Cho A = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
Chứng minh A < 1
Cho A = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{99}\)Chứng minh rằng A < 1
\(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{99}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{98}}\)
\(2A-A=1-\frac{1}{2^{99}}\)
=> \(A=1-\frac{1}{2^{99}}
Cho B=\(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh B<1
\(B=\frac{1}{2}+\frac{1^2}{2^2}+\frac{1^3}{2^3}+........+\frac{1^{99}}{2^{99}}\)
\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...........+\frac{1}{2^{99}}\right)\)
=>B=\(1-\frac{1}{2^{98}}\Rightarrow B
1) Cho a, b, c > 0. Chứng minh: \(\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2\ge\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
2) Cho \(a,b,c\in R\).
a) Chứng minh: \(\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a+b+c+1\right)^2\)
b) Chứng minh: \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
3) Cho \(a,b,c\in R\)Chứng minh: \(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
2) Theo nguyên lí Dirichlet, trong ba số \(a^2-1;b^2-1;c^2-1\) có ít nhất hai số nằm cùng phía với 1.
Giả sử đó là a2 - 1 và b2 - 1. Khi đó \(\left(a^2-1\right)\left(b^2-1\right)\ge0\Leftrightarrow a^2b^2-a^2-b^2+1\ge0\)
\(\Rightarrow a^2b^2+3a^2+3b^2+9\ge4a^2+4b^2+8\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\ge4\left(a^2+b^2+2\right)\)
\(\Rightarrow\left(a^2+3\right)\left(b^2+3\right)\left(c^2+3\right)\ge4\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\) (2)
Mà \(4\left[\left(a^2+b^2+1+1\right)\left(1+1+c^2+1\right)\right]\ge4\left(a+b+c+1\right)^2\) (3)(Áp dụng Bunhicopxki và cái ngoặc vuông)
Từ (2) và (3) ta có đpcm.
Sai thì chịu
Xí quên bài 2 b:v
b) Không mất tính tổng quát, giả sử \(\left(a^2-\frac{1}{4}\right)\left(b^2-\frac{1}{4}\right)\ge0\)
Suy ra \(a^2b^2-\frac{1}{4}a^2-\frac{1}{4}b^2+\frac{1}{16}\ge0\)
\(\Rightarrow a^2b^2+a^2+b^2+1\ge\frac{5}{4}a^2+\frac{5}{4}b^2+\frac{15}{16}\)
Hay \(\left(a^2+1\right)\left(b^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{3}{4}\right)\)
Suy ra \(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(a^2+b^2+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+\frac{1}{4}+c^2+\frac{1}{2}\right)\)
\(\ge\frac{5}{4}\left(\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c+\frac{1}{2}\right)^2=\frac{5}{16}\left(a+b+c+1\right)^2\) (Bunhiacopxki) (đpcm)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)
Cách nữa cho bài 2:
2a) Ta có: \(4\left(a^2+1+2\right)\left(1+1+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2\)
Hay \(4\left(a^2+3\right)\left(2+\frac{\left(b+c\right)^2}{2}\right)\ge4\left(a+b+c+1\right)^2=VP\)
Như vậy ta quy bài toán về chứng minh: \(\left(b^2+3\right)\left(c^2+3\right)\ge4\left(2+\frac{\left(b+c\right)^2}{2}\right)\)
\(\Leftrightarrow b^2c^2+b^2+c^2+1\ge4bc\Leftrightarrow\left(bc-1\right)^2+\left(b-c\right)^2\ge0\)(đúng)
Đẳng thức xảy ra khi a = b = c = 1
b) Áp dụng BĐT Bunhiacopxki:\(\left(a^2+\frac{1}{4}+\frac{1}{4}+\frac{1}{2}\right)\left(\frac{1}{4}+b^2+c^2+\frac{1}{2}\right)\ge\frac{1}{4}\left(a+b+c+1\right)^2\)
\(\Rightarrow\frac{5}{4}\left(a^2+1\right)\left(b^2+c^2+\frac{3}{4}\right)\ge\frac{5}{16}\left(a+b+c+1\right)^2\)
Từ đó ta có thể quy bài toán về chứng minh: \(\left(b^2+1\right)\left(c^2+1\right)\ge\frac{5}{4}\left(b^2+c^2+\frac{3}{4}\right)\)
...
Bài 3:Sửa đề a, b, c >0
Có: \(\frac{a^3}{b^2}+\frac{a^3}{b^2}+b\ge3\sqrt[3]{\frac{a^6}{b^3}}=\frac{3a^2}{b}\)
Tương tự: \(\frac{2b^3}{c^2}+c\ge\frac{3b^2}{c};\frac{2c^3}{a^2}+a\ge\frac{3c^2}{a}\)
Cộng theo vế 3 BĐT trên: \(2\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)+a+b+c\ge3\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(=2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)\)
\(\ge2\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)+a+b+c\)
Từ đó ta có đpcm.
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
chứng minh B <1
\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)\)
\(B=1-\frac{1}{2^{99}}< 1\left(đpcm\right)\)
Cho B= \(1+\frac{-1}{2}+\left(\frac{-1}{2}\right)^2+\left(\frac{-1}{2}\right)^3+\left(\frac{-1}{2}\right)^4+...+\left(\frac{-1}{2}\right)^{99}\)
Chứng minh rằng B < \(\frac{2}{3}\)
Cho \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
CMR: \(2^{99}\times A⋮2^{99}-1\)
\(A=\frac{1}{2}+\frac{1}{2^2}+.............+\frac{1}{2^{99}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+...........+\frac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(1+\frac{1}{2}+.......+\frac{1}{2^{98}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.......+\frac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=1-\frac{1}{2^{99}}\)
\(\Leftrightarrow2^{99}.A=2^{99}-1\left(đpcm\right)\)
B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh B<1
\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{98}+\left(\frac{1}{2}\right)^{99}\)
Chứng minh: B<1
B = \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)
B = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
2B = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
2B - B = \(1-\frac{1}{2^{99}}\)
=> B = \(1-\frac{1}{2^{99}}
Bài 1:
a, Cho S=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\) .Chứng minh rằng \(\frac{2}{5}< S< \frac{8}{9}\)
b, Tìm x thuộc z để phân số \(\frac{x^2-5x-1}{x+2}\)có giá trị là số nguyên
c, Chứng minh rằng \(\left(\frac{7}{65}+1\right)\left(\frac{7}{84}+1\right)\left(\frac{7}{105}+1\right)\left(\frac{7}{124}+1\right)...\left(\frac{7}{153+1}\right)\left(\frac{7}{560}+1\right)< 2\)
d, Chứng minh rằng \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\frac{5}{3^5}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)