giai hpt:
\(\left\{{}\begin{matrix}3x-y^2-2\sqrt{(x-2)(y+1)}=-5\\-2x+y^2+y=6\end{matrix}\right.\)
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NV
1
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2019
Cộng vế với vế của 2 pt ta được:
\(x+y-2\sqrt{\left(x-2\right)\left(y+1\right)}=1\)
\(\Leftrightarrow x-2-2\sqrt{\left(x-2\right)\left(y+1\right)}+y+1=0\) (1)
- Nếu \(\left\{{}\begin{matrix}x-2< 0\\y+1< 0\end{matrix}\right.\) \(\Rightarrow VT< 0\Rightarrow\) pt (1) vô nghiệm
- Nếu \(\left\{{}\begin{matrix}x-2\ge0\\y+1\ge0\end{matrix}\right.\) pt (1) trở thành:
\(\left(\sqrt{x-2}-\sqrt{y+1}\right)^2=0\Leftrightarrow\sqrt{x-2}=\sqrt{y+1}\)
\(\Leftrightarrow x-2=y+1\Rightarrow x=y+3\)
Thay vào pt dưới ta được:
\(-2\left(y+3\right)+y^2+y=6\Leftrightarrow y^2-y-12=0\Rightarrow\left[{}\begin{matrix}y=4\Rightarrow x=7\\y=-3< -1\left(l\right)\end{matrix}\right.\)
Vậy hệ pt có cặp nghiệm duy nhất \(\left(x;y\right)=\left(7;4\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2019
1/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x\le2\\y\le-1\end{matrix}\right.\)
Cộng vế với vế ta được:
\(x-2+y+1-2\sqrt{\left(x-2\right)\left(y+1\right)}=0\) (1)
- Nếu \(\left\{{}\begin{matrix}x\ge2\\y\ge-1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(\sqrt{x-2}-\sqrt{y+1}\right)^2=0\Rightarrow\sqrt{x-2}=\sqrt{y+1}\Leftrightarrow x=y+3\)
Thay vào pt dưới:
\(-2\left(y+3\right)+y^2+y=6\Leftrightarrow y^2-y-12=0\Rightarrow\left\{{}\begin{matrix}y=4\\x=7\end{matrix}\right.\)
- Nếu \(\left\{{}\begin{matrix}x\le2\\y\le-1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2-x+\left(-y-1\right)+2\sqrt{\left(2-x\right)\left(-y-1\right)}=0\)
\(\Leftrightarrow\left(\sqrt{2-x}+\sqrt{-y-1}\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}2-x=0\\-y-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
Thay vào pt dưới ta thấy ko thỏa mãn \(\Rightarrow\) loại
Vậy hệ có cặp nghiệm duy nhất \(\left(x;y\right)=\left(7;4\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2019
2/ \(x^4+2x^2y+y^2=4x^2y+y-4\Leftrightarrow\left(x^2+y\right)^2=4x^2y+y-4\)
Thay pt trên vào dưới:
\(16x^2=4x^2y+y-4\Leftrightarrow4x^2\left(y-4\right)+y-4=0\)
\(\Leftrightarrow\left(y-4\right)\left(4x^2+1\right)=0\Leftrightarrow y-4=0\)
\(\Rightarrow y=4\Rightarrow x^2+4=4x\Rightarrow\left(x-2\right)^2=0\Rightarrow x=2\)
Vậy hệ có cặp nghiệm duy nhất: \(\left(x;y\right)=\left(2;4\right)\)
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
1 tháng 1 2020
a, #Góp ý từ nhiều người nhưng họ không giải nên t làm giùm
ĐK: \(x\le3\)
\(\left\{{}\begin{matrix}x^2+y^2+1=2\left(xy-x+y\right)\left(1\right)\\x^3+3y^2+5x-12=\left(12-y\right)\sqrt{3-x}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x^2+y^2+1-2xy+2x-2y=0\)
\(\Leftrightarrow\left(x-y+1\right)^2=0\) \(\Leftrightarrow x-y+1=0\Leftrightarrow y=x+1\) Thay vào (2)
\(\left(2\right)\)\(\Leftrightarrow x^3+3\left(x+1\right)^2+5x-12=\left[12-\left(x+1\right)\right]\sqrt{3-x}\)
\(\Leftrightarrow x^3+3x^2+11x-9=\left(11-x\right)\sqrt{3-x}\)
\(\Leftrightarrow x^3+3x^2+8x=\left(11-x\right)\sqrt{3-x}+3\left(3-x\right)\)
\(\Leftrightarrow x^3+3x^2+8x=\left(3-x\right)\sqrt{3-x}+8\sqrt{3-x}+3\left(3-x\right)\)
\(\Leftrightarrow x^3+3x^2+8x=\sqrt{\left(3-x\right)^3}+3\sqrt{\left(3-x\right)^2}+8\sqrt{3-x}\)
\(\Leftrightarrow x=\sqrt{3-x}\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x-3=0\end{matrix}\right.\) \(\Rightarrow x=\frac{-1+\sqrt{13}}{2}\left(tm\right)\Rightarrow y=\frac{1+\sqrt{13}}{2}\)
Vậy...
