Tìm x,y € Z
a) x2+y2=19
b) 78-7y2=2x2
c) x2 +y2= x-y+7
d) x2-2x+y2=4y-2
Ai nhanh,đúng,lời giải đầy đủ mình sẽ tích 3 tích nha
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M
15 tháng 8 2018
\(\frac{3}{8}.x+1=\frac{7}{8}\) \(2012:x+23=526\)
\(\Leftrightarrow\frac{3}{8}.x=\frac{7}{8}-1\) \(\Leftrightarrow2012:x=503\)
\(\Leftrightarrow\frac{3}{8}.x=\frac{-1}{8}\) \(\Leftrightarrow x=4\)
\(\Leftrightarrow x=\frac{-1}{3}\)
15 tháng 8 2018
Câu a bạn viết lại đề đi nha khó nhìn quá
Câu b
2012 : x + 23= 526
2012 : x = 526-23
2012: x=503
x=2012 : 503
x =4
8 tháng 9 2021
\(a,\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)+\left(2z^2+4z+2\right)=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,\Leftrightarrow\left(4x^2+8xy+4y^2\right)+\left(x^2-2x+1\right)+\left(y^2+2y+1\right)=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
MH
8 tháng 9 2021
a,9x^2+y^2+2z^2−18x+4z−6y+20=0
⇔9(x−1)^2+(y−3)^2+2(z+1)^2=0
⇔x=1;y=3;z=−1
b,5x^2+5y^2+8xy+2y−2x+2=0
⇔4(x+y)2+(x−1)2+(y+1)2=0
⇔x=−y;x=1y=−1⇔x=1y=−1
c,5x^2+2y^2+4xy−2x+4y+5=0
⇔(2x+y)^2+(x−1)^2+(y+2)^2=0
⇔2x=−y;x=1;y=−2
⇔x=1;y=−2
d,x^2+4y^2+z^2=2x+12y−4z−14
⇔(x−1)^2+(2y−3)^2+(z+2)^2=0
⇔x=1;y=3/2;z=−2
e: Ta có: x^2−6x+y2+4y+2=0
⇔x^2−6x+9+y^2+4y+4−11=0
⇔(x−3)^2+(y+2)^2=11
Dấu '=' xảy ra khi x=3 và y=-2
8 tháng 12 2023
1: \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
2: \(x^2-y^2+x-y\)
\(=\left(x^2-y^2\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+1\right)\)
3: \(3x-3y+x^2-y^2\)
\(=\left(3x-3y\right)+\left(x^2-y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
4: \(5x-5y+x^2-y^2\)
\(=\left(5x-5y\right)+\left(x^2-y^2\right)\)
\(=5\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(5+x+y\right)\)
5: \(x^2-5x-y^2-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
6: \(x^2-y^2+2x-2y\)
\(=\left(x^2-y^2\right)+\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+2\right)\)
7: \(x^2-4y^2+x+2y\)
\(=\left(x^2-4y^2\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y\right)+\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+1\right)\)
8: \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
9: \(x^2-4y^2+2x+4y\)
\(=\left(x^2-4y^2\right)+\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y+2\right)\)
5 tháng 9 2021
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
5 tháng 9 2021
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
