a,b,c nguyên dương và a+b+c=1. chứng minh rằng \(\left(1+\dfrac{1}{a}\right)\cdot\left(1+\dfrac{1}{b}\right)\cdot\left(1+\dfrac{1}{c}\right)>=64\)
Bài 1:Tính
a, A=\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot....\cdot\dfrac{9999}{10000}\)
b,B=\(\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot....\cdot\left(1-\dfrac{1}{1326}\right)\)
c,C=\(\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\dfrac{1}{99\cdot101}\right)\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Cho a, b, c là các số dương thỏa mãn a + b + c = 1. Tìm Min:
\(A=\dfrac{\left(1+a\right)\cdot\left(1+b\right)\cdot\left(1+c\right)}{\left(1-a\right)\cdot\left(1-b\right)\cdot\left(1-c\right)}\)
P = \(1+\dfrac{1}{2}\cdot\left(1+2\right)+\dfrac{1}{3}\cdot\left(1+2+3\right)+\dfrac{1}{4}\cdot\left(1+2+3+4\right)+...+\dfrac{1}{16}\cdot\left(1+2+3+...+16\right)\)
Cho a + b + c = 2016 và \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}.\) Tính S = \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)
Giải giúp mình nha.
Làm lại cho you đây -_- vừa nãy bấm mt nhầm,đời t nhọ vãi
1)\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+....+16\right)\)
\(P=1+\dfrac{1+2}{2}+\dfrac{1+2+3}{3}+\dfrac{1+2+3+4}{4}+...+\dfrac{1+2+3+...+16}{16}\)
Xét thừa số tổng quát: \(\dfrac{1+2+3+...+t}{t}=\dfrac{\left[\left(t-1\right):1+1\right]:2.\left(t+1\right)}{t}=\dfrac{\dfrac{t}{2}\left(t+1\right)}{t}=\dfrac{\dfrac{t^2}{2}+\dfrac{t}{2}}{t}=\dfrac{t\left(\dfrac{t}{2}+\dfrac{1}{2}\right)}{t}=\dfrac{t}{2}+\dfrac{1}{2}\)
Như vậy: \(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)
\(P=1+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+....+\dfrac{17}{2}\)
\(P=\dfrac{2+3+4+5+...+17}{2}\)
\(P=\dfrac{152}{2}=76\)
2) \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{1}{3}\)
\(\Rightarrow2016\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{2016}{a+b}+\dfrac{2016}{b+c}+\dfrac{2016}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a}{b+c}+\dfrac{c+a}{c+a}+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow1+\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{c+a}=\dfrac{2016}{3}\)
\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{2016}{3}-1-1-1=\dfrac{2007}{3}\)
\(P=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
Xét thừa số tổng quát: \(\dfrac{1+2+3+..+n}{n}=\dfrac{\left[\left(n-1\right):1+1\right]:2.\left(n+1\right)}{n}=\dfrac{\dfrac{n}{2}\left(n+1\right)}{n}=\dfrac{\dfrac{n^2}{2}+\dfrac{n}{2}}{n}=\dfrac{n\left(\dfrac{n}{2}+\dfrac{1}{2}\right)}{n}=\dfrac{n}{2}+\dfrac{1}{2}\)
Như vậy:
\(P=1+\left(\dfrac{2}{2}+\dfrac{1}{2}\right)+\left(\dfrac{3}{2}+\dfrac{1}{2}\right)+\left(\dfrac{4}{2}+\dfrac{1}{2}\right)+...+\left(\dfrac{16}{2}+\dfrac{1}{2}\right)\)
\(P=1+\dfrac{2}{2}+\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{1}{2}+\dfrac{4}{2}+\dfrac{1}{2}+...+\dfrac{16}{2}+\dfrac{1}{2}\)
\(P=1+\left(\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{16}{2}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+...+\dfrac{1}{2}\right)\)
\(P=1+\dfrac{2+3+4+...+16}{2}+\dfrac{15}{2}\)
\(P=1+\dfrac{\left[\left(16-2\right):1+1\right]:2.\left(16+2\right)}{2}+\dfrac{15}{2}\)
\(P=1+210+\dfrac{15}{2}=218,5\)
Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
Cho a, b, c là các số dương biết abc = 1. Chứng minh rằng: \(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}\ge\dfrac{1}{2}\)
