Answer 1

Cau 2

TA CO :\(\frac{1}{A}\)+\(\frac{1}{B}\)+\(\frac{1}{c}\)=\(\frac{1}{2019}\)<=>\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)=\(\frac{1}{a+b+c}\)<=>(\(\frac{1}{a}\)+\(\frac{1}{b}\))+(\(\frac{1}{c}\)-\(\frac{1}{a+b+c}\)=0<=>\(\frac{a+b}{ab}\)+\(\frac{a+b}{c\left(a+b+c\right)}\)=0<=>(a+b)(\(\frac{1}{ab}\)+\(\frac{1}{c\left(a+b+c\right)}\))=0<=>\(\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}\)=0<=>\(\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\) gia su\(\left(a+b\right)=0\)=>c=2016 khi do\(\frac{a^{2019}+b^{2019}}{\left(ab\right)^{2019}}+\frac{1}{c^{2019}}=\frac{1}{c^{2019}}=\frac{1}{a^{2019}+b^{2019+c^{2019}}}\)cac truong hop kia tuong tu 

Answer 2

\(\left(a+b+c\right)^2=a^2+b^2+c^2\)<=>\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\),=>\(2\left(ab+bc+ac\right)=0\)<=>\(ab+bc+ac=0\)<=>\(ab=-\left(bc+ac\right),bc=-\left(ab+ac\right),ac=-\left(ab+bc\right)\)voi \(ab=-\left(bc+ac\right)\)=>\(c^2+2ab=c^2+ab-bc-ac=\left(a-c\right)\left(b-c\right)\)tuong tu \(a^2+2bc=\left(a-c\right)\left(a-b\right)\)              \(b^2+2ac=-\left(b-c\right)\left(a-b\right)\)=>\(P=\frac{a^2}{\left(a-c\right)\left(a-b\right)}-\frac{b^2}{\left(b-c\right)\left(a-b\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)\(=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)

Answer 3

Câu 1: 

a) \(x^2+y^2+z^2+4080597=2\left(14x+y+2020z\right)\)

\(\Leftrightarrow x^2+y^2+z^2+4080597-2\left(14x+y+2020z\right)=0\)

\(\Leftrightarrow\left(x^2-2.14x+196\right)+\left(y^2-2y+1\right)+\left(z^2-2.2020z+4080400\right)=0\)

\(\Leftrightarrow\left(x-14\right)^2+\left(y-1\right)^2+\left(z-2020\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-14=0\\y-1=0\\z-2020=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=14\\y=1\\z=2020\end{cases}}\)

Vậy nghiệm của phương trình là \(\left(x;y;z\right)=\left(14;1;2020\right)\)

Answer 4

GIÚP ĐC NHIÊU THÌ GIÚP THÔI CX ĐC Ạ

Answer 5

Đề khó quá nhỉ để mình giúp cho nha, nhưng mình chỉ giúp được bạn 1 tờ thôi thông cảm cho mình nha 45 phút là xong thôi cố gắng chờ đợi nha!!!!!!!!!!!!!!!!! 

      Mình làm xong nhớ nhé!!!!!!!!!!!!!!!!!!!!!!!!

Answer 6

* Câu bất

Ta có: \(16=xy+\left(x+y\right)+1\le\frac{\left(x+y\right)^2}{4}+\left(x+y\right)+1\)\(\Leftrightarrow\left(x+y\right)^2+4\left(x+y\right)-60\ge0\Leftrightarrow x+y\ge6\)

\(\Rightarrow P\ge\frac{\left(x+y\right)^2}{2}\ge\frac{6^2}{2}=18\)

Đẳng thức xảy ra khi x = y = 3

Answer 7

thanks tất cả các bạn nhé