Cho \(S=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}\)
So sánh S với\(\dfrac{1}{2}\)
SO SÁNH A VÀ B
A=\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\cdot\dfrac{1}{2}-2\cdot\dfrac{1}{4}-...-2\cdot\dfrac{1}{100}\)
\(A=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\dfrac{1}{1}-\dfrac{1}{2}-...-\dfrac{1}{50}\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}=B\)
\(\Rightarrow A=B\)
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
\(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}:\dfrac{1}{1-2}+\dfrac{1}{2-3}+...+\dfrac{1}{99-100}\)
Cho S= \(\dfrac{1}{50}\)+\(\dfrac{1}{51}\)+\(\dfrac{1}{52}\)+...+\(\dfrac{1}{98}\)+\(\dfrac{1}{99}\). Khẳng định nào sau đây là đúng:
A. S >\(\dfrac{1}{2}\) | B. S <\(\dfrac{1}{2}\) | C.S > 1 | D. S < 1 |
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)
a)Chứng minh \(\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+..........+\dfrac{1}{100}< 1\)
b)So sánh A=\(\dfrac{10^{19}+1}{10^{20}+1}\) B=\(\dfrac{10^{20}+1}{10^{21}+1}\)
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
So sánh A và B :
\(A=1.3.5.7.....99\)
\(B=\dfrac{51}{2}.\dfrac{52}{2}.\dfrac{53}{2}.....\dfrac{100}{2}\)
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
Câu 1:
a, Phân tích đa thức thành nhân tử: 4(x2 +15x +50)( x2 +18x +72) - 3x2
b, Tìm các cặp số nguyên (x; y) thỏa mãn: y2 +2(x2 +1) = 2y(x+1)
Câu 2: Cho \(S=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{99}+\dfrac{1}{100}\). So sánh S với \(\dfrac{1}{2}\)
\(4\left(x^2-15x+50\right)\left(x^2-18+72\right)-3x^2\)
\(=4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\)
\(=4\left[\left(x+5\right)\left(x+12\right)\right]\left[\left(x+10\right)\left(x+6\right)\right]-3x^2\)
\(=4\left(x^2+17x+60\right)\left(x^2+16x+60\right)-3x^2\)
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
\(A=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}\)
Tham khảo: (mk chx chắc lắm đâu nha)