tìm giá trị lớn nhất của biểu thức
\(M=\dfrac{\left|x-y\right|+\left|x+y\right|+\left|xy-1\right|+\left|xy+1\right|}{\sqrt{\left(x^2+1\right)\left(y^2+1\right)}}\)
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MH
22 tháng 11 2023
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
LS
Lê Song Phương
CTVHS
22 tháng 11 2023
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.
13 tháng 9 2022
\(A=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-xy\sqrt{x}-xy\sqrt{y}}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\sqrt{x}\left(1-y\right)+x\left(1-y\sqrt{y}\right)-y\left(1-\sqrt{y}\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{\left(1-\sqrt{y}\right)\left[x\sqrt{x}\left(1+\sqrt{y}\right)+x+x\sqrt{y}+xy-y\right]}{\left(1+\sqrt{x}\right)\left(1-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\sqrt{x}+x\sqrt{xy}+x+x\sqrt{y}+xy-y}{\left(1+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x\left(\sqrt{x}+1\right)+x\sqrt{y}\left(\sqrt{x}+1\right)+y\left(x-1\right)}{\left(1+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+x\sqrt{y}+y\sqrt{x}-y}{\sqrt{x}+\sqrt{y}}=\sqrt{x}-\sqrt{y}+\sqrt{xy}\)
Để A=2 thì x=2; y=2
6 tháng 4 2016
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
Ta có Max \(f\left(t\right)=f\left(\frac{1}{9}\right)=\frac{6\sqrt{10}}{10}+\frac{1}{9}\), \(t\in\left(0;\frac{1}{9}\right)\)
HN
9 tháng 10 2017
\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
\(=\sqrt{xy}+\sqrt{x}-\sqrt{y}\)
Ta có: \(P=\sqrt{xy}+\sqrt{x}-\sqrt{y}=2\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{y}+1\right)=3\)
\(\Rightarrow\left(\sqrt{x}-1,\sqrt{y}+1\right)=\left(1,3;3,1\right)\)
\(\Rightarrow\left(x,y\right)=\left(4,4;16,0\right)\)
26 tháng 5 2022
\(A=\dfrac{\left(a+b\right)\left(-x-y\right)-\left(a-y\right)\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{a\left(-x-y\right)+b\left(-x-y\right)-a\left(b-x\right)+y\left(b-x\right)}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ax-ay-bx-by-ab+ax+by-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-ay-bx-ab-xy}{abxy\left(xy+ay+ab+by\right)}\)
\(=\dfrac{-xy+ay+ab+by}{abxy\left(xy+ay+ab+by\right)}=\dfrac{-1}{abxy}\)
Với \(a=\dfrac{1}{3};b=-2;x=\dfrac{3}{2};y=1\)
\(\Rightarrow A=\dfrac{-1}{\dfrac{1}{3}.\left(-2\right).\dfrac{3}{2}.1}=-1\)
6 tháng 2 2018
\(P=\dfrac{x}{\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{y}\right)}-\dfrac{y}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)}-\dfrac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\left(\sqrt{x}+1\right)-y\left(1-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x\sqrt{x}+x-y+y\sqrt{y}-yx\sqrt{x}-xy\sqrt{y}}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{x+y+\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=\dfrac{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)+2\sqrt{xy}-xy-1}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\\ P=1-\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=2\\ \Rightarrow\dfrac{\left(\sqrt{xy}-1\right)^2}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}=1\\ \Leftrightarrow\left(\sqrt{xy}-1\right)^2=\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)\\ \Leftrightarrow xy-2\sqrt{xy}+1=\sqrt{x}-\sqrt{y}+1-\sqrt{xy}\\ \Leftrightarrow\sqrt{x}-\sqrt{y}-xy+\sqrt{xy}=0\)
tự giải quyết tiếp nhá :)) h có việc :)) nếu còn ko bt thì mai làm nốt cho :))
