\(\left(x-3y\right)^2+\left(y-1\right)^2+\left(z+2\right)^2=0\Rightarrow\hept{\begin{cases}x-3y=0\\y-1=0\\z+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=3y\\y=1\\z=-2\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=1\\z=-2\end{cases}}}\)

Thế vào A ta được \(2\left(3\right)+2\left(1\right)+\left(-2\right)=6\)

1.

\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

\(\Rightarrow\dfrac{1}{2a+c}=\dfrac{1}{2b}=\dfrac{1}{b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)

Bài 2 đề sai

Ở phân thức thứ 2 không thể là \(\dfrac{y+3x-x}{x}\)

Bài 2:

\(P=\dfrac{x+3y}{y}\cdot\dfrac{y+3z}{z}\cdot\dfrac{z+3x}{x}=\dfrac{\left(x+3y\right)\left(y+3z\right)\left(z+3x\right)}{xyz}\)

Với \(x+y+z=0\)

\(\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}\\ \Leftrightarrow\dfrac{x+3y+x+y}{z}=\dfrac{y+3z+y+z}{x}=\dfrac{z+3x+x+z}{y}\\ \Leftrightarrow\dfrac{2\left(x+2y\right)}{z}=\dfrac{2\left(y+2z\right)}{x}=\dfrac{2\left(z+2x\right)}{y}\\ \Leftrightarrow\dfrac{2\left(y-z\right)}{z}=\dfrac{2\left(z-x\right)}{x}=\dfrac{2\left(x-y\right)}{y}\\ \Leftrightarrow\dfrac{2y-2z}{z}=\dfrac{2z-2x}{x}=\dfrac{2x-2y}{y}\\ \Leftrightarrow\dfrac{2y}{z}-2=\dfrac{2z}{x}-2=\dfrac{2x}{y}-2\\ \Leftrightarrow\dfrac{2y}{z}=\dfrac{2z}{x}=\dfrac{2x}{y}\\ \Leftrightarrow\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x}{y}\Leftrightarrow x=y=z=0\left(\text{trái với GT}\right)\)

Với \(x+y+z\ne0\)

\(\Leftrightarrow\dfrac{x+3y-z}{z}=\dfrac{y+3z-x}{x}=\dfrac{z+3x-y}{y}=\dfrac{3\left(x+y+z\right)}{x+y+z}=3\\ \Leftrightarrow\left\{{}\begin{matrix}x+3y-z=3z\\y+3z-x=3x\\z+3x-y=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+3y=4z\\y+3z=4x\\z+3x=4y\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{4x\cdot4y\cdot4z}{xyz}=64\)

Ta có: \(\frac{1}{\left(3x+1\right)\left(y+z\right)+x}=\frac{1}{3x\left(y+z\right)+x+y+z}\le\frac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}\)

\(=\frac{1}{3x\left(y+z\right)+3\sqrt[3]{1}}=\frac{1}{3x\left(y+z\right)+3}=\frac{1}{3\left(xy+zx+1\right)}=\frac{1}{3}\cdot\frac{1}{\frac{1}{y}+\frac{1}{z}+1}\)

Tương tự ta chứng minh được:

\(\frac{1}{\left(3y+1\right)\left(z+x\right)+y}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\) ; \(\frac{1}{\left(3z+1\right)\left(x+y\right)+z}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{x}+\frac{1}{y}+1}\)

Cộng vế 3 BĐT trên lại:

\(A\le\frac{1}{3}\cdot\left(\frac{1}{\frac{1}{x}+\frac{1}{y}+1}+\frac{1}{\frac{1}{y}+\frac{1}{z}+1}+\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\right)\)

\(\Leftrightarrow3A\le\frac{1}{\left(\frac{1}{\sqrt[3]{x}}\right)^3+\left(\frac{1}{\sqrt[3]{y}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{y}}\right)^3+\left(\frac{1}{\sqrt[3]{z}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{z}}\right)^3+\left(\frac{1}{\sqrt[3]{x}}\right)^3+1}\)

Đặt \(\left(\frac{1}{\sqrt[3]{x}};\frac{1}{\sqrt[3]{y}};\frac{1}{\sqrt[3]{z}}\right)=\left(a;b;c\right)\) khi đó:

\(3A\le\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{c^3+a^3+1}\)

\(=\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+1}+\frac{1}{\left(b+c\right)\left(b^2-bc+c^2\right)+1}+\frac{1}{\left(c+a\right)\left(c^2-ca+a^2\right)+1}\)

\(\le\frac{1}{\left(a+b\right)\left(2ab-ab\right)+1}+\frac{1}{\left(b+c\right)\left(2bc-bc\right)+1}+\frac{1}{\left(c+a\right)\left(2ca-ca\right)+1}\)

\(=\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}+\frac{1}{ca\left(c+a\right)+1}\)

\(=\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}+\frac{abc}{ca\left(c+a\right)+abc}\)

\(=\frac{c}{a+b+c}+\frac{a}{b+c+a}+\frac{b}{c+a+b}\)

\(=\frac{a+b+c}{a+b+c}=1\)

Dấu "=" xảy ra khi: \(a=b=c\Leftrightarrow x=y=z=1\)

Vậy Max(A) = 1 khi x = y = z = 1

Câu hỏi của Pham Van Hung - Toán lớp 9 - Học toán với OnlineMath

Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)

\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-1\right)^2\ge0\forall y\)

\(2\left(x+y\right)^2\ge0\forall x,y\)

Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)

Dấu '=' xảy ra khi 

\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)

Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được: 

\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)

\(=0^{2016}+1^{2017}+0^{2018}=1\)

Vậy: M=1

- Bạn làm được bài này chưa bạn?

xin bài này , 5 phút sau làm 

đặt \(\hept{\begin{cases}A=3x-2y-2\sqrt{y+2012}+1=0\\B=3y-2z-.....\\C=3z-2x.....\end{cases}}.\)

vì a=b=c=0 

Suy ra A+B+C=0

A+B+c= \(\left(x\right)+\left(y\right)+\left(z\right)-2\sqrt{y+2012}-2\sqrt{z-2013}-2\sqrt{x-2}\) " rút gọn làm tắt "

đến đây ta thêm  3-3 , 2012-2012 , 2013-2013 , 2-2 vào biểu thức rồi dùng hằng đẳng thức ta được

\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2+2013-2012+2-3=0\)

\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2=0\) rút gọn

\(\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y+2012}=1\\\sqrt{z-2013}=1\end{cases}}\)

thay vào P ta được 

\(P=\left(3-4\right)^{2011}+\left(-2011+2012\right)^{2012}+\left(2014-2013\right)^{2013}\)

\(P=-1+1+1=1\)

Xét \(x+y+z=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+z=-x\\z+x=-y\\x+y=-z\end{matrix}\right.\)

\(\Rightarrow A=\left(2-1\right)\left(2-1\right)\left(2-1\right)=1\)

Xét \(x+y+z\ne0\) thì ta có:

\(\Rightarrow\left\{{}\begin{matrix}5x=y+z+3x\\5y=z+x+3y\\5z=x+y+3z\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=y+z\\2y=z+x\\2z=x+y\end{matrix}\right.\)

\(\Rightarrow A=\left(2+2\right)\left(2+2\right)\left(2+2\right)=64\)

Vậy \(\left[{}\begin{matrix}A=1\\A=64\end{matrix}\right.\)

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