ta có x²+y²-6x+18+6y=0

⇔(x²-6x+9)+(y²+6y+9)=0

⇔(x-3)²+(y+3)²=0

vì (x-3)²≥0 với mọi x;(y+3)²≥0 với mọi y

⇒ (x-3)²+(y+3)²≥0 với mọi x,y

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)

ta có A=x²⁰¹⁷.y²⁰¹⁸+x²⁰¹⁸.y²⁰¹⁷+\(\dfrac{1}{9}y\)

^{A=\(x^{2017}\cdot y^{2017}\cdot\left(x+y\right)+\dfrac{1}{9}y\)}

^{thay x=3; y=-3 vào A ta có giá trị biểu thức A là}

^{A=\(3^{2017}\cdot\left(-3\right)^{2017}\cdot\left(-3+3\right)+\dfrac{1}{9}\cdot\left(-3\right)\)}

^{A=\(-\dfrac{1}{3}\)}

^{Vậy A=\(-\dfrac{1}{3}\)} khi x=3;y=-3

Chúc bạn học tốt

\(x^2-6x+9+y^2+6y+9=0\Leftrightarrow\left(x-3\right)^2+\left(y+3\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}x-3=0\\y+3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)

Thay vào A:

\(A=x^{2017}.y^{2018}+x^{2018}.y^{2017}+\dfrac{y}{9}=x^{2017}.y^{2017}\left(x+y\right)+\dfrac{y}{9}\)

\(\Rightarrow A=3^{2017}.\left(-3\right)^{2017}\left(3-3\right)+\dfrac{-3}{9}=-\dfrac{1}{3}\)

Áp dụng bđt Cauchy-Schwarz:\(A=\frac{1}{1-2\left(ab+bc+ac\right)}+\frac{1}{abc}=\frac{1}{\left(a+b+c\right)^2-2\left(ab+bc+ac\right)}+\frac{a+b+c}{abc}\)

\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\ge\frac{1}{a^2+b^2+c^2}+\frac{9}{ab+bc+ac}\)

\(=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab+bc+ac}+\frac{1}{ab+bc+ac}+\frac{7}{ab+bc+ac}\ge\frac{9}{\left(a+b+c\right)^2}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}=30\)

\("="\Leftrightarrow a=b=c=\frac{1}{3}\)

Khôi Bùi , Akai Haruma, Nguyen, Ribi Nkok Ngok

giúp mk vs!

theo mình lúc đọc sơ qua đề bài và nghĩ 1 chút thì bạn tách cái 2y(2y-1) ra rồi nhóm sao cho nó có dạng (y-1)^2 là ok rồi

hình như đề bài thiếu dữ kiện

à đúng r EF=20cn

\(2x^2-2x=0\)

\(2x\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

Vậy......

2x^2-2x=0

<=>2x(x-1)=0

<=>2x=0 hay x-1=0

<=>x=0 hay x=1

Cho P=4

\(\Rightarrow P=\frac{2\left(2x+3\right)}{\left(x+3\right)\left(x-1\right)}=4\)

\(\Rightarrow2\left(2x+3\right)=4\left(x+3\right)\left(x-1\right)\)

\(4x+6=4x^2+12x-4x-12\)

\(4x+6-4x^2-12x+4x+12=0\)

\(\Rightarrow-4x^2-4x+18=0\)

\(\Rightarrow2\left(3-2x\right)\left(x+3\right)=0\)

Vì \(\left(x+3\right)\ne0\) (đkxđ)

=>\(\left(3-2x\right)=0\) \(\Leftrightarrow x=\frac{3}{2}\)

Vậy............

Phân thức thứ 2 nè

Ta có: \(P=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-c\right)\left(b-a\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{1}{\left(a-b\right)\left(a-c\right)}-\frac{1}{\left(a-b\right)\left(b-c\right)}-\frac{1}{\left(a-c\right)\left(c-b\right)}\)

\(=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}-\frac{a-c}{\left(a-c\right)\left(a-b\right)\left(b-c\right)}+\frac{a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\frac{b-c-a+c+a-b}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}=\frac{0}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}=0\)(đpcm)