Xét \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3+3x+3-6x+3x^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{1-3x+3x^2-x^3}{1-3x+3x^2}\)

\(=\frac{1-3x+3x^2}{1-3x+3x^2}=1\)

Thay vào ta tính được:

\(A=\left[f\left(\frac{1}{2020}\right)+f\left(\frac{2019}{2020}\right)\right]+...+\left[f\left(\frac{1009}{2020}\right)+f\left(\frac{1011}{2020}\right)\right]+f\left(\frac{1010}{2020}\right)\)

\(A=1+...+1+f\left(\frac{1010}{2020}\right)\) (với 1009 số 1)

\(A=1009+f\left(\frac{1}{2}\right)=1009+\frac{\left(\frac{1}{2}\right)^3}{1-3\cdot\frac{1}{2}+3\cdot\left(\frac{1}{2}\right)^2}\)

\(A=1009+\frac{1}{2}=\frac{2019}{2}\)

Vậy \(A=\frac{2019}{2}\)

Tks bạn nhé

hello ae xin chào

Ta xét : \(f\left(x\right)+f\left(1-x\right)=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{1-3\left(1-x\right)+3\left(1-x\right)^2}\)

\(=\frac{x^3}{1-3x+3x^2}+\frac{\left(1-x\right)^3}{3x^2-3x+1}=\frac{\left(x+1-x\right)\left(x^2+x^2-2x+1+x^2-x\right)}{3x^2-3x+1}=\frac{3x^2-3x+1}{3x^2-3x+1}=1\)

Áp dụng ta có : 

\(A=\left[f\left(\frac{1}{2012}\right)+f\left(\frac{2011}{2012}\right)\right]+\left[f\left(\frac{2}{2012}\right)+f\left(\frac{2010}{2012}\right)\right]+...+\left[f\left(\frac{1006}{2012}\right)+f\left(\frac{1006}{2012}\right)\right]\)

\(=1+1+...+1\)(Có tất cả 1006 số 1)

\(=1006\)

sai rồi bạn ơi

Ta có:\(f\left(x\right)-1=\left(x-1\right)^3\)

\(=>A+\frac{1}{2}=\left(\frac{1}{112}-1\right)^3+\left(\frac{2}{112}-1\right)^3+\left(\frac{3}{112}-1\right)^3+...\left(\frac{111}{112}-1\right)^3\)

\(A+\frac{1}{2}=-\frac{1^3+2^3+3^3+...+111^3}{112^3}=-\frac{\frac{111^2\left(111+1\right)^2}{4}}{112^3}=-\frac{111^2}{4\cdot112}=-\frac{12321}{448}\)

\(A=-\frac{12321}{448}-\frac{1}{2}=-\frac{12545}{448}\)

à nhầm :v

cho \(a\)và \(1-a\), ta có:

\(f\left(1-a\right)=\frac{\left(1-a\right)^3}{3\left(1-a\right)^2-3\left(1-a\right)+1}=\frac{\left(1-a-1\right)^3}{3-6a+a^2-3+3a+1}+1=1-\frac{a^3}{3a^3-3a+1}=1-f\left(a\right)\)

hay \(f\left(a\right)+f\left(1-a\right)=1\)

\(=>A=f\left(\frac{1}{112}\right)+f\left(\frac{111}{112}\right)+f\left(\frac{2}{112}\right)+f\left(\frac{110}{112}\right)+...+f\left(\frac{55}{112}\right)+f\left(\frac{57}{112}\right)+f\left(\frac{56}{112}\right)-\frac{1}{2}\)

\(=>A=55+f\left(\frac{1}{2}\right)-\frac{1}{2}=55\) vì  \(f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=2f\left(\frac{1}{2}\right)=1\)nên \(f\left(\frac{1}{2}\right)=\frac{1}{2}\)

Vậy \(A=55\)

a, Ta có : y^2 lớn hơn hoặc bằng 0 với mọi y

=> -y^2 nhỏ hơn hoặc bằng 0 với mọi y 

=>-2-y^2 nhỏ hơn hoặc bằng -2 với mọi y

=> H nhỏ hơn hoặc -2 với mọi y

Dấu "=" xảy ra <=>y^2=0 <=>y=0

Vậy GTLN của H là -2 tại y=0