Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\Rightarrow\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}-\frac{a+b+c}{a+b+c}=0\)
\(\Rightarrow\left(a+b+c\right).\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)
xét: \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\left(\text{vì a+b+c khác 0}\right)\)
\(\text{ta có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\frac{ab+bc+ac}{abc}-\frac{1}{a+b+c}=0\)
\(\Rightarrow\frac{\left(ab+bc+ac\right).\left(a+b+c\right)-abc}{abc.\left(a+b+c\right)}=0\)
\(\Rightarrow\left(ab+bc+ac\right).\left(a+b+c\right)-abc=0\)
\(\Rightarrow\left(b+a\right).\left(c+a\right).\left(c+b\right)=0\)
\(\Rightarrow\hept{\begin{cases}b=-a\\a=-c\\c=-b\end{cases}}\)
\(M=\left(-b^{101}+b^{101}\right).\left(-c^{2017}+c^{2017}\right).\left(b^{2019}+-b^{2019}\right)=0\)
p/s: dài nhỉ =)
Vì a ; b ; c dương \(\Rightarrow a+b+c\ne0\)
Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=0;b-c=0;c-a=0\Leftrightarrow a=b=c\)
Vậy \(A=\left(1-\frac{a}{b}\right)\left(2018-\frac{b}{c}\right)\left(2019-\frac{c}{a}\right)=\left(1-1\right).\left(...\right)=0\)
Sửa lại đề : CM : \(\frac{1}{b^2+c^2}+\frac{1}{a^2+b^2}+\frac{1}{c^2+a^2}\le\frac{a^3+b^3+c^3}{2abc}+3\)
Ta có :
\(\frac{1}{b^2+c^2}=\frac{a^2+b^2+c^2}{b^2+c^2}=\frac{b^2+c^2}{b^2+c^2}+\frac{a^2}{b^2+c^2}=1+\frac{a^2}{b^2+c^2}\)
Mà \(b^2+c^2\ge2bc\) nên \(\frac{1}{b^2+c^2}\le1+\frac{a^2}{2bc}\)(1)
CM tương tự ta cũng có : \(\hept{\begin{cases}\frac{1}{a^2+b^2}\le1+\frac{c^2}{2ab}\left(2\right)\\\frac{1}{c^2+a^2}\le1+\frac{b^2}{c^2+a^2}\left(3\right)\end{cases}}\)
Cộng vế với vế của (1);(2);(3) tại ta được :
\(\frac{1}{b^2+c^2}+\frac{1}{a^2+b^2}+\frac{1}{c^2+a^2}\le\frac{a^2}{2bc}+\frac{c^2}{2ab}+\frac{b^2}{2ac}+3=\frac{a^3+b^3+c^3}{2abc}+3\)
=> đpcm
Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)
\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
Em tham khảo cách làm tại link: Câu hỏi của Cao Chi Hieu - Toán lớp 9 - Học toán với OnlineMath
\(a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow2.\left(a+b+c\right)=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{1}{b}}+2\sqrt{c.\frac{1}{c}}\)
\(=2+2+2=6\)
\(\Rightarrow a+b+c\ge3\)
\(P=a+b^{2019}+c^{2020}\)
\(=a+\left(b^{2019}+1.2018\right)+\left(c^{2020}+1.2019\right)-4037\)
\(\ge a+2019.\sqrt[2019]{b^{2019}.1^{2018}}+2020.\sqrt[2020]{c^{2020}.1^{2019}}-4037\)(BDT Cauchy-Schwarz)
\(=a+2019b+2020c-4037\)
Do \(a\le b\le c\)nên
\(\Rightarrow P\ge a+2019b+2020c\)
\(\ge a+\left(\frac{2017}{3}+\frac{4040}{3}\right)b+\left(\frac{2020}{3}+\frac{4040}{3}\right)c-4037\)
\(\ge a+\frac{2017}{3}a+\frac{4040}{3}b+\frac{2020}{3}a+\frac{4040}{3}c-4037\)
\(=\frac{4040}{3}.\left(a+b+c\right)-4037\)
\(\ge4040-4037=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)