\(P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{99}{5^{99}}\)

\(5P=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{99}{5^{98}}\)

\(\Rightarrow4P=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}-\frac{99}{5^{99}}=A-\frac{99}{5^{99}}\)

\(A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{98}}\)

\(5A=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{97}}\)

\(\Rightarrow4A=5-\frac{1}{5^{98}}< 5\Rightarrow A< \frac{5}{4}\)

\(4P=A-\frac{99}{5^{99}}< A< \frac{5}{4}\Rightarrow P< \frac{5}{16}\)

bài 1. ta có

\(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow b^2+ab+\frac{a^2}{4}+c^2+ac+\frac{a^2}{4}+d^2+ad+\frac{a^2}{4}+\frac{a^2}{4}\ge0\)

\(\Leftrightarrow\left(b+\frac{a}{2}\right)^2+\left(c+\frac{a}{2}\right)^2+\left(d+\frac{a}{2}\right)^2+\frac{a^2}{4}\ge0\) luôn đúng

Bài 2

ta có \(\frac{a^5}{b^5}+1+1+1+1\ge\frac{5.a}{b}\) (bất đẳng thức cauchy)

Tương tự ta có \(\frac{b^5}{c^5}+4\ge\frac{5b}{c};\frac{c^5}{a^5}+4\ge\frac{5c}{a}\)

\(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\)

Mà dễ dàng chứng minh \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge3\)

Nên ta có \(\Rightarrow\frac{a^5}{b^5}+\frac{b^5}{c^5}+\frac{c^5}{a^5}\ge5\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-12\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

bài 1 : \(^{a^2+B^2+C^2+D^2}\)>hoặc =ab+ac+ad 

\(^{a^2+b^2+c^2}\)- ab-ac-ad>hoặc = 0

\((\frac{1}{4}^{a^2-ab+b^2})+(\frac{1}{4}^{a^2-ac+c^2})+(\frac{1}{4}^{a^2-ad+d^2})\)>hoặc =0

\((\frac{1}{2}a-b)^2+(\frac{1}{2}a-c)^2+(\frac{1}{2}a-d)^2>=0\)

Vì \((\frac{1}{2}a-b)^2>=0\)với mọi \(A,b\varepsilon n\)

=> đpcm tự kết luận

Đề sai, đề đúng phải là \(VT< \frac{1}{20}\)

Dễ dàng chứng minh đề sai, ta có:

\(\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}>\frac{1}{5^2}+\frac{1}{5^3}=\frac{6}{125}>\frac{1}{24}\)

Còn chứng minh \(VT< \frac{1}{20}\) thì như sau:

\(A=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)

\(\Rightarrow5A=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2005}}\)

\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{2006}}\)

\(\Rightarrow5A-\frac{1}{5}+\frac{1}{5^{2006}}=A\)

\(\Rightarrow4A=\frac{1}{5}-\frac{1}{5^{2006}}< \frac{1}{5}\)

\(\Rightarrow A< \frac{1}{20}\)

\(A=\frac{1}{5^2}+\frac{2}{5^3}+.....+\frac{99}{5^{100}}\)

\(\Leftrightarrow5A=\frac{1}{5}+\frac{2}{5^2}+......+\frac{99}{5^{99}}\)

\(\Leftrightarrow5A-A=\left(\frac{1}{5}+\frac{2}{5^2}+....+\frac{99}{5^{99}}\right)-\left(\frac{1}{5^2}+\frac{2}{5^3}+...+\frac{99}{5^{100}}\right)\)

\(\Leftrightarrow4A=\frac{1}{5}+\frac{1}{5^2}+......+\frac{1}{5^{99}}-\frac{99}{5^{100}}\)

Đặt : \(H=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{99}}\)

\(\Leftrightarrow5H=1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\)

\(\Leftrightarrow5H-H=\left(1+\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{98}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\right)\)

\(\Leftrightarrow4H=1-\frac{1}{5^{99}}\)

\(\Leftrightarrow H=\frac{1}{4}-\frac{1}{4.5^{99}}< \frac{1}{4}\)

\(\Leftrightarrow4A< B< \frac{1}{4}\)

\(\Leftrightarrow A< \frac{1}{16}\left(đpcm\right)\)

5A=1/5=2/5^2+......+11/5^11
4A=1/5+1/5^2+......+1/5^11-11/5^12
20A=1+1/5+1/5^2+.....+1/5^10-11/5^11
16A=1-1/5^11+11/5^12-11/5^11
vi 1-1/5^11<1;11/5^12-11/5^11<0
16A<1
A<1/16 
k cho minh nhe

Bonking

bn tham khảo đây nhé :

Câu hỏi của Khanh Mai Lê - Toán lớp 6 - Học toán với OnlineMath

mình tính siêu đúng

...

bạn alan walker cho mk biết web đi