Cho \(x,y>0.\)Rút gọn \(P=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}}}.\)
cho x,y >0 rút gọn A=\(\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{x^4+y^4+\frac{x^4y^4}{\left(x^2+y^2\right)^2}}}\)
\(A=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+\left(y^4+x^2y^2\right)^2+x^4y^4}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2x^4y^4+2x^2y^6+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2\right)^2+2\left(x^4+x^2y^2\right)y^4+y^8}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\sqrt{\frac{\left(x^4+x^2y^2+y^4\right)^2}{\left(x^2+y^2\right)^2}}}\)
\(=\sqrt{\frac{x^2y^2}{x^2+y^2}+\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+x^2y^2+y^4}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{x^4+2x^2y^2+y^4}{x^2+y^2}}=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+\frac{\left(x^2+y^2\right)^2}{x^2+y^2}}\)
\(=\sqrt{\frac{x^2y^2}{\left(x+y\right)^2}+x^2+y^2}=\sqrt{\frac{\left(x^2+xy\right)^2+\left(y^2+xy\right)^2+x^2y^2}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy\right)^2+2x^2y^2+2xy^3+y^4}{\left(x+y\right)^2}}=\sqrt{\frac{\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+y^4}{\left(x+y\right)^2}}\)
\(=\sqrt{\frac{\left(x^2+xy+y^2\right)^2}{\left(x+y\right)^2}}=\frac{x^2+xy+y^2}{x+y}\)
giải hệ pt :
\(\hept{\begin{cases}3x^2+6xy+9y^2+\left(x+2y\right)^2\sqrt{x+2y}-3\left(x+2y\right)\sqrt{x+2y}-4\left(x+2y\right)+4\sqrt{x+2y}=0\\\left(\frac{\sqrt[3]{x^2-y^2}}{\sqrt[4]{x}}+\sqrt[4]{\frac{x}{y}}\right)^{2017}+\left(\sqrt[3]{\frac{x}{y}}-\sqrt[4]{\frac{y}{x}}\right)^{2018}=1\end{cases}}\)
1)\(\begin{cases}x^2-y\left(x+y\right)+1=0\\\left(x^2+1\right)\left(x+y-2\right)+y=0\end{cases}\)
2)\(\begin{cases}x^2-4x+y^4+4y^2=2\\xy^2+2y^2+6x=23\end{cases}\)
3)\(\begin{cases}2x+\frac{1}{x+y}=3\\4x^2+4y^2+4xy+\frac{3}{\left(x+y\right)^2}=7\end{cases}\)
4)\(\begin{cases}y^6+x^9+3y^4+3y^2=8\\4y^2-3x^3y^2+x^3=2\end{cases}\)
5)\(\begin{cases}\sqrt{x+y}-2\sqrt{x-y}=1\\x+\sqrt{x^2+y^2}=8\end{cases}\)
6) \(\begin{cases}x+y-2=\frac{y}{x^2+1}\\x^2+y^2+xy=y-1\end{cases}\)
7) \(\begin{cases}4x-1=\sqrt{\left(2x+y\right).\left(2y+1\right)}\\\sqrt{x+2y+1}-\sqrt{x+y-1}=\sqrt{x-1}\end{cases}\)
8) \(\begin{cases}\left(x+y\right).\left(x+4y^2+y\right)+3y^4=0\\\sqrt{x+2y^2+1}-y^2+y+1=0\end{cases}\)
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Rút gọn: \(P=\frac{\sqrt{x^2y^2}}{xy}+\frac{\sqrt{\left(x-y\right)^2}}{x-y}.\left(\frac{\sqrt{x^2}}{x}-\frac{\sqrt{y^2}}{y}\right)\)
Rút gọn biểu thức:
\(2y.\sqrt{\frac{7}{y}}\left(y>0\right)\) \(2y.\sqrt{\frac{-7}{y}}\left(y< 0\right)\)\(x-2-\sqrt{4-4x+x^2}\left(x\le2\right)\)giải hệ phương trình:
1, \(\left\{{}\begin{matrix}2y\left(4y^2+3x^2\right)=x^4\left(x^2+3\right)\\2012^x\left(\sqrt{2y-2x+5}-x+1\right)=4024\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)
Ta xét các trường hợp sau:
Trường hợp 1:
\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:
\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)
\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)
Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)
Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:
\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)
Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)
+ Nếu như:
\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)
+ Nếu như:
\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)
Trường hợp 2:
\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:
\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)
Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)
Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v
3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)
\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)
Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)
\(\Leftrightarrow2b=1-\frac{1}{a}\)
Thay vào (1) ta được :
\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)
\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)
Giải pt được \(a=1\)
Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)
Ta có hệ :
\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)
Vậy...
Rút gọn
\(\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\left(\frac{x^4+4x^2y^2+y^4-4}{x^2+y+xy+x}\right):\frac{1}{2x^2+y+z}\)
cho BIỂU THỨC:
P =\(\left[\left(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\right):\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\right]:\frac{x+1}{2x^2y+2}\)
RÚT GỌN P
35Cho biểu thức
P=\(\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]:\frac{\sqrt{x^3}+y\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{xy^3}+\sqrt{x^3y}}\)
a) Rút gọn P
b)Cho xy=16 . Tìm Min P
34 Cho biểu thức
P=\(\frac{x}{\sqrt{xy}-2y}-\frac{2\sqrt{x}}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}-\frac{1-x}{1-\sqrt{x}}\)
a) Rút gọn P
b)Tính P biết 2x^2+y^2-4x-2xy+4=0