cho \(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\). CM: \(x^{2019}+y^{2019}=0\)
cho x,y ,z là các số dương thỏa mãn:xy+yz+zx=2019
Tính gtrị bt\(P=x\sqrt{\frac{\left(y^2+2019\right).\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right).\left(x^2+2019\right)}{y^{2^{ }}+2019}}+z\sqrt{\frac{\left(x^2+2019\right).\left(y^2+2019\right)}{z^2+2019}}\)
Có \(y^2+2019=y^2+xy+yz+zx=y\left(x+y\right)+z\left(x+y\right)=\left(y+z\right)\left(x+y\right)\)
\(x^2+2019=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
\(z^2+2019=z^2+xy+yz+xz=z\left(z+y\right)+x\left(y+z\right)=\left(z+x\right)\left(y+z\right)\)
Có \(P=x\sqrt{\frac{\left(y^2+2019\right)\left(z^2+2019\right)}{x^2+2019}}+y\sqrt{\frac{\left(z^2+2019\right)\left(x^2+2019\right)}{y^2+2019}}+z\sqrt{\frac{\left(x^2+2019\right)\left(y^2+2019\right)}{z^2+2019}}\)
=\(x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(z+y\right)}{\left(x+z\right)\left(y+x\right)}}+y\sqrt{\frac{\left(z+x\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(y+z\right)\left(x+y\right)}}+z\sqrt{\frac{\left(x+z\right)\left(x+y\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
=\(x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=\(x\left|y+z\right|+y\left|x+z\right|+z\left|x+y\right|\)
=\(x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\) (vì x,y,z >0)
= xy+xz+xy+yz+xz+yz
=2(xy+xz+yz)=2.2019(vì xy+xz+yz=2019)
=4038
Vậy P=4038
Cho \(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\)
Tính x + y
\(\left(x+\sqrt{x^2+2019}\right)\left(\sqrt{x^2+2019}-x\right)=x^2+2019-x^2=2019\)
\(\Rightarrow\sqrt{x^2+2019}-x=y+\sqrt{y^2+2019}\left(2\right)\)
Tương tự \(\sqrt{y^2+2019}-y=x+\sqrt{x^2+2019}\left(1\right)\)
Lấy (2) - (1) được: -2x = 2y
<=> -x = y
<=> x + y = 0
Cho x,y là 2 số t/m : \(\left(x+\sqrt{x^2+\sqrt{2019}}\right)\)\(\left(y+\sqrt{y^2+\sqrt{2019}}\right)=\sqrt{2019}\)
Tính B = x + y biết :
\(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\)
cho x,y là số thực thõa mãn
\(\left(x+\sqrt{x^2+2019}\right)\left(y+\sqrt{y^2+2019}\right)=2019\)
tính x+y
Từ giả thiết suy ra
\(x+\sqrt{x^2+2019}=\frac{2019}{y+\sqrt{y^2+2019}}\)
mà \(x+\sqrt{x^2+2019}=\frac{2019}{\sqrt{x^2+2019}-x}\)(nhân liên hợp)
\(\Rightarrow\)\(y+\sqrt{y^2+2019}=\sqrt{x^2+2019}-x\)(1)
Tương tự, có \(\sqrt{y^2+2019}-y=x+\sqrt{x^2+2019}\)(2)
Trừ từng vế của (1), (2) ta có
2y=\(-\)2x\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
a,Cho \(\left(x-2019+\sqrt{\left(x-2019\right)^2+2020}\right)\left(y-2019+\sqrt{\left(y-2019\right)^2+2020}\right)=2020\)Tính : D = x + y
b, Cho \(\frac{-3}{2}\le x\le\frac{3}{2},x\ne0,a=\sqrt{3+2x}-\sqrt{3-2x}\)
Tính : \(G=\frac{\sqrt{6+2\sqrt{9-4x^2}}}{x}\) theo a.
Em cảm ơn mọi người nhiều ạ.
Cho \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)tính \(A=x^{2019}+y^{2019}\)
Ta xét \(\left(x+\sqrt{x^2+1}\right)\left(x-\sqrt{x^2+1}\right)=x^2-\left(x^2+1\right)=-1.\)
Mà \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Rightarrow x-\sqrt{x^2+1}=-\left(y+\sqrt{y^2+1}\right)\)
\(\Leftrightarrow x+y=\sqrt{x^2+1}-\sqrt{y^2+1}.\)(1)
Xét \(\left(y+\sqrt{y^2+1}\right)\left(y-\sqrt{y^2+1}\right)=y^2-\left(y^2+1\right)=-1\)
Mà \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)
\(\Rightarrow y-\sqrt{y^2+1}=-\left(x+\sqrt{x^2+1}\right).\)
\(\Leftrightarrow x+y=\sqrt{y^2+1}-\sqrt{x^2+1}\)(2)
Cộng 2 vế của (1) và (2) Ta được
\(2\left(x+y\right)=0\Leftrightarrow x=-y\)Thế vào A
\(A=x^{2019}+y^{2019}=\left(-y\right)^{2019}+y^{2019}=0\)
Cho\(\left(x+\sqrt{x^2+2017}\right)\cdot\left(y+\sqrt{y^2+2017}\right)=2017\)
Tính A=\(x^{2019}+y^{2019}\)
Giải hệ phương trình:
\(\hept{\begin{cases}x^2+y^2=1\\\sqrt[2019]{x}-\sqrt[2019]{y}=\left(\sqrt[2020]{y}-\sqrt[2020]{x}\right)\left(xy+x+y+2021\right)\end{cases}}\)
xét x=y,x>y và x<y chú ý tới điều kiện x,y thuộc -1;1 nữa