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a: A(x)=0
=>2x-6=0
hay x=3
b: B(x)=0
=>3x-6=0
hay x=2
c: M(x)=0
\(\Rightarrow x^2-3x+2=0\)
=>x=2 hoặc x=1
d: P(x)=0
=>(x+6)(x-1)=0
=>x=-6 hoặc x=1
e: Q(x)=0
=>x(x+1)=0
=>x=0 hoặc x=-1
\(5A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{11}{5^{11}}.\)
\(4A=5A-A=\dfrac{1}{5}+\dfrac{1}{5^2}+\dfrac{1}{5^3}+...+\dfrac{1}{5^{11}}-\dfrac{11}{5^{12}}=B-\dfrac{11}{5^{12}}.\)
\(5B=1+\dfrac{1}{5}+\dfrac{1}{5^2}+...+\dfrac{1}{5^{10}}.\)
\(4B=5B-B=1-\dfrac{1}{5^{11}}\)
\(\Rightarrow4A=\dfrac{1}{4}\left(1-\dfrac{1}{5^{11}}\right)-\dfrac{1}{5^{12}}< \dfrac{1}{4}\Rightarrow A< \dfrac{1}{16}\)
Câu 2 :
b) \(\frac{x}{3}=\frac{-2}{9}\)
=> x = \(\frac{-2}{9}.3\) = \(\frac{-2}{3}\)
c) \(0,5x-\frac{2}{3}x=\frac{7}{12}\)
=> \(\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
=> \(-\frac{1}{6}\)x = \(\frac{7}{12}\)
=> x = \(\frac{7}{12}:\frac{-1}{6}\)
=> x =\(\frac{-7}{2}\)
Đề 1 câu 5 :
\(3B=3^2+3^3+3^4+...+3^{201}\)
\(\Rightarrow2B=3B-B=3^{201}-3\)
\(\Rightarrow2B+3=\left(3^{201}-3\right)+3=3^{201}\)
Do đó n = 201
Ta có:
\(3^{n+2}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=3^{n+1}\left(3+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}\cdot4+2^{n+2}\cdot3\)
\(=3^n\cdot3\cdot2\cdot2+2^{n+1}\cdot3\cdot2\)
\(=3^n\cdot6\cdot2+2^{n+1}\cdot6\)
\(=6\left(3^n\cdot2+2^{n+1}\right)⋮6\)
Vậy \(3^{n+2}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)