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HQ
6 tháng 4 2017
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!
N
1
14 tháng 12 2020
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, \(x^3-y^3+2x^2-2y^2=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+2\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left[\left(x^2+xy+y^2\right)+2\left(x+y\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
6 tháng 9 2023
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
6 tháng 8 2019
A=\(\left(x-y\right)^2+\left(x+y\right)^2=x^2-2xy+y^2+x^2+2xy+y^2=2x^2+2y^2\)
B=\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=\left(2y\right).\left(2x\right)\)
C=\(\left(2a+b\right)^2-\left(2a-b\right)^2=\left(2a+b-2a+b\right)\left(2a+b+2a-b\right)=\left(2b\right).\left(4a\right)\)
D=\(\left(2x-1\right)^2-2\left(2x-3\right)^2+4=4x^2-4x+1-4x+6+4=4x^2-8x+11\)
E=\(\left(x+3y\right)^2-\left(x-3y\right)^2=\left(x+3y-x+3y\right)\left(x+3y+x-3y\right)=\left(6y\right).\left(2x\right)\)
F=\(\left(2x+y\right)^2-\left(2x-y\right)^2=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)=\left(2y\right).\left(4x\right)\)
G=\(\left(x-2y\right)^2+4\left(x-2y\right)y+4y^2=x^2-4xy+4y^2+4xy-8y^2+4y^2=x^2\)
H=\(\left(x-y\right)^2-4\left(x-y\right)\left(x+2y\right)+4\left(x+2y^{ }\right)^2=x^2-2xy+y^2-4\left(x^2+2xy-xy-2y^2\right)+4x+8y=x^2-2xy+y^2-4x^2-8xy+4xy+8y^2+4x+8y=3x^2+12xy-9y^2+4x+8y\)
SS
15 tháng 9 2019
Ta có:
a) A= (x-y)^2 + (x+y)^2
A= x^2 -2xy + y^2 + x^2 + 2xy + y^2
A= 2x^2+ 2y^2
b) B= (x+y)^2 -( x-y)^2
B= (x+y-x+y)(x+y+x-y)
B= 2y.2x= 4xy
c) C= (2a+b)^2 -( 2a-b)^2
C= (2a+b-2a+b)(2a+b+2a-b)
C= 2b.4a
C= 8ab
d) D= (2x-1)^2 -2(2x-3)^2+4
D= 4x^2 -4x+1 -2( 4x^2 -12x + 9) +4
D= 4x^2 -4x+1 -8x^2 + 24x -18 +4
D= -4x^2 + 20x-13
e) E= (x+3y)^2-(x-3y)^2
E= (x+3y-x+3y)(x+3y+x-3y)
E= 6y.2x= 12xy
f) F= (2x+y)^2-(2x-y)^2
F=(2x+y-2x+y)(2x+y+2x-y)
F= 2y.4x= 8xy
g) G= (x-2y)^2 + 4(x-2y)y + 4y^2
G= (x-2y)^2 + 2(x-2y)2y + (2y)^2
G= (x-2y+2y)^2
G= x^2
h) H= (x-y)^2 -4(x-y)(x+2y)+ 4(x+2y)^2
H= (x-y)^2 - 2(x-y)2(x+2y) + [2(x+2y)]^2
H= (x-y- 2x-4y)^2
H= (-x-5y)^2
Lưu ý (-A-B)^2 = ( A+ B)^2
=> H= (x+5y)^2
30 tháng 8 2021
Ta có: \(\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y^2-x^2}\right):\dfrac{2y}{x-y}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x-y\right)\left(x+y\right)}:\dfrac{2y}{x-y}\)
\(=\dfrac{4y^2+4xy}{2\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x-y}{2y}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x+y\right)\cdot2y}\)
\(=1\)
11 tháng 7 2016
\(M=x^2\left(x+y-2\right)-y\left(x+y-2\right)+y+x-2+1\)
\(=1\)
\(N=x^2\left(x-2\right)-xy^2+2xy+2\left(x+y-2\right)+2\)
Ta có : \(x+y-2=0\Rightarrow x+2=-y\)
\(\Rightarrow N=-x^2y-xy^2+2xy+2\)
\(N=-xy\left(x+y-2\right)+2=2\)
\(P=x^3\left(x+y-2\right)+x^2y\left(x+y-2\right)-x\left(x+y-2\right)+3=3\)
TN
2 tháng 4 2017
Đặt \(\hept{\begin{cases}\frac{1}{x^2}=a\\\frac{1}{y^2}=b\\\frac{1}{z^2}=c\end{cases}}\Rightarrow abc=1\) và ta cần chứng minh
\(\frac{1}{2a+b+3}+\frac{1}{2b+c+3}+\frac{1}{2c+a+3}\le\frac{1}{2}\left(1\right)\)
Áp dụng BĐT AM-GM ta có:
\(2a+b+3=\left(a+b\right)+\left(a+1\right)+2\ge2\left(\sqrt{ab}+\sqrt{a}+2\right)\)
\(\Rightarrow\frac{1}{2a+b+3}\le\frac{1}{2\left(\sqrt{ab}+\sqrt{a}+1\right)}=\frac{1}{2}\cdot\frac{1}{\sqrt{ab}+\sqrt{a}+1}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{2b+c+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{bc}+\sqrt{b}+1};\frac{1}{2c+a+3}\le\frac{1}{2}\cdot\frac{1}{\sqrt{ac}+\sqrt{c}+1}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT_{\left(1\right)}\le\frac{1}{2}\left(\frac{1}{\sqrt{ab}+\sqrt{a}+1}+\frac{1}{\sqrt{b}+\sqrt{bc}+1}+\frac{1}{\sqrt{c}+\sqrt{ac}+1}\right)\le\frac{1}{2}=VP_{\left(2\right)}\left(abc=1\right)\)
\(x^2+2x-y^2+2y\\=(x^2-y^2)+(2x+2y)\\=(x-y)(x+y)+2(x+y)\\=(x+y)(x-y+2)\)