cho x,y> 0 thỏa mãn xy+x+y=1. Tính tổng
\(S=2x\sqrt{\frac{1+y^2}{1+x^2}}+2y\sqrt{\frac{1+x^2}{1+y^2}}+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
Cho các số dương x,y,z thỏa mãn: xy+yz+zx=1
Tính tổng:
\(S=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Thay \(xy+yz+xz=1\) ta có: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
Cho x,y>0 và xy+x+y=1.
Tính P = \(2x\sqrt{\frac{1+y^2}{1+x^2}}+2y\sqrt{\frac{1+x^2}{1+y^2}}+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)
cho x,y,z là các số thực không âm thỏa mãn x+y+z=1.Tìm min
\(T=\left[\frac{\sqrt[3]{x+y+2z}\left(\sqrt{xy+z}+\sqrt{2x^2+2y^2}\right)}{3\sqrt[6]{xy}}\right]\left(x^2+y^2+z^2\right)-2\sqrt{2x^2-2x+1}\)
dễ dàng phân tích được
\(\sqrt{2x-y}=\frac{\left(x^2-x-xy\right)}{\left(y+1\right)}\)
\(\left(y+1\right)=\frac{\left(x^2-x-xy\right)}{\sqrt{2x-y}}\)
\(\left(y+1\right)\sqrt{2x-y}=\frac{\left(x^2-x-xy\right)^2}{\sqrt{2x-y}\left(y+1\right)}\)
thay vào "pt" 1 ta được
\(\left(x^2-x-xy\right)\left(\frac{x^2-x-xy-1}{\sqrt{2x-y}\left(y+1\right)}\right)=0\)
\(x^2-x-xy=0\Leftrightarrow x^2=x\left(1+y\right)\Leftrightarrow x=1+y\)
thay x=y+1 vào pt2 ta được
\(\left(y+1\right)^2+y^2-2y\left(y+1\right)-3\left(y+1\right)+2=0\)
\(\left(y^2+y^2-2y^2\right)+\left(2y-2y-3y\right)+\left(1-3+2\right)=0\)
\(-3y=0\Leftrightarrow y=0\)
thay \(y=0\)
Cho x,y>0 tm xy+x+y=1. Tính
\(S=x\sqrt{\frac{2\left(1+y^2\right)}{1+x^2}}+y\sqrt{\frac{2\left(1+x^2\right)}{1+y^2}}+\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{2}}\)
Cho x,y,z>0 thỏa mãn xyz=1. Tìm min \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
bạn vào trang này nhé có bài như thến này đấy
//123doc.org//document/3173507-ren-luyen-chuyen-de-tim-maxmin-on-thi-thpt-quoc-gia.htm
tính diện tích hình vẽ dưới đây
cho 3 số dương x,y,z thỏa mãn xy+yz+xz =1
tính T =\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+y^2\right)\left(1+x^2\right)}{1+z^2}}\)
Ta có:
\(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào T ta được:
\(T=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)=2\left(xy+yz+xz=1\right)\)
Ta có \(1+x^2=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\).
Tương tự ta cũng có \(1+y^2=\left(x+y\right)\left(y+z\right)\) và \(1+z^2=\left(z+x\right)\left(y+z\right)\).
Thu gọn được \(T=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
Cho 3 số dương x,y,z thỏa mãn điều kiện xy+yz+zx=1. Tính
\(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Ta co: \(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=y+z\)
Thê vào ta được
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\left(xy+yz+zx\right)=2\)
cho x,y,z thỏa mãn xy+yz+xz=1. Tính giá trị của biểu thức:
\(A=x.\sqrt{\frac{\left(1+y^2\right).\left(1+z^2\right)}{1+x^2}}+y.\sqrt{\frac{\left(1+z^2\right).\left(1+x^2\right)}{1+y^2}}+z.\sqrt{\frac{\left(1+x^2\right).\left(1+y^2\right)}{1+z^2}}\)
Bài này hình như x,y,z>0
Ta có: \(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\sqrt{\frac{\left(y^2+xy+yz+zx\right)\left(z^2+xy+yz+zx\right)}{\left(x^2+xy+yz+zx\right)}}=x\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=x\sqrt{\left(y+z\right)^2}\)
Tương tự: \(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\sqrt{\left(x+z\right)^2}\)
\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\sqrt{\left(x+y\right)^2}\)
Cộng từng vế, ta có:
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)
\(\Leftrightarrow A=2\left(xy+yz+zx\right)=2\)
\(\hept{\begin{cases}1+y^2=y^2+xy+yz+zx=\left(x+y\right)\left(y+z\right)\\1+z^2=\left(z+x\right).\left(z+y\right)\\1+x^2=\left(x+y\right)\left(x+z\right)\end{cases}}\)
Thế vào \(A=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
\(=2\left(\left|xy\right|+\left|yz\right|+\left|zx\right|\right)\)
Nếu x,y,z\(\ge0\Rightarrow A=2\)
Nếu x,y,z\(< 0\)\(\Rightarrow A=-2\)