a)     \({\cos ^2}\alpha  + {\sin ^2}\alpha  = 1\)

b)     \(\tan \alpha .\cot \alpha  = \frac{{\sin \alpha }}{{\cos \alpha }}.\frac{{\cos \alpha }}{{\sin \alpha }} = 1\)

c)     \(\frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }} = {\tan ^2}\alpha  + 1\)

d)     \(\frac{1}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha  + {{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = \frac{{{{\sin }^2}\alpha }}{{{{\sin }^2}\alpha }} + \frac{{{{\cos }^2}\alpha }}{{{{\sin }^2}\alpha }} = 1 + {\cot ^2}\alpha \)

2A=1+1/2+................+1/2^49+1/2^50

A=1+1/2^50=> A>1

tinh 2A ROI RUT GON .ROI LAY 2A tru di A THI RA KET QUA

le đinh dat chỉ bậy tính thế thì ai tính ko được mà céc cũng phải mất 1 ngày mứ ra kiểu tính mò nớ đó

Ta có:

\(P=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{2500}\right)\)

\(P=\frac{3}{4}.\frac{8}{9}...\frac{2499}{2500}\)

\(P=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{49.51}{50.50}\)

\(P=\left(\frac{1}{2}.\frac{2}{3}...\frac{49}{50}\right).\left(\frac{3}{2}.\frac{4}{3}...\frac{51}{50}\right)\)

\(P=\frac{1}{50}.\frac{51}{2}=\frac{51}{100}\)

TRẦN MINH HOÀNG Dề bảo so sánh với 1/2 cơ mà

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? crisdevergamemer

\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)

\(P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)

\(P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)

\(P=\frac{1.51}{50.2}\)

\(P=\frac{51}{100}>\frac{1}{2}\)

Kết luận: \(P>\frac{1}{2}\)

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{50}}\)

      \(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)

       \(2A-A=1-\frac{1}{2^{50}}\)

     \(A=1-\frac{1}{2^{50}}< 1\)

       \(\Rightarrow A< 1\)