Ta có bđt \(ab^2+bc^2+ca^2\le\frac{1}{3}\left(a+b+c\right)\left(a^2+b^2+c^2\right)=a^2+b^2+c^2\)

\(P=2017\left(\frac{a^3}{1+b^2}+\frac{b^3}{1+c^2}+\frac{c^3}{1+a^2}\right)\)

Ta có: \(\frac{a^3}{1+b^2}+\frac{a\left(1+b^2\right)}{4}\ge2\sqrt{\frac{a^3}{1+b^2}.\frac{a\left(1+b^2\right)}{4}}=a^2\)

Tương tự suy ra \(\frac{a^3}{1+b^2}+\frac{b^3}{1+c^2}+\frac{c^3}{1+a^2}\ge\left(a^2+b^2+c^2\right)-\frac{1}{4}\left(a+b+c\right)-\frac{1}{4}\left(ab^2+bc^2+ca^2\right)\)

\(\ge\left(a^2+b^2+c^2\right)-\frac{3}{4}-\frac{1}{4}\left(a^2+b^2+c^2\right)=\frac{3}{4}\left(a^2+b^2+c^2\right)-\frac{3}{4}\ge\frac{3}{4}.3-\frac{3}{4}=\frac{3}{2}\)

\(F=\frac{x+5}{\sqrt{x}+2}=\frac{\left(\sqrt{x}+2\right)\sqrt{x}-\left(\sqrt{x}+2\right)2+9}{\sqrt{x}+2}=\sqrt{x}-2+\frac{9}{\sqrt{x}+2}=\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\)

theo bdt co si ta co:\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}\ge2\sqrt{\sqrt{x}+2.\frac{9}{\sqrt{x}+2}}=6\)

\(\sqrt{x}+2+\frac{9}{\sqrt{x}+2}-4\ge6-4=2\)

=>Fmin=2 xay ra dau = khi\(\sqrt{x}+2=\frac{9}{\sqrt{x}+2}\)

a)  \(2x^2+x+5=0\)

Xét:  \(\Delta=1^2-4.2.5=-39< 0\)

=> pt vô nghiệm

b)  \(2x^2-2x+8=0\)

Xét:  \(\Delta=2^2-4.2.8=-60< 0\)

=>  pt vô nghiệm

Hình bạn tự vẽ nha.

a, Ta có: BC là đường trung trực của \(\Delta ABC\)\(\Rightarrow BM=MC,\widehat{DMC}=90^o\)

\(\Delta ABC,\widehat{BAC}=90^o\)có AM là trung tuyến của \(\Delta ABC\)\(\Rightarrow AM=BM=MC=\frac{BC}{2}\)

\(\Delta AMC\)có: \(AM=MC\left(cmt\right)\Rightarrow\Delta AMC\)cân tại M

b, \(\Delta ABC\)và \(\Delta MDC\)có:

\(\widehat{BAC}=\widehat{DMC}=90^o\)

\(\widehat{C}\)chung

\(\Rightarrow \Delta ABC \sim \Delta MDC (g-g)\)

c, \(\Delta BEC\)có: \(EM\perp BC\left(gt\right)\)

                           \(AC\perp AB\left(gt\right)\)

                            \(EM \cap AC \) \(=\left\{D\right\}\)

\(\Rightarrow D\)là trực tâm của \(\Delta BEC\)\(\Rightarrow BD\perp CE\)