Violympic toán 9
Các câu hỏi tương tự
Cho 2 số thực a, b thỏa mãn xy + \(\sqrt{\left(x^2+1\right)\left(y^2+1\right)}=1\)
CMR: \(x\sqrt{1+y^2}+y\sqrt{1+x^2}=0\)
Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
Đọc tiếp
Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
Cho x,y,z thỏa mãn xy+yz+xz = 1.Tính
\(S=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Giải hệ phương trình:
1. left{{}begin{matrix}x+32sqrt{left(3y-xright)left(y+1right)}sqrt{3y-2}-sqrt{dfrac{x+5}{2}}xy-2y-2end{matrix}right.
2. left{{}begin{matrix}sqrt{2y^2-7y+10-xleft(y+3right)}+sqrt{y+1}x+1sqrt{y+1}+dfrac{3}{x+1}x+2yend{matrix}right.
3. left{{}begin{matrix}sqrt{4x-y}-sqrt{3y-4x}12sqrt{3y-4x}+yleft(5x-yright)xleft(4x+yright)-1end{matrix}right.
4. left{{}begin{matrix}9sqrt{dfrac{41}{2}left(x^2+dfrac{1}{2x+y}right)}3+40xx^2+5xy+6y4y^2+9x+9end{matrix}right.
5. left{{}begin{mat...
Đọc tiếp
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
a) Cho x,y,z thỏa mãn x+y+z+xy+yz+zx=6. Tìm Min \(P=x^2+y^2+z^2\)
giải hệ pt : 1) \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}+\sqrt{2-\dfrac{1}{y}}=2\\\dfrac{1}{\sqrt{y}}+\sqrt{2-\dfrac{1}{x}}=2\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^4+x^2y^2+y^4=21\end{matrix}\right.\)
Cho x, y, z dương thỏa mãn xyz=1. Tìm GTLN của \(\dfrac{1}{\sqrt{\left(x+y\right)^2+\left(x+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(y+z\right)^2+\left(y+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(z+x\right)^2+\left(z+1\right)^2+4}}\)
Cho 3 số dương x, y, z thỏa mãn điều kiện xy + yz + zx = 1. Tính tổng:
\(S=\sqrt[x]{\frac{\left(1+y^2\right)\left(1+z^2\right)}{\left(1+x^2\right)}}+\sqrt[y]{\frac{\left(1+x^2\right)\left(1+z^2\right)}{\left(1+y^2\right)}}+\sqrt[z]{\frac{\left(1+x^2\right)\left(1+y^2\right)}{\left(1+z^2\right)}}\)
Với x,y là các số dương thỏa mãn \(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)=\sqrt{2000}}\)
Tính giá trị biểu thức \(S=x\sqrt{1+y^2}+y\sqrt{1+X^2}\)
Cho x,y thỏa mãn x>1, y<0 và \(\frac{\left(x+y\right)\left(x^3-y^3\right)\sqrt{4x-2\sqrt{4x-1}}}{\left(1-\sqrt{4x-1}\right)\left(x^2y^2+xy^3+y^4\right)}=-8\). Vậy \(\frac{x}{y}=\)