\(x^2-2\)

\(=x^2-\left(\sqrt{2}\right)^2\)

\(=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

\(y^2-13\)

\(=y^2-\left(\sqrt{13}\right)^2\)

\(=\left(y-\sqrt{13}\right)\left(y+\sqrt{13}\right)\)

a) Ta xét: Tam giác ADE có: AD = AE

=> Tam giác ADE cân tại A

\(\Rightarrow\widehat{AED}=\widehat{ACB}\)

=> DE//BC

Ta xét: Tứ giác DECB có: DE//BC

\(\Rightarrow\widehat{ABC}=\widehat{ACB}\)

=> BDEC là hình thang cân

b) \(\widehat{ABC}=\frac{1}{2}\left(180^o-50^o\right)=65^o\)

\(\widehat{ACB}=\widehat{ABC}=65^o\)

\(\widehat{DEC}=180^o-65^o=115^o\)

\(\widehat{EDB}=\widehat{EDC}=115^o\)

Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) với \(x=a^2+2bc;y=b^2+2ac;z=c^2+2ab\)

Ta có : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=\frac{9}{\left(a+b+c\right)^2}\)

\(\Rightarrow\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)( Vì a + b + c = 1)

\(\left(4n+3\right)^2-25\)

\(=\left(4n+3\right)^2-5^2\)

\(=\left(4n+3-5\right)\left(4n+3+5\right)\)

\(=\left(4n-2\right)\left(4n+8\right)\)

xl chia hết cho 8

\(A=x^2-4x+4+y^2-8y+16-14=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

giúp tui vs

huhu