cho biết xy/x^2+y^2=312018/2017.Tính C=x^2-2xy+y^2x^2+2xy+y^2
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a) rút gọn 2 đơn thức sau:
A=x^2+y^2-2xy+2x+2xy+3
B=2x^2+y^2-xy+2x+xy+1
b) tính A+B và A-B
b) \(A+B=x^2+y^2+2x+3+2x^2+y^2+2x+1=3x^2+2y^2+4x+4\)
\(A-B=x^2+y^2+2x+3-2x^2-y^2-2x-1=-x^2+2\)
a) Ta có: \(A=x^2+y^2-2xy+2x+2xy+3\)
\(=x^2+y^2+2x-\left(2xy-2xy\right)+3\)
\(=x^2+y^2+2x+3\)
Ta có: \(B=2x^2+y^2-xy+2x+xy+1\)
\(=2x^2+y^2+2x+\left(xy-xy\right)+1\)
\(=2x^2+y^2+2x+1\)
Tính giá trị nhỏ nhất của biểu thức
P=X^2 + Y^2 + XY + X + Y
Q=X^2 + XY + Y^2 - 3X - 3Y + 2017
F=X^2 + 2Y^2 + 3Z^2 - 2XY + 2XZ - 2X - 2Y - 8Z + 1998
M=(X+1)^2 + (X-3)^2 + (Y-2)^2 + 4
Thực hiện phép tính sau
A.(2x^2y-3xy+4xy^2)÷(2xy)
B.1/xy-x^2-1/y^2-xy
C.[x/xy-y^2- 2x-y/x^2-xy]:(1/x-1/y)
a: \(=x-\dfrac{3}{2}+2y\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
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
Ai giúp mik nhé:
x-2y=xy
x(y+2)-y+1=0
x^2-2xy+x-2y=2017
(2+2x)(y+5)=0
\(\left(2+2x\right)\left(y+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2+2x=0\\y+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\y=-5\end{cases}}\)
Cho x+y-2=0
Tính giá trị C=x3+x2y-2x2-x2y-xy2+2xy+2y+2x-2
\(C=x^3+x^2y-2x^2-x^2y-xy^2+2xy+2y+2x-2\)
\(C=x^2\left(x+y-2\right)-xy\left(x+y-2\right)+2\left(x+y-2\right)+2\)
Thay x+y-2 =0 vào C ta được:
\(C=x^2\cdot0-xy\cdot0+2\cdot0+2=2\)
\(C=x^3+x^2y-2x^2-x^2y-xy^2+2xy+2y+2x-2\)
\(=\left(x^3+x^2y-2x^2\right)-\left(x^2y+xy^2-2xy\right)+\left(2y+2x-4\right)+2\)
\(=x^2\left(x+y-2\right)-xy\left(x+y-2\right)+2\left(x+y-2\right)+2\)
Thay \(x+y-2=0\)vào biểu thức ta được: \(C=2\)
Tìm x, y thuộc Z để:
a) xy + x - y = 2
b) x - 2xy + y = 0
c) x. (x - 2) - (2 - x)y - 2. (x - 2) = 3
d) (2x - y). (4x2 + 2xy + y2) + (2x + y). (4x2 - 2xy + y2) - 16x. (x2 - y) = 32
e) x2 - 2xy + 2y2 - 2x + 6y +5 = 0
g) x2 + 2xy + 7x + 7y + 2y2 = 0
Bài 1 : Tính giá trị biểu thức sau , biết x+y-2=0
a ) M = x^3+x^2y+2x^2-xy-y^2+3y+x-1
b ) N= x^3-2x^2-xy^2+2xy+2y+2x-2
c ) P = x^4+2x^3y-2x^3+x^2y^2-2x^2y-x*(x+y )+2x+3
Biến đổi mỗi đa thức theo hướng làm xuất hiện thừa số x+y-2 \(M=x^3+x^2y-2x^2-xy-y^2+3y+x-1\)
\(M=x^3+x^2y-2x^2-xy-y^2+\left(2y+y\right)+x-\left(-2+1\right)\)
\(M=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-2\right)+1\)
\(M=\left(x^2.x+x^2.y-2x^2\right)-\left(x.y+y.y-2y\right)+\left(x+y-2\right)+1\)
\(M=x^2.\left(x+y-2\right)-y.\left(x+y-2\right)+\left(x+y-2\right)+1\)
\(M=x^2.0+y.0+0+1\)
\(M=1\)
\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-2\)
\(N=x^3+x^2y-2x^2-xy^2+x^2y+2xy+2y+2x-\left(-4+2\right)\)
\(N=\left(x^3+x^2y-2x^2\right)-\left(x^2y+xy^2-2xy\right)+\left(2x+2y-4\right)+2\)
\(N=\left(x^2x+x^2y-2x^2\right)-\left(xyx+xyy-2xy\right)+\left(2x+2y-4\right)+2\)
\(N=x^2\left(x+y-2\right)-xy\left(x+y-2\right)+2\left(x+y-2\right)+2\)
\(N=x^2.0-xy.0+2.0+2\)
\(N=2\)
\(P=x^4+2x^3y-2x^3+x^2y^2-2x^2y-x\left(x+y\right)+2x+3\)
\(P=\left(x^4+x^3y-2x^3\right)+\left(x^3y+x^2y^2-2x^2y\right)-\left(x^2+xy-2x\right)+3\)\(P=\left(x^3x+x^3y-2x^3\right)+\left(x^2y.x+x^2yy-2x^2y\right)-\left(xx+xy-2x\right)+3\)
\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3\)
\(P=x^3.0+x^2y.0-x.0+3\)
\(P=3\)
Tích mình nha!
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
\(e)xy-2y^2+8y-3x=13\)
\(\Leftrightarrow xy-2y^2+2y+6y-3x-6=7\)
\(\Leftrightarrow y\left(x-2y+2\right)+3\left(-x+2y-2\right)=7\)
\(\Leftrightarrow y\left(x-2y+2\right)-3\left(x-2y+2\right)=7\)
\(\Leftrightarrow\left(x-2y+2\right)\left(y-3\right)=7\)
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