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\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=3\\\frac{x^2}{y^2}+\frac{y^2}{x^2}+2=49\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}+\frac{y}{x}=3\\\left(\frac{x}{y}+\frac{y}{x}\right)^2=49\end{matrix}\right.\)
Hệ vô nghiệm
\(x^2+y+\frac{3}{4}\ge x^2+\frac{1}{4}+y+\frac{1}{2}\ge2\sqrt{x^2\cdot\frac{1}{4}}+\left(y+\frac{1}{2}\right)\ge x+y+\frac{1}{2}\)
\(\Rightarrow VT\ge\left(x+y+\frac{1}{2}\right)^2=\left[\left(x+\frac{1}{4}\right)+\left(y+\frac{1}{4}\right)\right]^2\ge4\left(x+\frac{1}{4}\right)\left(y+\frac{1}{4}\right)\)
\(=\left(2x+\frac{1}{2}\right)\left(2y+\frac{1}{2}\right)\)
Dấu "=" xảy ra tại \(x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
\(PT\Leftrightarrow x^2y^2+y^3+x^3+\frac{3}{4}\left(x^2+y^2\right)+xy+\frac{3}{4}\left(x+y\right)+\frac{9}{16}=4xy+x+y+\frac{1}{4}.\)
\(\Leftrightarrow x^2y^2+\left(x+y\right)^3-3xy\left(x+y\right)+\frac{3}{4}\left[\left(x+y\right)^2-2xy\right]+\frac{1}{4}\left(x+y\right)-3xy+\frac{5}{16}=0\)
Đặt \(x+y=a,xy=b\)
\(\Rightarrow b^2+a^3-3ab+\frac{3}{4}\left(a^2-2b\right)+\frac{a}{4}-3b+\frac{5}{16}=0\)
\(\Leftrightarrow16b^2+16a^3-48ab+12a^2-24b+4a-48b+5=0\)
\(\Leftrightarrow16b^2+16a^3-48ab+12a^2-72b+4a+5=0\)
Đến đây phân tích thành nhân tử hay sao ấy, chưa nghĩ ra :P
\(M=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}-\frac{4x^2}{x^2-1}\right):\frac{4\left(x^2-3\right)}{x\left(1-x\right)}\)
\(=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}+\frac{4x^2}{1-x^2}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\left(\frac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\frac{\left(1-x\right)^2}{\left(1+x\right)\left(1-x\right)}+\frac{4x^2}{\left(1+x\right)\left(1-x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\left(\frac{\left(1+x\right)^2-\left(1-x\right)^2+4x^2}{\left(1-x\right)\left(1+x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\frac{\left(1+x+1-x\right)\left(1+x-1+x\right)+4x^2}{\left(1-x\right)\left(1+x\right)}.\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)
\(=\frac{2.2x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{4x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{4x\left(1+x\right)}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)
\(=\frac{x}{1}.\frac{x}{\left(x^3-3\right)}\)
\(=\frac{x^2}{x^3-3}\)
Xét: \(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}\)\(=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{\left(x^2+y^2\right)\left(x+y\right)\left(x-y\right)}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)(1)
Tương tự, ta có: \(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)(2); \(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)(3)
Cộng theo vế của 3 đẳng thức (1), (2), (3), ta được:
\(\left[\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]\)\(-\left[\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\right]=0\)
\(\Rightarrow\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(=\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Mà \(A=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)nên \(2A=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)\(\ge\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{\frac{\left(y^2+z^2\right)^2}{2}}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{\frac{\left(z^2+x^2\right)^2}{2}}{\left(z^2+x^2\right)\left(z+x\right)}\)
