Question

1. Tính:

C= \(\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right)\left(\dfrac{1}{4^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\)

2. Tìm x:

\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-2}{5}\right)^2\)

Help me!!!

Answer 1

Bài 2:

\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-2}{5}\right)^2\)

\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3.\left(\dfrac{2}{5}\right)^2\)

\(\Rightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^5\)

Vì \(\dfrac{2}{5}\ne\pm1;\dfrac{2}{5}\ne0\) nên \(x>5\)

Vậy \(x>5\) thoả mãn yêu cầu đề bài.

Chúc bạn học tốt!!!

Answer 2

Bài 1:

\(C=\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\)

\(C=\left(\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right).....\left(\dfrac{1}{\left(100-1\right)\left(100+1\right)}\right)\)

\(C=\dfrac{1}{1.3}\dfrac{1}{2.4}.....\dfrac{1}{99.101}=\dfrac{1}{101!}\)

Chúc bạn học tốt!!!

Answer 3

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Answer 4

Bài1:

\(C=\left(\dfrac{1}{2^2-1}\right)\left(\dfrac{1}{3^2-1}\right).....\left(\dfrac{1}{100^2-1}\right)\\ =\left(\dfrac{1}{\left(2-1\right)\left(2+1\right)}.\dfrac{1}{\left(3-1\right)\left(3+1\right)}.....\dfrac{1}{\left(100-1\right)\left(100+1\right)}\right)\\ =\dfrac{1}{1.3}.\dfrac{1}{2.4}.....\dfrac{1}{99.101}\\ =\dfrac{1}{99!.101!}\)

Bài2:

\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(-\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^3.\left(\dfrac{2}{5}\right)^2\\ \Leftrightarrow\left(\dfrac{2}{5}\right)^x>\left(\dfrac{2}{5}\right)^5\\ \Leftrightarrow x< 5\)

Do \(\dfrac{2}{5}\ne0;\dfrac{2}{5}\ne\pm1\)

Vậy x<5

Toshiro Kiyoshi giải sai :))