Một cửa hàng ngày đầu bán được 3 tạ 16 kg gạo, ngày sau bán được hơn ngày đầu 3,5 yến. Hỏi cả hai ngày bán đươc bao nhiêu tạ gạo ?

các bạn giải giúp mình với trong vòng từ 5h đến 6h nhé

\(a^3+a^3+b^3\ge3\sqrt[3]{a^6b^3}=3a^2b\)

\(b^3+b^3+a^3\ge3b^2a\)

\(\Rightarrow3\left(a^3+b^3\right)\ge3\left(a^2b+b^2a\right)\Leftrightarrow\left(a^3+b^3\right)\ge\left(a^2b+b^2a\right)\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)

a) A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}+\frac{b}{\left(b-a\right)\left(b-c\right)}+\frac{c}{\left(c-a\right)\left(c-b\right)}\)

=> A = \(\frac{a}{\left(a-b\right)\left(a-c\right)}-\frac{b}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-c\right)\left(b-c\right)}\)

=> A = \(\frac{a\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{b\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

\(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(b-a\right)\left(b-c\right)\left(c-a\right)}\)

\(+\frac{c^2\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)

\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(+\frac{c^2\left(a-b\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)

\(A=1-\frac{1}{n+1}\)

a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)

           \(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)

           \(A=1-\frac{1}{n+1}\)

           \(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)

           \(A=\frac{n}{n+1}\)

Học tốt nha^^

2017 = 2016 + 1 = x + 1

suy ra 2017x¹⁵ = x¹⁶ + x¹⁵

2017x¹⁴ = x¹⁵ + x¹⁴

.... 

từ đó ta dễ tính ra A