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NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 4 2021
\(\Leftrightarrow2x^2y+y=4x^2+5\)
\(\Leftrightarrow y\left(2x^2+1\right)=4x^2+5\)
\(\Leftrightarrow y=\dfrac{4x^2+5}{2x^2+1}=2+\dfrac{3}{2x^2+1}\)
y nguyên \(\Rightarrow\dfrac{3}{2x^2+1}\) nguyên \(\Rightarrow2x^2+1=Ư\left(3\right)\)
Mà \(2x^2+1\ge1\Rightarrow\left[{}\begin{matrix}2x^2+1=1\\2x^2+1=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=5\\x=1\Rightarrow y=3\\x=-1\Rightarrow y=3\end{matrix}\right.\)
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
\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)
NH
30 tháng 7 2018
\(\left(4x-5\right)\left(2x-3\right)\left(x-1\right)=9\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-5=9\\2x-3=9\\x-1=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=6\\x=10\end{matrix}\right.\)
Vậy \(x=\left\{3,5;6;10\right\}\)
6 tháng 8 2022
d: Sửa đề: \(\left(4x-5\right)^2\cdot\left(2x-3\right)\left(x-1\right)=9\)
a: \(\Leftrightarrow\left(2x^2+x\right)^2-3\left(2x^2+x\right)-\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)\left(2x^2+x-3\right)-\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow\left(2x^2+x-3\right)\left(2x^2+x-1\right)=0\)
\(\Leftrightarrow\left(2x^2+3x-2x-3\right)\left(2x^2+2x-x-1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};1;-1;\dfrac{1}{2}\right\}\)
N
13 tháng 3 2017
bài 1:
b) đề như vầy hả :\(\left\{{}\begin{matrix}\left(x^2-1\right)y+\left(y^2-1\right)x=2\left(xy-1\right)\left(1\right)\\4x^2+y^2+2x-y-6=0\left(2\right)\end{matrix}\right.\)
\(Pt\left(1\right)\Leftrightarrow x^2y+xy^2-x-y-2xy+2=0\)
\(\Leftrightarrow xy\left(x+y\right)-\left(x+y\right)-2\left(xy-1\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(xy-1\right)-2\left(xy-1\right)=0\)
\(\Leftrightarrow\left(xy-1\right)\left(x+y-2\right)=0\Leftrightarrow\left[{}\begin{matrix}xy=1\\x+y=2\end{matrix}\right.\)
*xét \(xy=1\Leftrightarrow x=\dfrac{1}{y}\)thế vào Pt (2):\(\dfrac{4}{y^2}+y^2+\dfrac{2}{y}-y-6=0\)
\(\Leftrightarrow\dfrac{4+2y}{y^2}+\left(y+2\right)\left(y-3\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(\dfrac{2}{y^2}+y-3\right)=0\)
\(\Leftrightarrow\left(y+2\right)\left(y^3-3y^2+2\right)=0\)\(\Leftrightarrow\left(y+2\right)\left(y-1\right)\left(y^2-2y-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=-2\\y=1\\y=1-\sqrt{3}\\y=1+\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\\x=-\dfrac{1+\sqrt{3}}{2}\\x=\dfrac{-1+\sqrt{3}}{2}\end{matrix}\right.\)
* xét x+y=2(tương tự thay x=2-y vào Pt (2))
câu 2:
ta đưa về PT ẩn x:\(x^2-x\left(y+1\right)+y^2-y-2=0\)
Pt phải có nghiệm ,xét \(\Delta=\left(y+1\right)^2-4\left(y^2-y-2\right)\ge0\)
\(\Leftrightarrow y^2-2y-3\le0\Leftrightarrow\left(y+1\right)\left(y-3\right)\le0\)
\(\Leftrightarrow-1\le y\le3\).
vì x,y thuộc Z ,lần luợt thay các giá trị của y vừa tìm được vào PT ban đầu ta được các cặp (x,y) t/m là (0;-1);(-1;0);(2;0);(0;2);(3;2);(2;3)
bài 3:
DKXĐ:\(\left\{{}\begin{matrix}2x^2-x\ge0\\2x-x^2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le0\end{matrix}\right.\\0\le x\le2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{1}{2}\le x\le2\end{matrix}\right.\)
bình phương , self study
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 5 2021
Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại
Với pt sau:
Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm
Với \(x;y;z\ne0\)
Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3
Do đó:
\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)
Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)
LL
1 tháng 5 2021
a.\(\left\{{}\begin{matrix}4x+2y=14\\2x-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=18\\2x-2y=4\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\4-2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\-2y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;0\right\}\)
LL
1 tháng 5 2021
b.\(\left\{{}\begin{matrix}2x-4y=0\\3x+2y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-4y=0\\6x+4y=16\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}8x=16\\2x-4y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\4-4y=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=2\\-4y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
vậy hệ pt có ndn \(\left\{2;1\right\}\)
TK
16 tháng 5 2021
a) \(\sqrt{4x^2-4x+9}=3\)
Vì \(4x^2-4x+9=\left(2x-1\right)^2+8>0\)( Với mọi x )
Nên \(\sqrt{4x^2-4x+9}=3\)
⇔\(4x^2-4x+9=9\)
⇔\(4x^2-4x=0\)
⇔\(4x\left(x-1\right)=0\)
⇔\(\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)là nghiệm
hpt \(\Leftrightarrow\left\{{}\begin{matrix}y=5-2x\\y+x^2=4x\end{matrix}\right.\)
\(\Rightarrow5-2x+x^2=4x\)
\(\Leftrightarrow x^2-6x+5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Nếu \(x=1\Rightarrow y=5-2x=3\)
Nếu \(x=5\Rightarrow y=5-2x=-5\)
Vậy hpt đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(1;3\right);\left(5;-5\right)\right\}\)