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a) 5 + 45(2x - 1) = 10
45(2x - 1) = 10 - 5
45(2x - 1) = 5
2x - 1 = 5 : 45
2x - 1 = 1/9
2x = 1/9 + 1
2x = 10/9
x = 10/9 : 2
x = 5/9
b) 54 : (2ˣ⁻³ + 1) + 3 = 9
54 : (2ˣ⁻³ + 1) = 9 - 3
54 : (2ˣ⁻³ + 1) = 6
2ˣ⁻³ + 1 = 54 : 6
2ˣ⁻³ + 1 = 9
2ˣ⁻³ = 9 - 1
2ˣ⁻³ = 8
2ˣ⁻³ = 2³
x - 3 = 3
x = 3 + 3
x = 6
c) 14 + 36 : 3ˣ⁻⁵ = 18
36 : 3ˣ⁻⁵ = 18 - 14
36 : 3ˣ⁻⁵ = 4
3ˣ⁻⁵ = 36 : 4
3ˣ⁻⁵ = 9
3ˣ⁻⁵ = 3²
x - 5 = 2
x = 2 + 5
x = 7
a: =>45(2x-1)=5
=>2x-1=1/9
=>2x=10/9
=>x=5/9
b: =>\(\dfrac{54}{2^{x-3}+1}=6\)
=>\(2^{x-3}+1=9\)
=>\(2^{x-3}=8\)
=>x-3=3
=>x=6
c: \(14+36:3^{x-5}=18\)
=>\(\dfrac{36}{3^{x-5}}=18-14=4\)
=>\(3^{x-5}=9\)
=>x-5=2
=>x=7
a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)
b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)
\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)
Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
c, Câu hỏi của truong nguyen kim
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) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)