Vì \(ab+bc+ca=2020\)

\(\Rightarrow a^2+2020=a^2+ab+bc+ca\)

\(=\left(a^2+ab\right)+\left(bc+ca\right)=a\left(a+b\right)+c\left(a+b\right)\)

\(=\left(a+b\right)\left(a+c\right)\)

Tương tự ta có: \(b^2+2020=\left(b+a\right)\left(b+c\right)\)

                          \(c^2+2020=\left(c+b\right)\left(c+a\right)\)

\(\Rightarrow\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)

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

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

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

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

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

\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)( đpcm )

Ta có 

  \(a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)

\(b^2+ab+bc+ac=\left(a+b\right)\left(b+c\right)\)

\(c^2+ab+bc+ac=\left(a+c\right)\left(b+c\right)\)

   Thay ab + bc + ac = 2020 vào biểu thức \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)có

       \(\frac{a^2-bc}{a^2+2020}+\frac{b^2-ca}{b^2+2020}+\frac{c^2-ab}{c^2+2020}\)

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

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

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

\(=\frac{0}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=0\)

\(8^4.8=8^5=32768\)

8 mũ 4 nhân 8 =8 mũ 5=32768

= 135 và 130

= 135 VÀ 130

450 mét

hoktốt

Xét tứ giác MNPQ có :

\(\widehat{M}+\widehat{N}+\widehat{P}+\widehat{Q}=360^o\)

\(35^o+67^o+127^o+\widehat{Q}=360^o\)

\(229^o+\widehat{Q}=360^o\)

                 \(\widehat{Q}=360^o-229^o\)

                 \(\widehat{Q}=131^o\)

Vậy \(\widehat{Q}=131^o\)