Cho \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
Tính \(S=x^{2017}+y^{2017}\)
Cho (x+\(\sqrt{x^2+2017}\))*(y+\(\sqrt{y^2+2017}\)) = 2017
Tính giá trị biểu thức: P= x2017+y2017+2017
Cho ba số thực dương x, y, z thỏa mãn: xy+yz+zx=2017. chứng minh : \(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{3}{2}\)
xy +\(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}\) =\(\sqrt{2017}\) . Tính giá trị của BT : A=\(x\sqrt{y^2+1}+y\sqrt{x^2+1}\)
Giải phương trình:
a) \(2\sqrt{x^2-4}-3=6\sqrt{x-2}-\sqrt{x+2}\)
b) \(\frac{\sqrt{x-2016}-1}{x-2016}+\frac{\sqrt{y-2017}-1}{y-2017}+\frac{\sqrt{z-2018}-1}{z-2018}=\frac{3}{4}\)
c) \(\sqrt{3+\sqrt{3+x}}=x\)
d) \(\sqrt{6x^2+1}=\sqrt{2x-3}+x^2\)
e) \(\sqrt{x^2+3x+5}+\sqrt{x^2-2x+5}=5\sqrt{x}\)
f) \(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)
So sanh: x=\(\sqrt{2019}\) va y=\(2\sqrt{2018}-\sqrt{2017}\)
Tìm x,y \(\in Z\) sao cho\(\sqrt{y}+\sqrt{x}=\sqrt{2017}\)
B1
a,cho \(\left(x+\sqrt{2017+x^2}\right).\left(y+\sqrt{2017+y^2}\right)=2017\)
Tính P=2019x+2019y+2020
b,Cho a,b,c là 3 số dương tm:a+b+c=3
Tìm min P=\(\frac{1}{a^2+a}+\frac{1}{b^2+b}+\frac{1}{c^2+c}\)
cau a cho x,y,z\(\ne\)0 thoa man x+y+z=0. CM: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}=|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}|\) cau b tinh G=\(\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{1+\dfrac{1}{4^2}+\dfrac{1}{5^2}}+.....+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)