Rút gọn:
a) A=(5-x)(5+x)-x(4-x)-25
b) B=(x2+1)(x+1)-(x+1)3
c) C=(x+y-2)2-2(x+y-2)(y+x)+(x+y)2
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9 tháng 11 2021
a) \(\dfrac{x}{x-y}+\dfrac{2y^2}{x^2-y^2}-\dfrac{x}{x+y}=\dfrac{x\left(x+y\right)+2y^2-x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2+xy+2y^2-x^2+xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y^2+2xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{2y}{x-y}\)
b) \(B=\dfrac{x}{x-2}-\dfrac{4x}{x^2-4}-\dfrac{2}{x+2}=\dfrac{x\left(x+2\right)-4x-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x-4x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
c) \(\dfrac{5}{x+1}-\dfrac{10}{-x^2+x-1}-\dfrac{15}{x^3+1}=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3+1}=\dfrac{5\left(x^2-x+1\right)+10\left(x+1\right)-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{5x}{x^2-x+1}\)
5 tháng 10 2021
Bài 2:
b: Ta có: \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-4x-x^4+1\)
\(=-x^4+x^3-4x+1\)
c: Ta có: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2ab\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)\)
\(=b\left(2a+b-2c\right)\)
\(=2ab+b^2-2bc\)
16 tháng 11 2021
a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)
TP
2
4 tháng 9 2021
a: Ta có: \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)\)
\(=6x^2-2x-6x^2-2x+18x+6\)
=14x+6
b: Ta có: \(\left(2x-3\right)^2-\left(2x+1\right)\left(2x-1\right)+3\left(2x-3\right)\)
\(=4x^2-12x+9-4x^2+1+6x-9\)
\(=-6x+1\)
c: Ta có: \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left(x+y-1-x-y\right)^2\)
=1
4 tháng 9 2021
a) \(2x\left(3x-1\right)-\left(x-3\right)\left(6x+2\right)=6x^2-2x-6x^2-2x+18x+6=14x+6\)
b) \(\left(2x-3\right)^2-\left(1+2x\right)\left(2x-1\right)+3\left(2x-3\right)=4x^2-12x+9-4x^2+1+6x-9=-6x+1\)
c) \(\left(x+y-1\right)^2-2\left(x+y-1\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)
25 tháng 12 2021
\(a.\left(3x-1\right)^2+\left(x+3\right)\left(2x-1\right)\)
\(=9x^2-6x+1-2x^2+x-6x+3\)
\(=7x^2-11x+4\)
C
1
13 tháng 7 2021
a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)
\(=64-x^3+\left(x-4\right)^3\)
\(=64-x^3+x^3-12x^2+48x-64\)
\(=-12x^2+48x\)
b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)
\(=27x^3+8-27x^3+8\)
=16
c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)
\(=x^3+1-x\left(x^2+2x+1\right)\)
\(=x^3+1-x^3-2x^2-x\)
\(=-2x^2-x+1\)
4 tháng 1 2022
a: \(=x^2+2xy+y^2+x^2-2xy+y^2=2x^2+2y^2\)
b: \(=\left(x+y+x-y\right)^2=\left(2x\right)^2=4x^2\)
d: \(=9x^2+6x+1-9x^2+6x-1=12x\)
a) \(A=\left(5-x\right)\left(5+x\right)-x\left(4-x\right)-25\\ =25-x^2-4x+x^2-25\\ =-4x\)
b) \(B=\left(x^2+1\right)\left(x+1\right)-\left(x+1\right)^3\\ =x^3+x+x^2+1-x^3-3x^2-3x-1\\ =-2x^2-2x\)
c) \(\left(x+y-2\right)^2-2\left(x+y-2\right)\left(y+x\right)+\left(x+y\right)^2\)
\(=x^2+y^2+4+2xy-4y-4x-2\left(xy+y^2-2y+x^2+xy-2x\right)+x^2+2xy+y^2\)
\(=x^2+y^2+4+2xy-4y-4x-2\left(2xy+y^2-2y+x^2-2x\right)+x^2+2xy+y^2\)
\(=x^2+y^2+4+2xy-4y-4x-4xy-2y^2+4y-2x^2+4x+x^2+2xy+y^2\)
\(=4\)
a) \(A=\left(5-x\right)\left(5+x\right)-x\left(4-x\right)-25=25-x^2-4x+x^2-25=-4x\)b) \(B=\left(x^2+1\right)\left(x+1\right)-\left(x+1\right)^3=\left(x+1\right)\left[x^2+1-\left(x+1\right)^2\right]=\left(x+1\right)\left(x^2+1-x^2-2x-1\right)=\left(x+1\right)\left(-2x\right)\)c) \(C=\left(x+y-2\right)^2-2\left(x+y-2\right)\left(y+x\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)