1 tháng 1 2020
Akai Haruma, No choice teen, Arakawa Whiter, Phạm Hoàng Lê Nguyên, Vũ Minh Tuấn, tth, HISINOMA KINIMADO, Nguyễn Việt Lâm
Mn giúp e vs ạ! thanks!
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
\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)
23 tháng 12 2021
Câu 1:
\(ĐK:x\ge2\)
Áp dụng BĐT cauchy ta có:
\(\left(x+1\right)+4\ge2\sqrt{4\left(x+1\right)}=4\sqrt{x+1}\\ \Leftrightarrow2\sqrt{x+1}\le\dfrac{x+5}{2}\)
Ta có \(\left(x-2\right)+1\ge2\sqrt{x-2}\Leftrightarrow\sqrt{x-2}\le\dfrac{x-1}{2}\)
\(\Leftrightarrow P\le\dfrac{x+5}{2}+\dfrac{x-1}{2}-x+2013=x+2-x+2013=2015\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow x=3\)
23 tháng 12 2021
Câu 2:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}10\sqrt{x}+15y^3=140\\4y^3-10\sqrt{x}=12\end{matrix}\right.\left(x\ge0\right)\\ \Leftrightarrow19y^3=152\\ \Leftrightarrow y^3=8\Leftrightarrow y=2\\ \Leftrightarrow2\sqrt{x}+24=28\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
Vậy \(\left(x;y\right)=\left(4;2\right)\)
Câu 3:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\my+2m+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=\dfrac{3-2m}{m+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\x=\dfrac{3-2m}{m+1}\end{matrix}\right.\\ \Leftrightarrow xy=\dfrac{5\left(3-2m\right)}{\left(m+1\right)^2}\)
Đặt \(xy=t\)
\(\Leftrightarrow m^2t+2mt+t=15-10m\\ \Leftrightarrow m^2t+2m\left(t+5\right)+t-15=0\)
PT có nghiệm nên \(\Delta'=\left(t+5\right)^2-t\left(t-15\right)\ge0\)
\(\Leftrightarrow10t+25+15t\ge0\Leftrightarrow t\ge-1\)
Vậy \(xy_{min}=-1\Leftrightarrow\dfrac{5\left(2m-3\right)}{\left(m+1\right)^2}=1\Leftrightarrow m^2-8m+16=0\Leftrightarrow m=4\)
VT
23 tháng 8 2018
Ta có hpt \(\left\{{}\begin{matrix}xy+3y-5x-15=xy\\2xy+30x-y^2-15y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}5x=3y-15\\6\left(3y-15\right)-y^2-15y=0\end{matrix}\right.\)
Ta có pt (2) \(\Leftrightarrow3y-y^2-80=0\Leftrightarrow y^2-3y+80=0\left(VN\right)\)
=> hpy vô nghiệm
VT
23 tháng 8 2018
c) Ta có hpt \(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left(xy+x+y\right)=30\\xy\left(x+y\right)+xy+x+y=11\end{matrix}\right.\)
Đặt j\(xy\left(x+y\right)=a;xy+x+y=b\), ta có hpt
\(\left\{{}\begin{matrix}ab=30\\a+b=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=5;b=6\\a=6;b=5\end{matrix}\right.\)
với a=5;b=6, ta có \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}xy=1;x+y=5\\xy=5;x+y=1\end{matrix}\right.\)
đến đây thì thế y hoặc x ra pt bậc 2, còn TH còn lại bn tự giải nhé !