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a, 230 + [32 + (x - 5)] =315
32 + (x-5) = 315 - 230
32 + (x-5) =85
x-5=85-32
x-5=53
x=53+5
x=58
a) 390-(x-7)=169:13
390-(x-7)=13
x-7=390-13
x-7=377
x=377-7
x=370
b)70-5.(x-3)=45
5.(x-3)=70-45
5.(x-3)=25
x-3=25:5
x-3=5
x = 5+3
x =8
c)(x-140):7=3^3-2^3.3
(x - 140) : 7 = 27 - 8 . 3
(x - 140) : 7 = 27 - 24
(x - 140) : 7 = 3
x - 140 = 3 . 7
x - 140 = 21
x = 21 + 140
x = 161
d) 2^x=32
=>2^x=2^5
=)x=5
e)6x^3 - 8 = 40
6x^3 = 48
x^3 = 8
x = 2
f)4x^3+15=47
4x^3 = 47 - 15
4x^3 = 32
x^3 = 32 : 4
x^3 = 8
x^3 = 23
x = 2
g)6x-5=5^48:5^46
6x-5=5^2
6x-5=25
6x =25+5
6x =30
x =30:6
x =5
h)10+2x=4^51:4^49
10+2x=4^2
10+2x=16
2x = 16 - 10
2x = 6
x = 6 : 2
x = 3
a) \(P=2^{100}-2^{99}-2^{98}-...-2^3-2^2-2\)
\(=2^{100}-\left(2+2^2+2^3+...+2^{99}\right)\)
\(A=2+2^2+2^3+...+2^{99}\)
\(2A=2^2+2^3+...+2^{100}\)
\(2A-A=\left(2^2+2^3+...+2^{100}\right)-\left(2+2^2+2^3+...+2^{99}\right)\)
\(A=2^{100}-2\)
\(P=2^{100}-\left(2^{100}-2\right)=2\)
Trả lời:
\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow1-\frac{1}{5x+6}=\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=1-\frac{2005}{2006}\)
\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2006}\)
\(\Rightarrow5x+6=2006\)
\(\Rightarrow5x=2000\)
\(\Rightarrow x=400\)
Vậy x = 400
Trả lời:
\(\frac{x}{2008}-\frac{1}{10}-\frac{1}{15}-\frac{1}{21}-...-\frac{1}{120}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}-\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)=\frac{5}{8}\)\(\frac{5}{8}\)
Đặt \(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\), ta được : \(\frac{x}{2008}-A=\frac{5}{8}\) (*)
\(\Rightarrow A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(\Rightarrow A=2\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\Rightarrow A=2\left(\frac{1}{4}-\frac{1}{16}\right)=2.\frac{3}{16}=\frac{3}{8}\)
Thay A vào (*) , ta có:
\(\frac{x}{2008}-\frac{3}{8}=\frac{5}{8}\)
\(\Rightarrow\frac{x}{2008}=1\)
\(\Rightarrow x=2008\)
Vậy x = 2008