\(\dfrac{a^3}{\left(b+1\right)\left(c+2\right)}+\dfrac{b+1}{12}+\dfrac{c+2}{18}\ge3\sqrt[3]{\dfrac{a^3\left(b+1\right)\left(c+2\right)}{216\left(b+1\right)\left(c+2\right)}}=\dfrac{a}{2}\)
Tương tự: \(\dfrac{b^3}{\left(c+1\right)\left(a+2\right)}+\dfrac{c+1}{12}+\dfrac{a+2}{18}\ge\dfrac{b}{2}\)
\(\dfrac{c^3}{\left(a+1\right)\left(b+2\right)}+\dfrac{a+1}{12}+\dfrac{b+2}{18}\ge\dfrac{c}{2}\)
Cộng vế:
\(VT+\dfrac{5}{36}\left(a+b+c\right)+\dfrac{7}{12}\ge\dfrac{1}{2}\left(a+b+c\right)\)
\(\Rightarrow VT\ge\dfrac{13}{36}\left(a+b+c\right)-\dfrac{7}{12}\ge\dfrac{13}{36}.3\sqrt[3]{abc}-\dfrac{7}{12}=\dfrac{1}{2}\) (đpcm)
Cho a,b,c là 3 số thực, thỏa mãn điều kiện:
a)\(\dfrac{a+b-c}{3\cdot c}=\dfrac{b+c-a}{3\cdot a}=\dfrac{c+a-b}{3\cdot b}\)
b)Tính giá trị biểu thức
\(P=\left(1+\dfrac{b}{a}\right)\cdot\left(1+\dfrac{a}{c}\right)\cdot\left(1+\dfrac{c}{b}\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}=\dfrac{a+b-c+b+c-a+c+a-b}{3a+3b+3c}=\dfrac{a+b+c+\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{3a+3b+3c}=\dfrac{a+b+c}{3\left(a+b+c\right)}=\dfrac{1}{3}\)
Khi đó:
\(\left\{{}\begin{matrix}\dfrac{a+b-c}{3c}=\dfrac{1}{3}\\\dfrac{b+c-a}{3a}=\dfrac{1}{3}\\\dfrac{c+a-b}{3b}=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b-3c=3c\\3b+3c-3a=3a\\3c+3a-3b=3b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6c\\3b+3c=6a\\3c+3a=6b\end{matrix}\right.\)Thay vào \(P\)
\(P=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\left(\dfrac{a+b}{a}\right)\left(\dfrac{c+a}{c}\right)\left(\dfrac{b+c}{b}\right)\)
\(27P=3\left(\dfrac{a+b}{a}\right).3\left(\dfrac{c+a}{c}\right).3\left(\dfrac{b+c}{b}\right)\)
\(27P=\left(\dfrac{3a+3b}{a}\right)\left(\dfrac{3c+3a}{c}\right)\left(\dfrac{3b+3c}{b}\right)\)
\(27P=\)\(\dfrac{6c}{a}.\dfrac{6b}{c}.\dfrac{6a}{b}=\dfrac{216abc}{abc}=216\Leftrightarrow P=\dfrac{216}{27}=8\)
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
1.Cho \(a,b,c,d\) là các số nguyên thỏa mãn \(a^3+b^3=2\left(c^3-d^3\right)\) . Chứng minh rằng a+b+c+d chia hết cho 3
2.Cho ba số dương a,b,c thỏa mãn abc=1. Chứng minh rằng \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)
thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
Tính các tích sau:
a) \(P=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{99}{100}\)
b) \(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)
\(\left(a\right):P=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{99}{100}\)
Nhận xét
thừa số tổng quát là \(\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\) với n =1 đến 10
\(P=\dfrac{1.3.2.4.3.5...9.11}{2^2.3^2...9^2.10^2}=\dfrac{\left(1.2.3...9\right)\left(3.4.5....11\right)}{\left(2.3.4....10\right)\left(2.3.4....10\right)}\)
\(P=\dfrac{1.2.3..9}{2.3.4..9.10}.\dfrac{3.4.5...10.11}{2.3.4....10}=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}\)
Mk bt giải bài này r nè! Thank ngonhuminh nhìu! Mk lm thêm câu Q, m.n xem thử mk lm cs đg k nha!!!!!
\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)
\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...\left(\dfrac{9}{9}-1\right)...\left(\dfrac{19}{9}-1\right)\)
\(Q=\left(\dfrac{1}{9}-1\right)\left(\dfrac{2}{9}-1\right)\left(\dfrac{3}{9}-1\right)...0...\left(\dfrac{19}{9}-1\right)\)
\(\Rightarrow Q=0\)