Câu 1: Tìm x biết
a) \(-\dfrac{2}{3}\)\(\left(x-\dfrac{1}{4}\right)\) = \(\dfrac{1}{3}\left(2x-1\right)\) b) \(\dfrac{1}{5}.2^x+\dfrac{1}{3}.2^{x+1}=\dfrac{1}{5}.2^7+\dfrac{1}{3}.2^8\)
Câu 2: a) Cho A = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}......\dfrac{9999}{10000}\)
So sánh A vs 0,01
b) Chứng tỏ rằng: \(\left[\left(1+2+3+....+n\right)-7\right]⋮̸10\)
Câu 1:
a) \(-\dfrac{2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Rightarrow-\dfrac{2}{3x}+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{1}{3}\)
\(\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{1}{2}\)
\(\Rightarrow x.\dfrac{4}{3}=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}:\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{3}{8}\)
2.
a) \(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}\)
Ta có:
\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}.....\dfrac{10000}{10001}=B\)
\(\Rightarrow A.A^2< A.B\)
\(\Rightarrow A^2< \dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.....\dfrac{9999}{10000}.\dfrac{10000}{10001}=\dfrac{1}{10001}\)
\(\Rightarrow A^2< \dfrac{1}{10001}< \dfrac{1}{10000}=\dfrac{1}{100}.\dfrac{1}{100}=\left(0,01\right)^2\)
\(\Rightarrow A< 0,01\)
Vậy A , 0,01
1
b) \(\dfrac{1}{5}.2^x+\dfrac{1}{3}.2^{x+1}=\dfrac{1}{5}.2^7+\dfrac{1}{3}.2^8\Rightarrow\dfrac{1}{5}.2^x+\dfrac{1}{3}.2^x.2=\dfrac{1}{5}.2^7+\dfrac{1}{3}.2^7.2\Rightarrow2^x.\left(\dfrac{1}{5}+\dfrac{1}{3}\right)=2^7.\left(\dfrac{1}{5}+\dfrac{1}{3}\right)\Rightarrow2^x=2^7\Rightarrow x=7\)
