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\(a,2\left(x-1\right)\left(x+1\right)+\left(x-1\right)^2+\left(x+1\right)^2\)
\(=2\left(x^2-1\right)+x^2-2x+1+x^2+2x+1\)
\(=2x^2-2+2x^2+2=4x^2\)
\(b,\left(x-y+1\right)^2+\left(1-y\right)^2+2\left(x-y+1\right)\left(y-1\right)\)
\(=\left(x-y+1\right)^2+2\left(x-y+1\right)\left(y-1\right)+\left(y-1\right)^2\)
\(=\left[\left(x-y+1\right)+\left(y-1\right)\right]^2\)
\(=\left[x-y+1+y-1\right]^2=x^2\)
đề cuối phải sửa cái cuối thành \(\left(3x+5\right)^2\)
\(c,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2=\left[3x+1-3x-5\right]^2=16\)
\(a,P=x^2-16-x^2+8x-16=8x-32\\ b,=3x^2-6xy+3y^2-2x^2-4xy-2y^2-x^2+y^2\\ =2y^2-10xy=2\cdot9-10\left(-3\right)\cdot2=78\)
\(a,\left(2x-1\right)^2-\left(x-3\right)\left(x+3\right)-1969\\ =4x^2-4x+1-x^2+9-1969\\ =3x^2-4x-1959\)
\(b,\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\\ =4x^2-9y^2-4x^2+4xy-y^2\\ =8y^2+4xy=4y\left(2y+x\right)\)
\(c,\left(x+3y\right)^2+\left(x+y\right)\left(x-y\right)+280\\ =x^2+6xy+9y^2+x^2-y^2+280\\ =2x^2+8y^2+6xy+280\)
a: \(\left(2x-1\right)^2-\left(x-3\right)\cdot\left(x+3\right)-1969\)
\(=4x^2-4x+1-x^2+9-1969\)
\(=3x^2-4x-1959\)
b: \(\left(2x-3y\right)\left(2x+3y\right)-\left(2x-y\right)^2\)
\(=4x^2-9y^2-4x^2+4xy-y^2\)
\(=-10y^2+4xy\)
\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)
a. (x + y)2 - (x - y)2
= (x + y - x + y)(x + y + x - y)
= 2y . 2x
= 4xy
b. (x + y)2 + (x - y)2 - 2(x + y)(x - y)
= (x2 + 2xy + y2) + (x2 - 2xy + y2) - 2(x2 - y2)
= x2 + 2xy + y2 + x2 - 2xy + y2 - 2x2 + 2y2
= x2 + x2 - 2x2 + 2xy - 2xy + y2 + y2 + 2y2
= 4y2
c. (x2 - 1)(x2 - x + 1)
= x4 - x3 + x2 - x2 + x - 1
= x4 - x3 + x - 1
\(a,A=\left(2x+y\right)^2-\left(2x-y\right)^2\\ =\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\\ =2y\cdot4x\\ =8xy\\ b,B=\left(x-2y\right)^2-4y\left(x-2y\right)+4y^2\\ =x^2-4xy+4y^2-4xy+8y^2+4y^2\\ =x^2+16y^2-8xy\\ =\left(x-4y\right)^2\)
a) \(\left(3-xy^2\right)^2-\left(2+xy^2\right)^2\)
\(=\left(3-xy^2+2+xy^2\right)\left(3-xy^2-2-xy^2\right)\)
\(=5.\left(-2xy^2\right)\)
\(=-10xy^2\)
b) \(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
c) \(\left(x-3\right)^3+\left(2-x\right)^3\)
\(=x^3-3x^2.3+3x.3^2-3^3+2^3-3.2^2.x+3.2.x^2-x^3\)
\(=x^3-9x^2+27x-27+8-12x+6x^2-x^3\)
\(=\left(x^3-x^3\right)+\left(-9x^2+6x^2\right)+\left(27x-12x\right)+\left(-27+8\right)\)
\(=-3x^2+15x-19\)
B1
a, \(=>A=\left(x+y+x-y\right)\left(x+y-x+y\right)=2x.2y=4xy\)
b, \(=>B=\left[\left(x+y\right)-\left(x-y\right)\right]^2=\left[x+y-x+y\right]^2=\left[2y\right]^2=4y^2\)
c,\(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\)\(\left(x+1\right)\left(x^2-x+1\right)\left(x-1\right)\left(x^2+x+1\right)=\left(x^3+1^3\right)\left(x^3-1^3\right)=x^6-1\)
d, \(\left(a+b-c\right)^2+\left(a-b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a-b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c+b-c\right)\left(a+b-c-b+c\right)\)
\(+\left(a-b+c+b-c\right)\left(a-b+c-b+c\right)\)
\(=a\left(a+2b-2c\right)+a\left(a-2b\right)\)
\(=a\left(a+2b-2c+a-2b\right)=a\left(2a-2c\right)=2a^2-2ac\)
B2:
\(\)\(x+y=3=>\left(x+y\right)^2=9=>x^2+2xy+y^2=9\)
\(=>xy=\dfrac{9-\left(x^2+y^2\right)}{2}=\dfrac{9-\left(17\right)}{2}=-4\)
\(=>x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=3\left(17+4\right)=63\)
Bài 1:
a) Ta có: \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=x^2+2xy+y^2-x^2+2xy+y^2\)
=4xy
b) Ta có: \(\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y-x+y\right)^2\)
\(=\left(2y\right)^2=4y^2\)
c) Ta có: \(\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^6-1\)
d) Ta có: \(\left(a+b-c\right)^2+\left(a+b+c\right)^2-2\left(b-c\right)^2\)
\(=\left(a+b-c\right)^2-\left(b-c\right)^2+\left(a+b+c\right)^2-\left(b-c\right)^2\)
\(=\left(a+b-c-b+c\right)\left(a+b-c+b-c\right)+\left(a+b+c-b+c\right)\left(a+b+c+b-c\right)\)
\(=a\cdot\left(a+2b-2c\right)+\left(a+2c\right)\left(a-2b\right)\)
\(=a^2+2ab-2ac+a^2-2ab+2ac-4bc\)
\(=2a^2-4bc\)
a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)
\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)
\(=2x^2-4xy+\dfrac{15}{4}y^2\)
b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)
\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(=2x^2+2x+13-2x^2+2\)
=2x+15
a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)
b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(=2x+15\)
Lời giải:
$x(x+y)-y(x+y)+x^2+y^2=(x-y)(x+y)+x^2+y^2$
$=x^2-y^2+x^2+y^2=2x^2$
Bài 1:
- a,(2+xy)^2=4+4xy+x^2y^2
- b,(5-3x)^2=25-30x+9x^2
- d,(5x-1)^3=125x^3 - 75x^2 + 15x^2 - 1
a, \(x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2\left(x^2+y^2\right)\)
b, \(\left(x+y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)\(=\left(2x\right)^2=4x^2\)
xin tiick
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+y^2+2xy+x^2+y^2-2xy\)
\(=2x^2+2y^2\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2\)
\(=2\left(x^2-y^2\right)+x^2+y^2-2xy+x^2+y^2+2xy\)
\(=2\left(x^2-y^2\right)+2\left(x^2+y^2\right)\)
\(=2\left(2x^2\right)=4x^2\)