\(=\frac{1}{2}\left(\frac{x^2+y^2}{x+y}+\frac{y^2+z^2}{y+z}+\frac{z^2+x^2}{z+x}\right)\)\(\ge\frac{1}{2}\left(\frac{\frac{\left(x+y\right)^2}{2}}{x+y}+\frac{\frac{\left(y+z\right)^2}{2}}{y+z}+\frac{\frac{\left(z+x\right)^2}{2}}{z+x}\right)\)\(=\frac{1}{4}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]=\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\)(Do theo giả thiết thì x + y + z = 1)
\(\Rightarrow A\ge\frac{1}{4}\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Bài này t làm rồi, "nhẹ" không ấy mà :|
Dự đoán khi \(x=y=z=\frac{1}{3}\Rightarrow A=\frac{1}{4}\). Ta c/m nó là GTNN của A
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(A=Σ\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\)
Cần chứng minh BĐT \(\frac{\left(x^2+y^2+z^2\right)^2}{Σ\left(x^2+y^2\right)\left(x+y\right)}\ge\frac{x+y+z}{4}\)
\(\Leftrightarrow4\left(x^2+y^2+z^2\right)^2\ge\left(x+y+z\right)Σ\left(2x^3+x^2y+x^2z\right)\)
\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+6x^2y^2-2x^2yz\right)\ge0\)
\(\LeftrightarrowΣ\left(2x^4-3x^3y-3x^3z+4x^2y^2\right)+Σ\left(2x^2y^2-2x^2yz\right)\ge0\)
\(\LeftrightarrowΣ\left(x^4-3x^3y+4x^2y^2-3xy^3+y^4\right)+Σ\left(x^2z^2-2z^2xy+y^2z^2\right)\ge0\)
\(\LeftrightarrowΣ\left(x-y\right)^2\left(x^2-xy+y^2\right)+Σz^2\left(x-y\right)^2\ge0\)
BĐT cuối đúng tức ta có \(A_{Min}=\frac{1}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
P/s: Nguồn lời giải Câu hỏi của Vo Trong Duy - Toán lớp 9 - Học toán với OnlineMath, rảnh quá ngồi gõ lại :V
\(\)đặt x+y=a , xy=b
=> \(\left\{{}\begin{matrix}a\left(1+\frac{1}{b}\right)=5\\\left(a^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}a=\frac{5}{1+\frac{1}{b}}\\\left(\left(\frac{5}{1+\frac{1}{b}}\right)^2-2b\right)\left(1+\frac{1}{b^2}\right)=49\end{matrix}\right.\)
=> giải đc b2+7b+1=0 => b => a => x,y
số xấu vler , chả biết sếp nào nghĩ ra bài này nữa
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\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=5\\x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=49\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=53\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=a\\y+\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=5\\a^2+b^2=53\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=-14\end{matrix}\right.\) \(\Rightarrow a;b\) là nghiệm của \(t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(-2;7\right);\left(7;-2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}x+\frac{1}{x}=-2\\y+\frac{1}{y}=7\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x+\frac{1}{x}=7\\y+\frac{1}{y}=-2\end{matrix}\right.\)
the mình, ta nên đặt x-1=a , 2-x=b sao cho a,b>0, ta đc a+b=1 thì biểu thức S có dạng:
S= 1/a2+ 1/b2 + 1/ab = (1/a2 + 1/b2 - 2/ab) + 3/ab =(1/a - 1/b)2 + 3/ab.
Ta có (a+b)2 >= 4ab nên thay a+b=1 vào ta được 1>= 4ab
suy ra 1/ab >= 4 suy ra tiếp 3/ab >=12
mà (1/a - 1/b)2 >=0 nên S >= 12
dấu bằng sảy ra khi a=b=1/2 nên x=3/2
Đặt x -2006 = y
pt <=> \(\frac{y^2-y\left(y-1\right)+\left(y-1\right)^2}{y^2+y\left(y-1\right)+\left(y-1\right)^2}=\frac{19}{49}\)
<=> \(\frac{y^2-y^2+y+y^2-2y+1}{y^2+y^2-y+y^2-2y+1}=\frac{19}{49}\)
<=> \(\frac{y^2-y+1}{3y^2-3y+1}=\frac{19}{49}\)
<=> \(49y^2-49y+49=57y^2-57y+19\)
<=> \(8y^2-8y-30=0\)
<=> \(4y^2-4y+15=0\)
Giải tiếp nha
