Chứng minh (với những giá trị của biến làm cho biểu thức có nghĩa)

a) \(\dfrac{\left(3\sqrt{xy}-6y-2x\sqrt{y}+4y\sqrt{x}\right)\left(3\sqrt{y}+2\sqrt{xy}\right)}{y\left(\sqrt{x}-2\sqrt{y}\right)\left(y-4x\right)}=1\)

b) \(\left(\sqrt{x}-\sqrt{y}-\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\dfrac{2\sqrt{xy}}{x-y}\right)=\sqrt{x}+\sqrt{y}\)

So sánh:

\(A=\sqrt{\dfrac{37}{4}-\sqrt{49+12\sqrt{5}}}\) với \(B=\sqrt{5}-\dfrac{3}{2}\)

Giúp với mình sắp cần rồi

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. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)

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!!

giải giúp mik bt này vs mn!

1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)

2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)

3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)

4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)

5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)

Rút gọn:

\(A=\dfrac{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}+\sqrt[3]{y^4}}{\sqrt[3]{x^2}+\sqrt[3]{xy}+\sqrt[3]{y^2}}\)

\(B=\dfrac{\sqrt[3]{xy}\left(\sqrt[3]{y^2}-\sqrt[3]{x^2}\right)+\left(\sqrt[3]{x^4}-\sqrt[3]{y^4}\right)}{\sqrt[3]{x^4}+\sqrt[3]{x^2y^2}-\sqrt[3]{x^3y}}.\sqrt[3]{x^2}\)

\(C=\left(\dfrac{x\sqrt[3]{x}-2x\sqrt[3]{y}+\sqrt[3]{x^2y^2}}{\sqrt[3]{x^2}-\sqrt[3]{xy}}+\dfrac{\sqrt[3]{x^2y}-\sqrt[3]{xy^2}}{\sqrt[3]{x}-\sqrt[3]{y}}\right).\dfrac{1}{\sqrt[3]{x^2}}\)