\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)

\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)

\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)

\(\Rightarrow A=B\)

tớ giải chi tiết hơn nhá:

A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

A=(\(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

A=\(\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)

A=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)

Vậy A=B

Chọn A nha

CÂU A nha

A

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)

\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)

\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)

Lời giải:

\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)

\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)

\(4\left(x^2-15x+50\right)\left(x^2-18+72\right)-3x^2\)

\(=4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\)

\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+10\right)\left(x+6\right)\right]-3x^2\)

\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)

Đặt \(x^2+16x+60=a\), ta có:

\(4\left(a+x\right)\left(a\right)-3x^2\)

\(=4a^2+4ax-3x^2\)

\(=4a^2-2ax+6ax-3x^2=2a\left(2a-x\right)+3x\left(2a-x\right)\)

\(=\left(2a-x\right)\left(2a+3x\right)\)

Thay a vào ta có: \(\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\)

\(=\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)

S_1=1/1.2+1/3.4+1/5.6+...+1/99.100

=1‐1/2+1/3‐1/4+1/5‐1/6+...+1/99‐1/100

=﴾1+1/3+1/5+...+1/99﴿‐﴾1/2+1/4+1/6+...+1/100﴿

=﴾1+1/2+1/3+1/4+1/5+1/6+...+1/99+1/100﴿‐2﴾1/2+1/4+1/6+...+1/100﴿

=﴾1+1/2+1/3+1/4+...+1/100﴿‐﴾1+1/2+1/3+..+1/50﴿

=1/51+1/52+1/53+..+1/100 ﴾1﴿

S_2=1/51+1/52+1/53+..+1/100 ﴾2﴿

Từ ﴾1﴿,﴾2﴿=> S_1/S_2=1

Vậy S > 1/2

Tham khảo: (mk chx chắc lắm đâu nha)