\(\text{Tính }A=x^2\left(x+1\right)-y^2\left(y-1\right)+xy-3xy\left(x-y+1\right)+2019\) \(\text{biết }x-y=\sqrt{12\sqrt{5}+29}\)
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\(Cho\text{ }x,y,z\text{ }\in R\text{ thỏa}\text{ }xyz=1.\text{Tìm Min:}\)
\(P=\left(\left|xy\right|+\left|yz\right|+\left|zx\right|\right)\left[15\sqrt{x^2+y^2+z^2}-7\left(x+y-z\right)\right]+1\)
\(\)\(\begin{cases}\left(\sqrt{2016+x^2}+x\right)\left(\sqrt{504+y^2}+y\right)=1008\\x\sqrt{6\text{x}-4\text{x}y+1}=8\text{x}y+6\text{x}+1\end{cases}\)\(\begin{cases}\sqrt{x+1}\left(x+5\right)-3\left(x-y+1\right)\sqrt{y}\left(y+4\right)=0\\x^2+6y+7=2\sqrt{3y+1}+3\sqrt{x+4y+5}\end{cases}\)
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giải hpt:1)begin{cases}text{x+y+xy(2x+y)5xy }text{x+y+xy(3x-y)4xy}end{cases} 2)begin{cases}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{cases} 3)begin{cases}sqrt{9x+frac{y}{x}}+2.sqrt{y+frac{2x}{y}}4left(frac{2x}{y^2}-1right)left(frac{y}{x^2}-9right)18end{cases}
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giải hpt:1)\(\begin{cases}\text{x+y+xy(2x+y)=5xy }\\\text{x+y+xy(3x-y)=4xy}\end{cases}\)
2)\(\begin{cases}\left(2x+y+1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{cases}\)
3)\(\begin{cases}\sqrt{9x+\frac{y}{x}}+2.\sqrt{y+\frac{2x}{y}}=4\\\left(\frac{2x}{y^2}-1\right)\left(\frac{y}{x^2}-9\right)=18\end{cases}\)
1. \(\begin{cases}x+y+xy\left(2x+y\right)=5xy\\x+y+xy\left(3x-y\right)=4xy\end{cases}\) \(\Leftrightarrow\begin{cases}2y-x=1\\x+y+xy\left(2x+y\right)=5xy\end{cases}\) (trừ 2 vế cho nhau)
\(\Leftrightarrow\begin{cases}x=2y-1\\\left(2y-1\right)+y+\left(2y-1\right)y\left(4y-2+y\right)=5\left(2y-1\right)y\end{cases}\) \(\Leftrightarrow\begin{cases}x=2y-1\\10y^3-19y^2+10y-1=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=1\end{cases}\)
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\(\text{Cho x,y,z }\in R\text{ thỏa mãn điều kiện }xyz=1\text{.Tìm Min:}\)
\(P=\left(\left|xy\right|+\left|yz\right|\left|zx\right|\right).\left[15\sqrt{x^2+y^2+z^2}-7\left(x+y-z\right)\right]+1\)
\(\left|xy\right|+\left|yz\right|+\left|zx\right|\)
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1)begin{cases}sqrt{4x^2+left(4x-9right)left(x-3yright)}+sqrt{3xy}9y4sqrt{left(x+2right)left(3y+2xright)}3x+9end{cases} 4)begin{cases}left(x^2+yright)sqrt{x-y+6}2x^2-x+3y-2sqrt{10x-xy-12}+1frac{y-x}{sqrt{y-4}+sqrt{6-x}}end{cases}2)begin{cases}x^2+left(y-6right)^2y+13x+27sqrt{9x^2+left(2x-3right)left(x-yright)}+4sqrt{xy}7yend{cases} 5)begin{cases}sqrt{4xy+left(3sqrt{xy}-7right)left(x-yright)}+2sqrt{xy}4yle...
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1)\(\begin{cases}\sqrt{4x^2+\left(4x-9\right)\left(x-3y\right)}+\sqrt{3xy}=9y\\4\sqrt{\left(x+2\right)\left(3y+2x\right)}=3x+9\end{cases}\) 4)\(\begin{cases}\left(x^2+y\right)\sqrt{x-y+6}=2x^2-x+3y-2\\\sqrt{10x-xy-12}+1=\frac{y-x}{\sqrt{y-4}+\sqrt{6-x}}\end{cases}\)
2)\(\begin{cases}x^2+\left(y-6\right)^2=y+13x+27\\\sqrt{9x^2+\left(2x-3\right)\left(x-y\right)}+4\sqrt{xy}=7y\end{cases}\) 5)\(\begin{cases}\sqrt{4xy+\left(3\sqrt{xy}-7\right)\left(x-y\right)}+2\sqrt{xy}=4y\\\left(2x+1\right)\left[12y-1+9\sqrt{xy}-x^2-x\right]=27\left(x+1\right)\end{cases}\)
3)\(\begin{cases}\sqrt{\left(x+2\right)\left(y+1\right)+\left(x-y+1\right)\sqrt{y^2+1}}+\sqrt{x+2}=y+\sqrt{y+1}+1\\\sqrt{3x+1}-\sqrt{y+1}=2x^2+4x-y-1\end{cases}\)
giải hệ phương trình
a,left{{}begin{matrix}left(x+sqrt{x^2+1}right)left(y+sqrt{y^2+1}right)1y+frac{y}{sqrt{x^2-1}}frac{35}{12}end{matrix}right.
