Tìm nghiệm nguyên của các phương trình
a/ 2x^2-xy-6y^2+13y-3x+7=0
b/ 3x^2+10xy+8y^2=21
c/ 2x^2+y^2+2z^2-2xy+2xz=12
d/ x^2+2y^2+3z^2+4t^2+2xy+2xz+2xt+4yz-2zt=10
e/ 3x^2y+5xy-8y-x^2-10x=4
Tìm x,y,z biết: a) x^2+y^2-4x+4y+8=0 b) 5x^2-4xy+y^2=0 c) x^2+2y^2+z^2-2xy-2y-4z+5=0 d) 3x^2+3y^2+3xy-3x+3y+3=0 e) 2x^2+y^2+2z^2-2xy-2xz+2yz-2z-2z-2x+2=0
a) x2+y2-4x+4y+8=0
⇔ (x-2)2+(y+2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)
b)5x2-4xy+y2=0
⇔ x2+(2x-y)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2x-y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
c)x2+2y2+z2-2xy-2y-4z+5=0
⇔ (x-y)2+(y-1)2+(z-2)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\\z-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\z=2\end{matrix}\right.\)
b: Ta có: \(5x^2-4xy+y^2=0\)
\(\Leftrightarrow x^2-\dfrac{4}{5}xy+y^2=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{2}{5}y+\dfrac{4}{25}y^2+\dfrac{21}{25}y^2=0\)
\(\Leftrightarrow\left(x-\dfrac{2}{5}y\right)^2+\dfrac{21}{25}y^2=0\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
d)3x2+3y2+3xy-3x+3y+3=0
⇔ 6x2+6y2+6xy-6x+6y+6=0
⇔ 3(x+y)2+3(x-1)2+3(y+1)2=0
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x-1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
Tìm x y biết
a)xy+3x-2y=11
b)2x^2-2xy+x-y=12
c)2xy-10y-x=13
e)xy-2y^2+8y-3x=13
f)xy-2y^2+8y-3x=13
\(a)xy+3x-2y=11\)
\(\Leftrightarrow xy+3x-2y-6=5\)
\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)
\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)
\(b)2x^2-2xy+x-y=12\)
\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)
\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)
\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)
\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)
Vì 2x+1 luôn lẻ
\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
\(c)2xy-10y-x=13\)
\(\Leftrightarrow x\left(2y-1\right)-2y.5+5=18\)
\(\Leftrightarrow x\left(2y-1\right)-5\left(2y-1\right)=18\)
\(\Leftrightarrow\left(2y-1\right)\left(x-5\right)=18\)
\(\Leftrightarrow2y-1;x-5\inƯ\left(18\right)\)
\(\RightarrowƯ\left(18\right)\in\left\{-1;1;-2;2;-3;3;-6;6;-9;9;-18;18\right\}\)
Vì 2y-1 luôn lẻ
=>2y-1 thuộc {-1;1;-3;3;-9;9}
=> Làm tương tự nhé
\(e)xy-2y^2+8y-3x=13\)
\(\Leftrightarrow xy-2y^2+2y+6y-3x-6=7\)
\(\Leftrightarrow y\left(x-2y+2\right)+3\left(-x+2y-2\right)=7\)
\(\Leftrightarrow y\left(x-2y+2\right)-3\left(x-2y+2\right)=7\)
\(\Leftrightarrow\left(x-2y+2\right)\left(y-3\right)=7\)
Tự khai triển như các câu trên.
Mình đg bận nên ko lm đc hết câu.
Phân tích đa thức thành nhân tử
1) 32X^4+1
2)X^8+3x^4+1
3)x^2+2xy-8y^2+2xz+14yz-3z^2
4)3x^2-22xy-4x+8y+7y^2+1
5)12x^2+5x-12y^2+12y-10xy-3
6)2x^2-7xy+3y^2+5xz-5yz+2z^2
7)x^2+3xy+2y^2+3xz+5yz=2z^2
8)x^2-8xy+15y^2+2x-4y-3
9)x^4-13x^2+36
10)4(x^2+15x+50)(x^2+18x+72)-3x^2
Phân tích đa thức thành nhân tử:
1, (x+y)^7-x^7-y^7
2, x^4+4x^2+5
3, x^2+2xy-8y^2+2xz+14yz-3z^2
4, 3x^2-22xy-4x+8y+7y^2+1
5, 12x^2+5x-12y^2+12y-10xy-3
6, 2x^2-7xy+3y^2+5xz-5yz+2z^2
7, x^2+3xy+2y^2+3xz+5yz+2z^2
8, x^2-8xy+15y^2+2x-4y-3
Phân tích đa thức thành nhân tử
a) \(x^2+2xy-8y^2+2xz+14yz+3z^2\)
b) \(3x^2-22xy-4x+8y+7y^2+1\)
c)\(12x^2+5x-12y^2+12y-10xy-3\)
d) \(2x^2-7xy+3y^2+5xz-5yz+2z^2\)
e) \(x^2+3xy+2y^2+3xz+5yz+2z^2\)
Tìm số nguyên x biết
a,3x+3y-2xy=7
b,xy+2x+y+11=0
c,xy+x-y=4
d,2x.(3y-2)+(3y-2)=12
e,3x+4y-xy=15
f,xy+3x-2y=11
g,xy+12=x+y
h,xy-2x-y=-6
i,xy+4x=25+5y
ii,2xy-6y+x=9
iii,xy-x+2y=3
k,2.x^2.y-x^2-2y-2=0
l,x^2.y-x+xy=6
Cho 2x^2 + 2y^2 + 2z^2 + 2xy + 2yz +2xz +10x +6y +34 =0 Tìm x, y,z
Ta có:
\(\left(x^2+2xy+y^2\right)+\left(y^2+2yz+z^2\right)+\left(z^2+2zx+x^2\right)+\left(x^2+10x+25\right)+\left(y^2+6y+9\right)+z^2=0\)\(\Leftrightarrow\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2+\left(x+5\right)^2+\left(y+3\right)^2+z^2=0\)
Không tồn tại x,y,z thỏa mãn đề bài
Tìm bậc của các đa thức sau:
a) \(x^3y^3+6x^2y^2+12xy-8
\)
b) \(x^2y+2xy^2-3x^3y+4xy^5\)
c) \(x^6y^2+3x^6y^3-7x^5y^7+5x^4y\)
d) \(2x^3+x^4y^5+3xy^7-x^4y^5+10-xy^7\)
e) \(0,5x^2y^3+3x^2y^3z^3-a.x^2y^3-x^4-x^2y^3\) với a là hằng số
a, bậc 6
b, bậc 6
c, bậc 12
d, bậc 9
e, bậc 8
a,x2-4x+4-y2+2y-1
b, x2+2xy-8y2+2xz+14yz-3z2
c,3x2-22xy-4x+8y+7y2+1