Bài 1: Căn bậc hai
Các câu hỏi tương tự
F = \(\dfrac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left[\left(\dfrac{1}{x}+\dfrac{1}{y}\right).\dfrac{1}{x+y+2\sqrt{xy}}+\dfrac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}.\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}\right)\right]\)
Cho 3 số dương x,y,z thỏa mãn x + y + z = xyz. Cmr:
\(A=\frac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\frac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{xz}+\frac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}=0\)
1. Tính:
sqrt{dfrac{x-1+sqrt{2x-3}}{x+2-sqrt{2x+3}}}
2. Chứng minh:
a) dfrac{left(3sqrt{xy}-6y.2xsqrt{y}+4ysqrt{x}right)left(3sqrt{y}+2sqrt{xy}right)}{yleft(sqrt{x}-2sqrt{y}right)left(y-4xright)}1
b) left(sqrt{x}-sqrt{y}-dfrac{sqrt{xy}}{sqrt{x}+sqrt{y}}right)left(dfrac{sqrt{x}}{sqrt{x}+sqrt{y}}+dfrac{y}{sqrt{x}-sqrt{y}}-dfrac{2sqrt{xy}}{xy}right)sqrt{x}+sqrt{y}
Đọc tiếp
1. Tính:
\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\)
2. Chứng minh:
a) \(\dfrac{\left(3\sqrt{xy}-6y.2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)
b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{y}{\sqrt{x}-\sqrt{y}}-\dfrac{2\sqrt{xy}}{xy}\right)=\sqrt{x}+\sqrt{y}\)
Cho các số thực dương x, y, z thỏa mãn : xyz=1.CMR:
\(\dfrac{1}{\left(\sqrt{xy}+\sqrt{x}+1\right)^2}+\dfrac{1}{\left(\sqrt{yz}+\sqrt{y}+1\right)^2}+\dfrac{1}{\left(\sqrt{xz}+\sqrt{z}+1\right)^2}\ge\dfrac{1}{3}\)
Giúp mk với , mk sắp thi r...
Cho x, y, z thỏa mãn xy+yz+xz=1
Hãy tính giá trị biểu thức A=\(\sqrt[x]{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+\sqrt[y]{\frac{\left(1+z^2\right)\left(1+x^2\right)}{\left(1+y^2\right)}}+\sqrt[z]{\frac{\left(1+x^2\left(1+y^2\right)\right)}{\left(1+z^2\right)}}\)
Cho 3 số x y z thỏa mãn x+y+z=xyz.Cm:\(\dfrac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\dfrac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+z^2}-\sqrt{1+x^2}}{zx}+\dfrac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{yz}=0\)
Cho x,y >0 thỏa mãn: \(xy+\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=\sqrt{2018}\)
Tính \(Á=x\sqrt{y^2+1}+y\sqrt{x^2+1}\)
cho x, y là hai số thực thỏa mãn : xy + \(\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\)=1
cm: \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=0\)
ghpt
1) \(\left\{{}\begin{matrix}3\left(2-x\right)\sqrt{2-y^2}=2-y+\dfrac{4}{x+1}\\\left(x^2+xy-x+y-2\right)\sqrt{2-y^2}+2=x+y\end{matrix}\right.\)