4/5=8/10=-8/-10

VẬY X=-88

đặt \(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2022}=k\Rightarrow\hept{\begin{cases}a=2020k\\b=2021k\\c=2022k\end{cases}}\)

Khi đó \(A=\frac{a-b+c}{a+2b-c}=\frac{2020k-2021k+2022k}{2020k+2\cdot2021k-2022k}=\frac{2021k}{4040k}=\frac{2021}{4040}\)

\(\frac{a}{2020}=\frac{b}{2021}=\frac{c}{2020}=\frac{a-b+c}{2020-2021+2022}=\frac{a-b+c}{2021}\)

\(\frac{a}{2020}=\frac{2b}{2021.2}=\frac{c}{2022}=\frac{a+2b-c}{2020+4042-2022}=\frac{a+2b-c}{4040}\)

\(\Rightarrow\frac{a-b+c}{2021}=\frac{a+2b-c}{4040}\Rightarrow A=\frac{a-b+c}{a+2b-c}=\frac{2021}{4040}\)

Ta có : f(1/2) = \(a.\left(\frac{1}{2}\right)^2+b.\frac{1}{2}+c=\frac{a}{4}+\frac{b}{2}+c\)

f(-2) = \(a.\left(-2\right)^2+b.\left(-2\right)+c=4a-2b+c\)

Khi đó f(1/2) + f(-2) = \(\frac{a}{4}+\frac{b}{2}+c+4a-2b+c=\frac{17a}{4}-\frac{3b}{2}+2c=\frac{17a}{4}-\frac{6b}{4}+\frac{8c}{4}\)

\(=\frac{17a-6b+8c}{4}=0\left(\text{Vì }17a-6b+8c=0\right)\)

=>  f(1/2) + f(-2) = 0 (đpcm)

Hàm số đồ thì sao không có biến x bạn. Sai đề rồi

Đặt A = \(\left|x-2013\right|+\left|x-1\right|\)

Ta có : \(\left|x-2013\right|\ge0\forall x\); \(\left|x-1\right|\ge0\forall x\)

=> A \(\ge\)0

Dấu ''='' xảy ra : <=> x = 2013 ; 1 

Vật GTNN A là 0 <=> x = 2013 ; 1

\(\left(x-1\right)\left(x^2+1\right)=0\)

TH1 : \(x-1=0\Leftrightarrow x=1\)

TH2 : \(x^2+1=0\Leftrightarrow x^2=-1\)vô lí 

\(x^2\ge0\forall x;-1< 0\)

Vậy x = 1