Giải hệ pt sau:
\(x^3-5x=y^3-5y\)
\(x^2+y^4=1\)
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BD
11 tháng 1 2021
Giải hệ pt sau \(\left\{{}\begin{matrix}x^2-xy+y^2=3\\z^2+yz+1=0\end{matrix}\right.\)
13 tháng 4 2017
\(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{2x+y+1}+2\sqrt[3]{7x+12y+8}=2xy+y+5\end{matrix}\right.\)
Xét \(pt\left(1\right)\) dễ dàng suy ra \(x+y\ge0\)
\(VT=\sqrt{\left(x-y\right)^2+\left(2x+y\right)^2}+\sqrt{\left(x-y\right)^2+\left(2y+x\right)^2}\)
\(\ge\left|2x+y\right|+\left|2y+x\right|\ge3\left(x+y\right)\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}x=y\\x,y\ge0\end{matrix}\right.\)
Thay vào \(pt\left(2\right)\) ta được:
\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)
\(\Leftrightarrow\left[\sqrt{3x+1}-\left(x+1\right)\right]+2\left[\sqrt[3]{19x+8}-\left(x+2\right)\right]=2x^2-2x\)
\(\Leftrightarrow\left(x-x^2\right)\left[\dfrac{1}{\sqrt{3x+1}+x+1}+2\cdot\dfrac{x+7}{\sqrt[3]{\left(19x+8\right)^2}+\left(x+2\right)\sqrt[3]{19x+8}+\left(x+2\right)^2}+2\right]=0\)
Do \(x;y\ge0\) nên pt trong ngoặc luôn dương
\(\Rightarrow x-x^2=0\Rightarrow x\left(1-x\right)=0\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Mà \(x=y\)\(\Rightarrow\left[{}\begin{matrix}x=y=0\\x=y=1\end{matrix}\right.\) là nghiệm của hpt
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 3 2021
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2xy=10\\xy\left(x+y\right)+5\left(x+y\right)=32\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u^2-2v=10\\uv+5u=32\end{matrix}\right.\)
\(\Rightarrow u\left(\dfrac{u^2-10}{2}\right)+5u=32\)
\(\Leftrightarrow u^3=64\Rightarrow u=4\Rightarrow v=3\)
\(\Rightarrow\left(x;y\right)=\left(1;3\right);\left(3;1\right)\)
BD
0
PH
1
H9
HT.Phong (9A5)
CTVHS
3 tháng 7 2023
a) \(\left\{{}\begin{matrix}2x+3y=5\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\4x-5y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3y=5\\11y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\cdot\dfrac{9}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+\dfrac{27}{11}=5\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=\dfrac{28}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{14}{11}\\y=\dfrac{9}{11}\end{matrix}\right.\)
Vậy: \(x=\dfrac{14}{11};y=\dfrac{9}{11}\)
28 tháng 9 2021
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
11 tháng 10 2021
a: \(\left\{{}\begin{matrix}x+4y=-11\\5x-4y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x=-10\\x+4y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=\dfrac{-11-x}{4}=\dfrac{-11+\dfrac{5}{3}}{4}=-\dfrac{7}{3}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}2x-y=7\\3x+5y=-22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-3y=21\\6x+15y=-66\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-18y=78\\2x-y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-13}{3}\\x=\dfrac{y+7}{2}=\dfrac{4}{3}\end{matrix}\right.\)
DT
7
3 tháng 2 2016
<=> xy+5x+3y+15=xy+8x+y+8 <=> 3x-2y=7 <=> 9x-6y=21 <=> x=3 <=> x=3
10xy+14x-15y-21=10xy+10x-12y-12 4x-3y=9 8x-6y=18 8.3-6y=18 y=1
x^3 - 5x = y^3 - 5y (1)
x^2 + y^4 = 1 (2)
Từ (1) => \(x^3-y^3-5x+5y=0\Rightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-5\left(x-y\right)=0\)
=> \(\left(x-y\right)\left(x^2+xy+y^2-5\right)=0\)
=> \(x=y\) hoặc \(x^2+xy+y^2=5\)
(+) thay x = y vào (2) ta có : \(x^4+x^2=1\)
đặt x^2 = t (ĐK t>= 0 )pt <=> t^2 + t = 1
giải ra t
(+) TH2 :\(x^2+xy+y^2=5\) (3)
ta có : \(x^2+y^4=1\Rightarrow0\le x;y\le1\)
=> \(0 (4)
Từ (3) và (4) => pt vo nghiệm