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Ta có \(\left(x+\sqrt{x^2+2003}\right).\left(y+\sqrt{y^2+2003}\right)=2003\)
\(\Rightarrow\frac{-2003}{x-\sqrt{x^2+2003}}.\frac{-2003}{y-\sqrt{y^2+2003}}=2003\)
\(\Leftrightarrow\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=2003\)
\(\Rightarrow\left(x+\sqrt{x^2+2003}\right).\left(y+\sqrt{y^2+2003}\right)=\left(x-\sqrt{x^2+2003}\right).\left(y-\sqrt{y^2+2003}\right)\)
\(\Leftrightarrow xy+x\sqrt{y^2+2003}+y\sqrt{x^2+2003}+\sqrt{\left(x^2+2003\right)\left(y^2+2003\right)}=xy-x\sqrt{y^2+2003}-y\sqrt{x^2+2003}+\sqrt{\left(x^2+2003\right)\left(y^2+2003\right)}\)
\(\Leftrightarrow x\sqrt{y^2+2003}=-y\sqrt{x^2+2003}\left(1\right)\)
Ta thấy pt (1)có 1 nghiệm \(x=y=0\)
\(\left(1\right)\Rightarrow\hept{\begin{cases}x^2\left(y^2+2003\right)=y^2\left(x^2+2003\right)\\x>0;y< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=y^2\\x>0;y< 0\end{cases}\Leftrightarrow}x=-y}\)
Vậy \(x+y=0\)
\(\Leftrightarrow x+y+z=2\sqrt{x-2}+2\sqrt{y+2003}+2\sqrt{z-2004}\)
\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)\)
\(+\left(z-2004-2\sqrt{z-2004}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)
Vì biểu thức trên là tổng của các số hạng không âm nên nó bằng 0 khi và chỉ khi các số hạng phải bằng 0
\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-2003}=1\\\sqrt{z-2004}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2004\\z=2005\end{cases}}}\)
\(ĐK:x\ge2,y\ge-2003,z\ge2004\)
Pt đã cho tương đương :
\(x+y+z-2\sqrt{x-2}-2\sqrt{y+2003}-2\sqrt{z-2004}=0\)
\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)+\left(z-2004-2\sqrt{z-2004}+1\right)\)\(=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=1\\y+2003=1\\z-2004=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=-2002\\z=2005\end{cases}}\)(Thỏa mãn)
a: \(=\dfrac{1}{x-y}\cdot x^2\cdot\left(x-y\right)=x^2\)
b: \(=\sqrt{27\cdot48}\cdot\left|a-2\right|=36\left(a-2\right)\)
c: \(=\left(\sqrt{2012}+\sqrt{2011}\right)^2\)
d: \(=\dfrac{8}{7}\cdot\dfrac{-x}{y+1}\)
e: \(=\dfrac{11}{12}\cdot\dfrac{x}{-y-2}=\dfrac{-11x}{12\left(y+2\right)}\)
hình như...
b) \(x+y+z+8=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)
\(\Leftrightarrow x-3+y-3+z-3+17=2\sqrt{x-3}+4\sqrt{y-3}+6\sqrt{z-3}\)
\(\Leftrightarrow\left(x-3-2\sqrt{x-3}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-3-6\sqrt{z-3}+9\right)+3=0\)
\(\Leftrightarrow\left(\sqrt{x-3}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-3}-3\right)^2+3=0\) (vô nghiệm, VT >/3)
Kl: ptvn
\(\hept{\begin{cases}a^2=x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\\b^2=y^2\left(1+x^2\right)+x^2\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}\end{cases}}\)
\(\Rightarrow a^2-b^2=1\)
\(\Rightarrow a^2=1+b^2\)
Ta có:\(\left(x+\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)=\dfrac{\left(x+\sqrt{x^2+2003}\right)\left(x-\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=\dfrac{\left(x^2-x^2-2003\right)\left(y^2-y^2-2003\right)}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=\dfrac{2003^2}{\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)}=2003\)
=>\(\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=2003\)
=>\(\left(x-\sqrt{x^2+2003}\right)\left(y-\sqrt{y^2+2003}\right)=\left(x+\sqrt{x^2+2003}\right)\left(y+\sqrt{y^2+2003}\right)\)
nhân phá và thu gọn ta được
\(x\sqrt{y^2+2003}=-y\sqrt{x^2+2003}\)(1)
Bình phương
=>x2y2+2003x2=x2y2+2003y2
<=>x2=y2
<=>x=y hoặc x=-y
Thay vào (1) thì
x=y <=>x=y=0
x=-y (luôn đúng)
=>x+y=0