Giải:

Gọi khối lượng hàng mỗi xe dự định chở là \(x\) (tấn) \(\left(x>1\right)\)

Số xe ban đầu dự định là \(\dfrac{100}{x}\)

Do lúc sau mỗi xe chỉ chở \(x-1\) tấn hàng nên số xe lúc sau là \(\dfrac{100}{x-1}\)

Số xe bổ sung thêm là \(5\) nên ta có:

\(\dfrac{100}{x-1}-\dfrac{100}{x}=5\Leftrightarrow\dfrac{20\left(x-x+1\right)}{x\left(x-1\right)}=1\)

\(\Leftrightarrow x^2-x=20\Leftrightarrow x^2-x-20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(TM\right)\\x=-4\left(L\right)\end{matrix}\right.\)

Vậy khối lượng hàng mỗi xe dự định chở là \(5\) tấn hàng.

Giải:

Ta có:

\(S=\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{17.20}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{17.20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{17}-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{20}\right)\)

\(=\dfrac{1}{3}.\dfrac{9}{20}=\dfrac{3}{20}\)

Vậy \(S=\dfrac{3}{20}\)

Đề phải cho \(a,b,c\) dương nữa

Giải:

Cộng hai vế của BĐT với \(3\) thì BĐT cần chứng minh trở thành:

\(\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{c+a}+1\right)+\left(\dfrac{c}{a+b}+1\right)\) \(\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) \(\ge\dfrac{9}{2}\)

Áp dụng BĐT \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) ta có:

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\) \(\dfrac{9}{a+b+b+c+c+a}\)

\(=\dfrac{9}{2\left(a+b+c\right)}\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

\(\ge\left(a+b+c\right).\dfrac{9}{2\left(a+b+c\right)}=\dfrac{9}{2}\) (Đúng)

\(\Rightarrow\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\) (Đpcm)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)

Ta có:

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

\(=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.\dfrac{6}{5}.\dfrac{7}{6}...\dfrac{99}{98}.\dfrac{100}{99}\)

\(=\dfrac{3.4.5.6.7...99.100}{2.3.4.5.6...98.99}\)

\(=\dfrac{\left(3.4.5.6.7...99\right).100}{2.\left(3.4.5.6...98.99\right)}\)

\(=\dfrac{100}{2}=50\)

Vậy \(T=50\)

Bài 1: b) Ta có: \(\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

\(=\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}\)

\(=\dfrac{a+b}{a+b}+\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b+c}{b+c}\) \(+\dfrac{a+c}{c+a}+\dfrac{b}{c+a}\)

\(=1+1+1+\left(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)\) \(=2010.\dfrac{1}{3}\)

\(\Rightarrow3+S=\dfrac{2010}{3}=670\Leftrightarrow S=667\)

Vậy \(S=667\)