Cho 3 số dương x,y,z thỏa mãn điều kiện: xy + yz + zx = 1
cmr
\(\frac{x}{1+x^2}+\frac{y}{1+y^2}-\frac{z}{1+z^2}=\frac{2xy}{\sqrt{\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)}}\)
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 3 số dương x, y, z thỏa mãn điều kiện xy + yz + zx = 1. Tính tổng:
\(S=\sqrt[x]{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+\sqrt[y]{\frac{\left(1+x^2\right)\left(1+z^2\right)}{\left(1+y^2\right)}}+\sqrt[z]{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
Ta có \(x^2+1=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
\(y^2+1=\left(y+z\right)\left(y+x\right)\)
\(z^2+1=\left(z+x\right)\left(z+y\right)\)
Khi đó
\(S=x.\sqrt{\left(y+z\right)^2}+y.\sqrt{\left(x+z\right)^2}+z.\sqrt{\left(x+y\right)^2}=2\left(xy+yz+xz\right)=2\)
Cho 3 số dương x;y;z thỏa mãn điều kiện xy+yz+zx=1
Tính:
\(A=x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+y^2\right)\left(1+x^2\right)}{1+z^2}}\)
Cho 3 số dương x,y,z thỏa mãn: xy+yz+zx=1. Chứng minh rằng:
\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\frac{\left(1+z^2\right)\left(1+y^2\right)}{1+z^2}}=2\)
Ta có:
\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A được:
\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)
\(=xy+xz+xy+yz+xz+zy\)
\(=2\left(xy+yz+xz\right)\)
\(=2\)(do xy+yz+xz=1)
=>Đpcm
Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.
Cho 3 số dương x,y,z thỏa mãn điều kiện: xy+yz+xz=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 có :
\(\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}=\frac{\left(xy+yz+xz+y^2\right)\left(xy+yz+xz+z^2\right)}{\left(xy+yz+xz+x^2\right)}=\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}=\left(y+z\right)^2\)
tương tự ta sẽ có :
\(A=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)
Cho ba số dương x,y,z thõa 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 có : \(xy+yz+zx=1\)
\(\Rightarrow\hept{\begin{cases}1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\\1+y^2=xy+yz+zx+y^2=\left(y+x\right)\left(y+z\right)\\1+z^2=xy+yz+zx+z^2=\left(z+x\right)\left(z+y\right)\end{cases}}\)
Do đó :
\(\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=\sqrt{\frac{\left(y+x\right)\left(y+z\right)\left(z+x\right)\left(z+y\right)}{\left(x+y\right)\left(x+z\right)}}=\sqrt{\left(y+z\right)^2}\)\(=y+z\)
\(x\sqrt{\frac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}=x\left(y+z\right)\)
Hoàn toàn tương tự :
\(y\sqrt{\frac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}=y\left(z+x\right)\)
\(z\sqrt{\frac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}=z\left(x+y\right)\)
Do đó :
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\)
\(=2\left(xy+yz+zx\right)=2\)
Từ xy+yz+zx=1 (giả thiết)
\(\Rightarrow\hept{\begin{cases}1+x^2=xy+yz+zx+x^2\\1+y^2=xy+yz+zx+y^2\\1+z^2=xy+yz+zx+z^2\end{cases}}\Leftrightarrow\hept{\begin{cases}1+x^2=y\left(x+z\right)+x\left(x+z\right)\\1+y^2=z\left(x+y\right)+y\left(x+y\right)\\1+z^2=y\left(x+z\right)+z\left(x+z\right)\end{cases}}.\)
\(\Leftrightarrow\hept{\begin{cases}1+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=\left(x+z\right)\left(y+z\right)\end{cases},}\)Khi đó thế vào biểu thức A đã cho ta có:
\(A=x\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+z\right)\left(x+y\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+y\right)\left(y+z\right)}}\)
\(+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(x+y\right)\left(y+z\right)}{\left(x+z\right)\left(y+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)(vì x,y,z là các số dương)
\(=2\left(xy+yz+xz\right)=2.1=2\)
cho 3 số thực dương x;y;z thỏa mãn x+y+z<=3/2. tìm GTNN của biểu thức:
\(p=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Áp dụng bđt AM-GM ta có
\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)
Ta có \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng bđt AM-GM ta có
\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)
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\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Không phải ngược đâu nha mọi người,dấu bằng không xảy ra nhé!
cho 3 số thực dương x;y;z thỏa mãn x+y+z<=3/2. tìm GTNN của biểu thức :
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Bài này thì AM-GM thôi
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Sử dụng BĐT AM-GM cho 3 số không âm ta có :
\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)
\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)
\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :
\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)
\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)
Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)
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: \(\hept{\begin{cases}1+x^2=xy+yz+xz+x^2=\left(x+z\right)\left(x+y\right)\\1+y^2=xy+yz+xz+y^2=\left(x+y\right)\left(y+z\right)\\1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\end{cases}}\)
\(\Rightarrow S=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)=2\left(xy+yz+xz\right)=2\)