x+1}+...+2^{2x+1918}\)
=>\(2\cdot A=2^{2x+1}+2^{2x+2}+...+2^{2x+1919}\)
=>\(A=2^{2x+1919}-2^{2x}\)
Theo đề, ta có; \(2^{2x+1919}-2^{2x}=2^{1923}-2^4\)
=>\(2^{2x}\cdot\left(2^{2019}-1\right)=2^4\left(2^{2019}-1\right)\)
=>2x=4
=>x=2
3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
\(1\dfrac{14}{15}-\dfrac{5}{17}-\dfrac{12}{17}-\dfrac{14}{15}\\ =\left(1\dfrac{14}{15}-\dfrac{14}{15}\right)-\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\\ =\left[1+\left(\dfrac{14}{15}-\dfrac{14}{15}\right)\right]-1=1-1=0\)
M = \(15.\left(\frac{1}{15.16}+\frac{1}{16.17}+...+\frac{1}{19.20}\right)\)
= \(15.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+...+\frac{1}{19}-\frac{1}{20}\right)\)
= \(15.\left(\frac{1}{15}-\frac{1}{20}\right)\)
= \(15.\frac{1}{60}\)= \(\frac{1}{4}\)\(< \frac{1}{3}\)
(=) \(M< \frac{1}{3}\)\(\left(đpcm\right)\)
Ta có: \(M=\frac{15}{15.16}+\frac{15}{16.17}+\frac{15}{17.18}+\frac{15}{18.19}+\frac{15}{19.20}\)
\(\Rightarrow M=15.\left(\frac{1}{15.16}+\frac{1}{16.17}+\frac{1}{17.18}+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(\Rightarrow M=15.\left(\frac{1}{15}-\frac{1}{16}+\frac{1}{16}-\frac{1}{17}+\frac{1}{17}-\frac{1}{18}+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(\Rightarrow M=15.\left(\frac{1}{15}-\frac{1}{20}\right)\)
\(\Rightarrow M=15.\frac{1}{60}=\frac{1}{4}\)
Ta thấy: \(\frac{1}{4}< \frac{1}{3}\Rightarrow M< \frac{1}{3}\)
Vậy \(M< \frac{1}{3}\)
Chúc bạn học tốt!
\(2^{2x+1}-15=17\\ =>2^{2x+1}=32=2^5\\ =>2x+1=5\\ =>2x=4\\ =>x=2\)
22x + 1 - 15 = 17
22x+1 = 17 + 15
22x+1 = 32
22x+1 = 25
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2.