Giai pt:\(\sqrt{x-2}+\sqrt{y+2003}+\sqrt{z-2004}=\frac{1}{2}\left(x+y+z\right)\)
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Giải phương trình:
\(\sqrt{x-2}+\sqrt{y+2003}+\sqrt{x-2004}=\frac{1}{2}\left(x+y+z\right)\)
Giai PT
a, \(x+\sqrt{2-x^2}=4y^2+4y+3\)
b,\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
tìm số x,y,x TM\(\frac{\sqrt{x-2002}-1}{x-2002}+\frac{\sqrt{y-2003}-1}{y-2003}+\frac{\sqrt{z-2004}-1}{z-2004}=\frac{3}{4}\)
\(\frac{\sqrt{x-2002}}{x-2002}-\frac{1}{x-2002}+\frac{\sqrt{y-2003}}{y-2003}-\frac{1}{y-2003}+\frac{\sqrt{z-2004}}{z-2004}-\frac{1}{z-2004}=\frac{3}{4}\)
\(1-\frac{1}{x-2002}+1-\frac{1}{y-2003}+1-\frac{1}{z-2004}=\frac{3}{4}\)
\(3-\frac{1}{x-2002}-\frac{1}{y-2003}-\frac{1}{z-2004}=\frac{3}{4}\)
\(\frac{1}{x-2002}+\frac{1}{y-2003}+\frac{1}{z-2004}=3-\frac{3}{4}=\frac{9}{4}\)
=> không có giá trị x,y,z thỏa mãn đề
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Cho 3 số dương x,y,z thỏa mãn x + y + z = xyz. Cmr:
\(A=\frac{\sqrt{\left(1+y^2\right)\left(1+z^2\right)}-\sqrt{1+y^2}-\sqrt{1+z^2}}{yz}+\frac{\sqrt{\left(1+z^2\right)\left(1+x^2\right)}-\sqrt{1+x^2}-\sqrt{1+z^2}}{xz}+\frac{\sqrt{\left(1+x^2\right)\left(1+y^2\right)}-\sqrt{1+x^2}-\sqrt{1+y^2}}{xy}=0\)
@Akai Haruma, Nguyen, Nguyễn Thị Ngọc Thơsvtkvtm
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Bạn tham khảo tại đây:
Câu hỏi của Vũ Sơn Tùng - Toán lớp 9 | Học trực tuyến
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Giải pt:
\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
giải pt: \(\sqrt{x-2009}+\sqrt{y-2008}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\)
Phương trình có vô số nghiệm
Nếu thay \(\sqrt{y-2008}\) bằng \(\sqrt{y+2008}\) thì phương trình có bộ nghiệm duy nhất: \(\left(x;y;z\right)=\left(2010;-2007;3\right)\)
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Cho \(x+y+z=\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)Chứng minh: \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}=\frac{2}{\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Cho x,y,z>0 thỏa mãn \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=2\\x+y+z=2\end{matrix}\right.\)
Tính P=\(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\left(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\right)\)
+ \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=4\Rightarrow x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
+ \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)\)
+ Tương tự : \(y+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\); \(z+1=\left(\sqrt{x}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+ \(P=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}+\sqrt{z}\right)^2\left(\sqrt{z}+\sqrt{x}\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=2\)
Cho x,y,z>0 tm : \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=2\\x+y+z=2\end{matrix}\right.\) .Tính:
P= \(\sqrt{\left(x+1\right).\left(y+1\right).\left(z+1\right)}.\left(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\right)\)
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)
\(\Leftrightarrow x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}=4\)
\(\Leftrightarrow2+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Khi đó ta có : \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow x+1=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(\Leftrightarrow x+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
Tương tự : \(y+1=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\);
\(z+1=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Ta lần lượt xét các biểu thức :
+) \(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\sqrt{\left[\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\right]^2}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+) \(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
Do đó ta có :
\(P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\cdot\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(P=2\)
Vậy...
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