Tìm x,y biết: \(\left(x-12+y\right)^2+\left(y+4-x\right)^2=0\).
Vậy x =...
y =...
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Những câu hỏi liên quan
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 7 2021
b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
24 tháng 7 2017
\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\forall x\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
đề sai câu b các câu sau áp dụng tương tự
24 tháng 7 2017
c/ Vì: \(\left(x-12+y\right)^{200}+\left(x-4-x\right)^{200}=0\)
mà \(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\forall x,y\\\left(x-4-y\right)^{200}\ge0\forall x,y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-4-y\right)^{200}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-12+y=0\\x-4-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=12\\x-y=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\)
ND
1
3 tháng 2 2018
Ta thấy \(\left(x+y-z\right)^2\ge0\); \(\left(x-y+2\right)^2\ge0\);\(\left(x+4\right)^2\ge0\)với mọi x,y,z
Suy ra \(\left(x+y-z\right)^2+\left(x-y+2\right)^2+\left(x+4\right)^2\ge0\)với mọi x,y,z
Mặt khác \(\left(x+y-z\right)^2+\left(x-y+2\right)^2+\left(x+4\right)^2=0\)
Nên \(\hept{\begin{cases}x+y-z=0\\x-y+2=0\\x+4=0\end{cases}\Rightarrow\hept{\begin{cases}x+y=z\\x+2=y\\x=-4\end{cases}\Rightarrow}\hept{\begin{cases}x+y=z\\y=-2\\x=-4\end{cases}\Rightarrow}\hept{\begin{cases}z=-6\\y=-2\\x=-4\end{cases}}}\)
Vậy.....
4 tháng 4 2022
vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)
mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
LP
2
5 tháng 1 2017
(x-12+y)2 + (y+4-x)2=0
x-12+y=0 => x+y=12
y+4-x=0 =>y-x=-4
=>x=8 , y=4
5 tháng 1 2017
vì (x-12+y)2 >0:(y+4-x)2 >0
=>x-12+y=y+4-x=0
=>x-12+x=y+4-y
=>2x-12=4
=>x=8 hoặc -8
thay x vào là ra y nha.
28 tháng 3 2018
=> \(\hept{\begin{cases}x^2+2xy+y^2-4x+4y=12\\x^2-2xy+y^2-2x-2y=3\end{cases}}\)
Rồi đến đây tự làm nhé
BT
28 tháng 3 2018
HPT <=> \(\hept{\begin{cases}\left(x+y\right)^2-4\left(x+y\right)+4=16\\\left(x-y\right)^2-2\left(x-y\right)+1=4\end{cases}}\)<=> \(\hept{\begin{cases}\left(x+y-2\right)^2=4^2\\\left(x-y-1\right)^2=2^2\end{cases}}\)
=> \(\hept{\begin{cases}x+y-2=\pm4\\x-y-1=\pm2\end{cases}}\)
Có các TH:
1/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{9}{2}\\y=\frac{3}{2}\end{cases}}\)
2/ \(\hept{\begin{cases}x+y-2=4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{7}{2}\end{cases}}\)
3/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=3\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{5}{2}\end{cases}}\)
4/ \(\hept{\begin{cases}x+y-2=-4\\x-y-1=-2\end{cases}}\)=> \(\hept{\begin{cases}x+y=-2\\x-y=-1\end{cases}}\)=> \(\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{1}{2}\end{cases}}\)
5 tháng 10 2021
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)
\(\left(x-12+y\right)^2+\left(y+4-x\right)^2=0\)
\(\Rightarrow\left(x-12+y\right)^2=0\) và \(\left(y+4-x\right)^2=0\)
+) \(\left(x-12+y\right)^2=0\Rightarrow x-12+y=0\)
\(\Rightarrow x+y=12\)
+) \(\left(y+4-x\right)^2=0\Rightarrow y+4-x=0\Rightarrow y-x=-4\)
\(\Rightarrow x=\left(12+4\right):2=8\)
\(\Rightarrow y=\left(12-4\right):2=4\)
Vậy \(x=8;y=4\)