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\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+..+\dfrac{100}{2^{100}}\\ \Rightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\\ \Rightarrow A=\dfrac{7}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\\ B=\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\\ \Rightarrow2B=\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}\\ \Rightarrow B=\dfrac{1}{4}-\dfrac{1}{2^{99}}\\ \Rightarrow A=\dfrac{7}{4}+\dfrac{1}{4}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\\ =2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
T làm biếng lắm; làm C thôi
\(A=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\\ \Rightarrow A< \dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\\ \Rightarrow A^2< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{100}{101}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}.\dfrac{100}{101}\\ =\dfrac{1}{101}< \dfrac{1}{100}\\ \Rightarrow A< \dfrac{1}{10}\)
Làm tương tự ta được A > 1/15
câu a
\(A=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{30}>\dfrac{20}{30}=\dfrac{2}{3}>\dfrac{1}{3}\)
\(A=\left(\dfrac{1}{11}+..+\dfrac{1}{15}\right)+\left(\dfrac{1}{16}+...+\dfrac{1}{30}\right)< 5.\dfrac{1}{10}+25.\dfrac{1}{15}=\dfrac{1}{2}+\dfrac{5}{3}=\dfrac{8}{6}=\dfrac{4}{3}< \dfrac{5}{2}\)
Bài 1:
a: \(\Leftrightarrow\dfrac{x+2}{2}=x-5\)
=>2x-10=x+2
=>x=12
b: \(\Leftrightarrow\left(x+2\right)^2=100\)
=>x+2=10 hoặc x+2=-10
=>x=-12 hoặc x=8
c: \(\Leftrightarrow\left(2x-5\right)^3=27\)
=>2x-5=3
=>2x=8
=>x=4
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)
\(2A=2\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\right)\)
\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+....+\dfrac{100}{2^{99}}\)
\(2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\right)\)\(A=2+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(A=\dfrac{11}{4}+\dfrac{1}{2^3}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
Đặt \(D=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\)
\(2D=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\)
\(2D-D=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{100}{2^{100}}\right)\)
\(D=2+\dfrac{3}{2^2}+\dfrac{1}{2^3}+....+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(D=\dfrac{11}{4}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
\(D=\dfrac{11}{4}+\dfrac{1}{2^3}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)