giải hpt: \(\left\{{}\begin{matrix}x\sqrt{5}-\left(1+\sqrt{3}y\right)=1\\\left(1-\sqrt{3}\right)+y\sqrt{5}=1\end{matrix}\right.\)
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NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 5 2020
\(\Leftrightarrow\left\{{}\begin{matrix}5x-\sqrt{5}\left(1+\sqrt{3}\right)y=\sqrt{5}\\\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)x+\sqrt{5}\left(1+\sqrt{3}\right)y=1+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-\sqrt{5}\left(1+\sqrt{3}\right)y=\sqrt{5}\\-2x+\sqrt{5}\left(1+\sqrt{3}\right)y=1+\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5x-\sqrt{3}\left(1+\sqrt{3}\right)y=\sqrt{5}\\3x=1+\sqrt{3}+\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{1+\sqrt{3}+\sqrt{5}}{3}\\y=\frac{x\sqrt{5}-1}{1+\sqrt{3}}=\frac{\sqrt{5}+\sqrt{15}+2}{1+\sqrt{3}}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2021
a.
ĐKXĐ: \(x;y\ge-1;xy\ge0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=\sqrt{xy}\\x+y+2\sqrt{xy+x+y+1}=14\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\ge0\end{matrix}\right.\) với \(u^2\ge4v\)
\(\Rightarrow\left\{{}\begin{matrix}u-3=\sqrt{v}\\u+2\sqrt{u+v+1}=14\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-6u+9\left(u\ge3\right)\\4\left(u+v+1\right)=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\4u+4\left(u^2-6u+9\right)+4=\left(14-u\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\3u^2+8u-156=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\\left[{}\begin{matrix}u=6\\u=-\dfrac{26}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=6\\v=9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=6\\xy=9\end{matrix}\right.\) \(\Rightarrow x=y=3\)
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 2 2021
b.
ĐKXĐ: \(x;y\ge1\)
Xét \(\sqrt{x-1}+\sqrt{y-1}=3\)
\(\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=\dfrac{11-x-y}{2}\)
Thế vào pt đầu:
\(x+y=5+\dfrac{11-x-y}{2}\)
\(\Leftrightarrow x+y=7\Rightarrow y=7-x\)
Thế xuống pt dưới:
\(\sqrt{x-1}+\sqrt{6-x}=3\)
\(\Leftrightarrow5+2\sqrt{\left(x-1\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\left(x-1\right)\left(6-x\right)=4\)
\(\Leftrightarrow...\)
25 tháng 11 2023
a:
ĐKXĐ: y+1>=0
=>y>=-1
\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}+7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\left(x^2-2x\right)+2\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7\left(x^2-2x\right)=-7\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x=-1\\3\cdot\left(-1\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x^2-2x+1=0\\2\sqrt{y+1}=-3+7=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\sqrt{y+1}=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1=0\\y+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\left(nhận\right)\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\cdot\sqrt{\left(2x-2\right)^2}+5\cdot\sqrt{\left(y+2\right)^2}=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20\left|x-1\right|-12\left|y+2\right|=28\\20\left|x-1\right|+25\left|y+2\right|=65\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-37\left|y+2\right|=-37\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left|y+2\right|=1\\4\left|x-1\right|=13-5=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|y+2\right|=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-1\in\left\{2;-2\right\}\\y+2\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;-1\right\}\\y\in\left\{-1;-3\right\}\end{matrix}\right.\)
c: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-1\\y< >-4\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=3-4=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=2-9=-7\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=-1+2\cdot\dfrac{19}{11}=\dfrac{27}{11}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x=\dfrac{2}{9}\end{matrix}\right.\)(nhận)
d:
ĐKXĐ: x<>1 và y<>-2
