Cho x,y biết \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
Chứng minh \(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)
Cho x; y; z là các số dương nhỏ hơn 1 thỏa mãn x + y + z + 2\(\sqrt{xyz}\)= 1. Chứng minh rằng \(\sqrt{x\left(1-y\right)\left(1-z\right)}+\sqrt{y\left(1-x\right)\left(1-z\right)}+\sqrt{z\left(1-x\right)\left(1-y\right)}=1+\sqrt{xyz}\)
\(\sqrt{x\left(1-y\right)\left(1-z\right)}=\sqrt{x\left(yz-y-z+1\right)}=\sqrt{x\left(yz-y-z+x+y+z+2\sqrt{xyz}\right)}\)
\(=\sqrt{x\left(yz+x+2\sqrt{xyz}\right)}=\sqrt{x^2+2x\sqrt{xyz}+xyz}=\sqrt{\left(x+\sqrt{xyz}\right)^2}\)
\(=x+\sqrt{xyz}\)
Tương tự: \(\sqrt{y\left(1-x\right)\left(1-z\right)}=y+\sqrt{xyz}\) ; \(\sqrt{z\left(1-x\right)\left(1-y\right)}=z+\sqrt{xyz}\)
\(\Rightarrow VT=x+y+z+3\sqrt{xyz}=1-2\sqrt{xyz}+3\sqrt{xyz}=1+\sqrt{xyz}\) (đpcm)
Cho các số x,y thỏa mãn điều kiện \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\). Chứng minh rằng:\(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)
áp dụng cauchy ngược dấu là xong nhé bạn :>> mình ko đánh đc sorry bạn
Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Cho x, y, z > 0
Chứng minh :
\(\sqrt{x\left(y+1\right)}+\sqrt{y\left(z+1\right)}+\sqrt{z\left(x+1\right)}\le\frac{3}{2}\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
chứng minh $\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}\leq \frac{3}{2}\sqrt{(x+1)(y+1)(z+1)}$ - Bất đẳng thức và cực trị - Diễn đàn Toán học
Cho x,y thỏa: \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
Chứng minh: \(\left(x+\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)
Chắc chắn đúng đề ạ!
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 \(x+y+z=\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)Chứng minh: \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}=\frac{2}{\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
\(\frac{\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{y}\left(\sqrt{y}+2\right)-2\sqrt{xy}+1}{\sqrt{x}\left(\sqrt{x}-2\sqrt{y}\right)+\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)
\(=\frac{x-2\sqrt{x}+y+2\sqrt{y}-2\sqrt{xy}+1}{x-2\sqrt{xy}+y-1}\)\(=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1}{\left(\sqrt{x}-\sqrt{y}\right)^2-1}\)
\(=\frac{\left(\sqrt{x}-\sqrt{y}-1\right)^2}{\left(\sqrt{x}-\sqrt{y}+1\right)\left(\sqrt{x}-\sqrt{y}-1\right)}=\frac{\sqrt{x}-\sqrt{y}-1}{\sqrt{x}-\sqrt{y}+1}\)
Cho x,y biết \(\left(x+\sqrt{1+y^2}\right)\left(y+\sqrt{1+x^2}\right)=1\)
CMR \(x\left(\sqrt{1+x^2}\right)\left(y+\sqrt{1+y^2}\right)=1\)