(x+2y ). (y +2x)
(x+y/2x-2y-x-y/2x+2y-2y^2/y^2-x^2):2y/x-y
Ta có: \(\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y^2-x^2}\right):\dfrac{2y}{x-y}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}:\dfrac{2y}{x-y}\)
\(=\dfrac{4y^2+4xy}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x+y\right)\cdot2y}\)
\(=1\)
tính
a) A+x^2+4xy+x^2-y^2=2y+3xy-5x^2y+2x^2y^2
b)A-(-2x^3)-y^2+32x^2-4xy-y=10z^2+y2x^2
c)A=-2x+5xy-3x^2y+2x^2y^2-2y^2x
B=xy-3x^2y+2x^2y^2-2y^2x
Mình viết lại cho dễ đọc.
a) A+ x2+4xy + x2- y2 = 2y +3xy- 5x2y +5x2y + 2x2y2
b) A- ( -2 x3) -y2+ 32x2- 4xy - y = 10z2 + y2z2
c) A= -2x + 5xy - 3x2y + 2x2y2 - 2 y2x
B= xy- 3x2y+ 2x2y + 2x2y2 - 2- y2x
Mà hình như nhìn đề thấy sao sao phải không ạ với lại chỗ 10z2 + y2x2 tớ nhìn ko hỉu nên viết thành 10z2 +y2x2 có gì thì mấy cậu sửa hộ ạ!
M)(x^2-2xy+y^2)(x-y) N)-(x-y)(x^2+xy-1) Ờ)-(x^2-2y)(x+y^2-1) P)(1/2x-1)(2x-3) Q)(x-1/2y)(x-1/2y) R)(x^2-2x+3)(1/2x-5)
m: (x-y)(x^2-2xy+y^2)
=(x-y)*(x-y)^2
=(x-y)^3
=x^3-3x^2y+3xy^2-y^3
n: =-(x^3+x^2y-x-x^2y-xy^2+y)
=-x^3+x+xy^2-y
o: =-(x^3+x^2y^2-x^2-2xy-2y^3+2y)
=-x^3-x^2y^2+x^2+2xy+2y^3-2y
p: (1/2x-1)(2x-3)
=1/2x*2x-1/2x*3-2x+3
=x^2-3/2x-2x+3
=x^2-7/2x+3
q: (x-1/2y)(x-1/2y)
=(x-1/2y)^2
=x^2-xy+1/4y^2
r: (x^2-2x+3)(1/2x-5)
=1/2x^3-5x^2-x^2+10x+3/2x-15
=1/2x^3-6x^2+11,5x-15
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!
Tính giá trị x + y - 2 = 0
M = x ^3 + x ^2y - 2x ^2 - xy - y ^2 + 3y + x - 1
N = x ^4 + 2x ^3y - 2x ^3 + x ^2y^2 - 2x ^2y - x ( x + y ) + 2x + 3
phân tích thành nhân tử
x^3+2x^2+x-xy
x^3-y^3+2x^2-2y^2
x^3+y^3+x^2y+y^2x+2^2x+2^2y
a, \(x^3+2x^2+x-xy=x\left(x^2+2x+1-y\right)\)
\(=x\left[\left(x+1\right)^2-y\right]\)
b, \(x^3-y^3+2x^2-2y^2=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+2\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
1) \(\left\{{}\begin{matrix}xy+x+y=x^2-2y^2\\x\sqrt{2y}-y\sqrt{x-1}=2x-2y\end{matrix}\right.\)
2) \(\left\{{}\begin{matrix}2x^2+y^2-3xy+3x-2y+1=0\\4x^2-y^2+x+4=\sqrt{2x+y}+\sqrt{x+4y}\end{matrix}\right.\)
Tính giá trị biểu thức
A=\(2x+2y+3xy\left(x+y\right)+5\left(x^3y_{ }^2+x^2y^3\right)\)
tại x+y=0
B=\(3xy\left(x+y\right)+2x^3y+2x^2y^2\)
tại x+y=0
a: \(A=2\left(x+y\right)+3xy\left(x+y\right)+5x^2y^2\left(x+y\right)=0\)
b: \(B=3xy\left(x+y\right)+2x^2y\left(x+y\right)=0\)
\(\Rightarrow\)A=2(x+y)+3xy(x+y)+5x2y2(x+y)
Thay x+y=0 vào A
\(\Rightarrow\)A=0
a, x^2 +2xy^2+y^3/ 2x^2 +xy -y^2=xy+x^2/2x-y
b, x^2 + 3xy +2y^2 /x^3 +2x^2y-xy^2 -2y^3= 1/2x-7