\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)
22 tháng 3 2020
\( a)\left\{ \begin{array}{l} x\sqrt 5 - \left( {1 + \sqrt 3 } \right)y = 1\\ \left( {1 - \sqrt 3 } \right)x + y\sqrt 5 = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\sqrt 5 - \left( {1 + \sqrt 3 } \right)y = 1\\ x = - \dfrac{{1 + \sqrt 3 - y\sqrt 5 - y\sqrt {15} }}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{{ - 1 - \sqrt 3 - \sqrt 5 }}{3}\\ y = - \dfrac{{ - 1 - \sqrt 3 - \sqrt 5 }}{3} \end{array} \right.\\ b)\left\{ \begin{array}{l} 0,2x + 0,1y = 0,3\\ 3x + y = 5 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} 0,2x + 0,1y = 0,3\\ y = 5 - 3x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 2\\ y = - 1 \end{array} \right.\\ c)\left\{ \begin{array}{l} \left( {3x + 2} \right)\left( {2y - 3} \right) = 6xy\\ \left( {4x + 5} \right)\left( {y - 4} \right) = 4xy \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \dfrac{4}{9}y - \dfrac{2}{3}\\ \left( {4x + 5} \right)\left( {y - 4} \right) = 4xy \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - \dfrac{{50}}{{19}}\\ y = - \dfrac{{84}}{{19}} \end{array} \right. \)
3 tháng 9 2023
1) đkxđ \(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\y\ge0\end{matrix}\right.\)
Xét biểu thức \(P=x^3+y^3+7xy\left(x+y\right)\)
\(P=\left(x+y\right)^3+4xy\left(x+y\right)\)
\(P\ge4\sqrt{xy}\left(x+y\right)^2\)
Ta sẽ chứng minh \(4\sqrt{xy}\left(x+y\right)^2\ge8xy\sqrt{2\left(x^2+y^2\right)}\) (*)
Thật vậy, (*)
\(\Leftrightarrow\left(x+y\right)^2\ge2\sqrt{2xy\left(x^2+y^2\right)}\)
\(\Leftrightarrow\left(x+y\right)^4\ge8xy\left(x^2+y^2\right)\)
\(\Leftrightarrow x^4+y^4+6x^2y^2\ge4xy\left(x^2+y^2\right)\) (**)
Áp dụng BĐT Cô-si, ta được:
VT(**) \(=\left(x^2+y^2\right)^2+4x^2y^2\ge4xy\left(x^2+y^2\right)\)\(=\) VP(**)
Vậy (**) đúng \(\Rightarrowđpcm\). Do đó, để đẳng thức xảy ra thì \(x=y\).
Thế vào pt đầu tiên, ta được \(\sqrt{2x-3}-\sqrt{x}=2x-6\)
\(\Leftrightarrow\dfrac{x-3}{\sqrt{2x-3}+\sqrt{x}}=2\left(x-3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(nhận\right)\\\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}=2\end{matrix}\right.\)
Rõ ràng với \(x\ge\dfrac{3}{2}\) thì \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x}}\le\dfrac{1}{\sqrt{\dfrac{2.3}{2}-3}+\sqrt{\dfrac{3}{2}}}< 2\) nên ta chỉ xét TH \(x=3\Rightarrow y=3\) (nhận)
Vậy hệ pt đã cho có nghiệm duy nhất \(\left(x;y\right)=\left(3;3\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 7 2020
3.
ĐKXĐ: ...
Trừ vế cho vế ta được:
\(2x-2y=y-x+\sqrt{y-2}-\sqrt{x-2}\)
\(\Leftrightarrow3\left(x-y\right)+\sqrt{x-2}-\sqrt{y-2}=0\)
\(\Leftrightarrow3\left(x-y\right)+\frac{x-y}{\sqrt{x-2}+\sqrt{y-2}}=0\)
\(\Leftrightarrow\left(x-y\right)\left(3+\frac{1}{\sqrt{x-2}+\sqrt{y-2}}\right)=0\)
\(\Leftrightarrow x=y\) (ngoặc to luôn dương)
Thay vào pt đầu:
\(2x-2=x+\sqrt{x-2}\)
\(\Leftrightarrow x-2=\sqrt{x-2}\Rightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y=2\\x=y=3\end{matrix}\right.\)
2 tháng 1 2023
Bài 2:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)
=>-4x-2y=3 và 8x+2y=-2
=>x=1/4; y=-2
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)
=>y=6 và x-2=5/4
=>x=13/4; y=6
c: =>x+y=24 và 3x+y=78
=>-2x=-54 và x+y=24
=>x=27; y=-3
d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)
=>y+2=1 và x-1=25
=>x=26; y=-1
20 tháng 1
a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
c: ĐKXĐ: y>2
\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)
Lời giải:
HPT \(\Leftrightarrow \left\{\begin{matrix} x\sqrt{5}-y\sqrt{3}=2\\ y\sqrt{5}=\sqrt{3}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\sqrt{5}=y\sqrt{3}+2\\ y=\sqrt{\frac{3}{5}}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\sqrt{5}=\frac{10+3\sqrt{5}}{5}\\ y=\sqrt{\frac{3}{5}}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{3+2\sqrt{5}}{5}\\ y=\sqrt{\frac{3}{5}}\end{matrix}\right.\)
Vậy.........