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Ta có: \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=a^3+b^3+c^3-3abc\)
\(\Rightarrow\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=2019\left(đpcm\right)\)
Ta có : \(\left(a+b+c\right)\left(a^2+b^2+^2-ab-ac-bc\right)\)
\(=a^3+b^3+c^3-3abc\)
\(\Leftrightarrow\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=2019\)
\(\Rightarrowđpcm\)
\(GT\Rightarrow a^2=b^2+c^2\ge\frac{1}{2}\left(b+c\right)^2\Rightarrow\frac{a}{b+c}\ge\frac{\sqrt{2}}{2}\)
\(M=8a^2\left(\frac{1}{b^2}+\frac{1}{c^2}\right)+\frac{b+c}{a}+2019\ge4a^2\left(\frac{1}{b}+\frac{1}{c}\right)^2+\frac{b+c}{a}+2019\)
\(M\ge4a^2\left(\frac{4}{b+c}\right)^2+\frac{b+c}{a}+2019=16\left(\frac{a}{b+c}\right)^2+\frac{b+c}{a}+2019\)
Đặt \(\frac{a}{b+c}=x\ge\frac{\sqrt{2}}{2}\)
\(\Rightarrow M\ge16x^2+\frac{1}{x}+2019=\sqrt{2}x^2+\frac{1}{2x}+\frac{1}{2x}+\left(16-\sqrt{2}\right)x^2+2019\)
\(\Rightarrow M\ge3\sqrt[3]{\frac{\sqrt{2}x^2}{4x^2}}+\left(16-\sqrt{2}\right).\frac{1}{2}+2019=2027+\sqrt{2}\)
Dấu "=" xảy ra khi \(b=c=\frac{a}{\sqrt{2}}\)
EM tham khảo phần đầu ở link: Câu hỏi của Đinh Nguyến Nhật Minh - Toán lớp 8 - Học toán với OnlineMath
Trong 3 số a,b, c có hai số đối nhau g/s 2 số đó là a và b kho đó a=-b
=> \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{\left(-b\right)^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=-\frac{1}{b^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}\)
và \(\frac{1}{a^{2019}+b^{2019}+c^{2019}}=\frac{1}{\left(-b\right)^{2019}+b^{2019}+c^{2019}}=\frac{1}{-b^{2019}+b^{2019}+c^{2019}}=\frac{1}{c^{2019}}\)
Do đó: \(\frac{1}{a^{2019}}+\frac{1}{b^{2019}}+\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019}+c^{2019}}\)
Ta có: \(\sqrt{a^2-ab+b^2}=\sqrt{\dfrac{1}{4}\left(a+b\right)^2+\dfrac{3}{4}\left(a-b\right)^2}\ge\sqrt{\dfrac{1}{4}\left(a+b\right)^2}=\dfrac{1}{2}\left(a+b\right)\)
Tương tự: \(\sqrt{b^2-bc+c^2}\ge\dfrac{1}{2}\left(b+c\right)\)
\(\sqrt{c^2-ca+a^2}\ge\dfrac{1}{2}\left(c+a\right)\)
\(P\ge\dfrac{1}{2}\left(a+b\right)+\dfrac{1}{2}\left(b+c\right)+\dfrac{1}{2}\left(c+a\right)=a+b+c=2019\)
Dấu "=" xảy ra <=> a = b = c = 673
Ta có: a2-ab+b2 = \(\dfrac{1}{4}\)(a+b)2+3(a-b)2\(\ge\)\(\dfrac{1}{4}\)(a+b)2
\(\Rightarrow\)\(\sqrt{a^2-ab+b^2}\ge\dfrac{1}{2}\)(a+b)
Dấu "=" xảy ra \(\Leftrightarrow\) a=b
CMTT ta có: \(\sqrt{b^2-bc+c^2}\)\(\ge\dfrac{1}{2}\)(b+c) \(\Leftrightarrow\) b=c
\(\sqrt{c^2-ca+c^2}\)\(\ge\dfrac{1}{2}\left(c+a\right)\Leftrightarrow\)c=a
\(\Rightarrow\) P\(\ge\) \(\dfrac{1}{2}2\left(a+b+c\right)\)= 2019
Vậy Pmin = 2019
Dấu "=" xảy ra\(\Leftrightarrow\)a=b=c=673
Ta có: \(a^2+2019=a^2+ab+bc+ca=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự ta có : \(b^2+2019=\left(a+b\right)\left(b+c\right)\)
\(c^2+2019=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(b+c\right)}\)\(=\frac{\left(a^2-bc\right)\left(b+c\right)+\left(b^2-ac\right)\left(a+c\right)+\left(c^2-ab\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)\(=\frac{a^2b-b^2c+a^2c-bc^2+ab^2-a^2c+b^2c-ac^2+ac^2+bc^2-a^2b-ab^2}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=0\)\(\Rightarrow dpcm\)
\(\text{Thay }ab+bc+ac=2019\text{ vào biểu thức trên, ta có: }\)
\(\frac{a^2-bc}{a^2+ab+bc+ac}+\frac{b^2-ac}{b^2+ab+bc+ac}+\frac{c^2-ab}{c^2+ab+bc+ac}\)
\(=\frac{\left(a^2-bc\right).\left(b+c\right)}{\left(a+c\right).\left(a+b\right).\left(b+c\right)}+\frac{\left(b^2-ac\right).\left(a+c\right)}{\left(a+b\right).\left(b+c\right).\left(a+c\right)}+\frac{\left(c^2-ab\right).\left(a+b\right)}{\left(a+c\right).\left(b+c\right).\left(a+b\right)}\)
\(=\frac{a^2b+a^2c-b^2c-bc^2+b^2a+b^2c-a^2c-ac^2+c^2a+c^2b-a^2b-ab^2}{\left(a+c\right).\left(a+b\right).\left(b+c\right)}=0\)
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