ĐKXĐ: \(x\ge0\)
\(\Leftrightarrow y\sqrt{x}+y-\sqrt{x}-1=-7\)
\(\Leftrightarrow y\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)=-7\)
\(\Leftrightarrow\left(y-1\right)\left(\sqrt{x}+1\right)=-7\)
Dạng pt nghiệm nguyên cơ bản, chỉ cần biện luận thêm \(\sqrt{x}+1\ge1\) để giảm số cặp ước cần xét cho gọn bài.
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!!
Bài tập: Chứng minh
a,\(\left(\sqrt{3}-\sqrt{2}\right).\sqrt{5+2\sqrt{6}}=1\)
b,\(\left[\dfrac{\sqrt{x}-\sqrt{y}}{x-y}+\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right].\dfrac{\sqrt{xy}+1}{\sqrt{x+\sqrt{y}}}\) (với x\(\ge\) 0; y\(\ge\) 0; x\(\ne\)y)
Tim x, y, z
1/ \(\sqrt{x-2}+\sqrt{y-2008}+\sqrt{z-2009}=\dfrac{1}{2}\left(x+y+z\right)\)
2/ \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{x-5}\)
3/ Tinh T = \(x^2+y^2+z^2-7\) biet x-y-z = \(2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45\)
4/ \(2x^2+9y^2-6xy-12y-6x+29=0\)
5/\(4x^2+3y-4x+4xy-10y+9=0\)
Cho x,y,z>0 và \(xy\sqrt{xy}+yz\sqrt{yz}+xz\sqrt{xz}=1\)
Tìm MinP= \(\Sigma\dfrac{x^6}{x^3+y^3}\)
Tìm x, y biết:
\(a)x+y+\sqrt{8y}+5=4\sqrt{x+1}+\sqrt{2}.\sqrt{xy+y}\)
\(b)x+y\le6\)và \(\frac{1}{x}+\frac{25}{y}=6\) với \(\left(x>0;y>0\right)\)
Giải hệ phương trình:
1, \(\left\{{}\begin{matrix}\left(17-3x\right)\sqrt{5-x}+\left(3y-14\right)\sqrt{4-y}=0\\2\sqrt{2x+y+5}+3\sqrt{3x+2y+11}=x^2+6x+13\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x\left(x+y\right)+\sqrt{x+y}=\sqrt{2y}\left(\sqrt{2y^3}+1\right)\\x^2y-5x^2+7\left(x+y\right)-4=6\sqrt[3]{xy-x+1}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\end{matrix}\right.\)
Tìm x, y thuộc Z thỏa mãn
\(y\sqrt{x}+y-\sqrt{x}+6=0\)
Chứng minh: \(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\) với \(x>0,y>0\)
Rút gọn biểu thức:
\(P=\frac{\frac{\left(x-y\right)^3}{\left(\sqrt{x}+\sqrt{y}\right)^3}+2x\sqrt{x}+y\sqrt{y}}{x\sqrt{x}+y\sqrt{y}}+\frac{3\left(\sqrt{xy}-y\right)}{x-y}\) ( Với x>0, y>0; \(x\ne y\))