b,left{{}begin{matrix}left(x^2+y^2right)left(x+y+1right)25left(y+1right)x^2+xy+2y^2+x-8y9end{matrix}right.
c,left{{}begin{matrix}2x^2+3xy-2y^2-5left(2x-yright)0x^2-2xy-3y^2+150end{matrix}right.
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giải hệ phương trình
\(a,\left\{{}\begin{matrix}\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\\y+\frac{y}{\sqrt{x^2-1}}=\frac{35}{12}\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}2x^2+3xy-2y^2-5\left(2x-y\right)=0\\x^2-2xy-3y^2+15=0\end{matrix}\right.\)
17 tháng 10 2015 lúc 11:04
\(\text{Tính }C=\frac{\left(1+\sqrt{3}\right)x^2y-\left(2-\sqrt{5}\right)xy^2}{x^3+y^3}\text{ với }x,y\ne0\text{ và }\frac{x}{4}=\frac{y}{7}\)
Đặt \(\frac{x}{4}=\frac{y}{7}\) = k => x = 4k; y = 7k ( k khác 0)
Thay vào C ta được: \(C=\frac{\left(1+\sqrt{3}\right)\left(4k\right)^2.7k-\left(2-\sqrt{5}\right).4k.\left(7k\right)^2}{\left(4k\right)^3+\left(7k\right)^3}=\frac{\left(112.\left(1+\sqrt{3}\right)-196.\left(2-\sqrt{5}\right)\right).k^3}{407k^3}\)
\(C=\frac{112+112\sqrt{3}-392+196\sqrt{5}}{407}=\frac{112\sqrt{3} +196\sqrt{5}-280}{407}\)
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giải hệ phương trình
a) left{{}begin{matrix}sqrt{2x^2+2y^2}+sqrt{frac{4}{3}left(x^2+xy+y^2right)}2left(x+yright)sqrt{3x+1}+sqrt{5x+4}3xy-y+3end{matrix}right.
b) left{{}begin{matrix}sqrt{5x^2+2xy+2y^2}+sqrt{2x^2+2xy+5y^2}3left(x+yright)sqrt{x+2y+1}+2sqrt[3]{12x+7y+8}2xy+x+5end{matrix}right.
c)left{{}begin{matrix}x^2+xy+x+30left(x+1right)^2+3left(y+1right)+2left(xy-sqrt{x^2y+2y}right)0end{matrix}right.
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giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
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caau a) binh phuong len ra no x=y tuong tu
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c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
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ai giúp t với
1:left{begin{matrix}xsqrt{12-y}+sqrt{yleft(12-x^2right)}12x^3-8x-12sqrt{y-2}end{matrix}right.
2:left{begin{matrix}left(1-yright)sqrt{x-y}+x2+left(x-y-1right)sqrt{y}2y^2-3x+6y+12sqrt{x-2y}-sqrt{4x-5y-3}end{matrix}right.
3:left{begin{matrix}yleft(x^2+2x+2right)xleft(y^2+6right)left(y-1right)left(x^2+2x+7right)left(x+1right)left(y^2+1right)end{matrix}right.
4:left{begin{matrix}x-2sqrt{y+1}3x^3-4x^2sqrt{y+1}-9x-8y-52-4xyend{matrix}right.
5:left{begin{matrix}frac{y-2x+sqrt{y}-x}{sq...
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ai giúp t với
1:\(\left\{\begin{matrix}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\\x^3-8x-1=2\sqrt{y-2}\end{matrix}\right.\)
2:\(\left\{\begin{matrix}\left(1-y\right)\sqrt{x-y}+x=2+\left(x-y-1\right)\sqrt{y}\\2y^2-3x+6y+1=2\sqrt{x-2y}-\sqrt{4x-5y-3}\end{matrix}\right.\)
3:\(\left\{\begin{matrix}y\left(x^2+2x+2\right)=x\left(y^2+6\right)\\\left(y-1\right)\left(x^2+2x+7\right)=\left(x+1\right)\left(y^2+1\right)\end{matrix}\right.\)
4:\(\left\{\begin{matrix}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9x-8y=-52-4xy\end{matrix}\right.\)
5:\(\left\{\begin{matrix}\frac{y-2x+\sqrt{y}-x}{\sqrt{xy}}+1=0\\\sqrt{1-xy}+x^2-y^2=0\end{matrix}\right.\)