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Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
1) Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2\)
\(=x^2+4x+4+x^2-6x+9\)
\(=2x^2-2x+13\)
2) Ta có: \(\left(4-x\right)^2-\left(x-3\right)^2\)
\(=\left(4-x-x+3\right)\left(4-x+x-3\right)\)
\(=-2x+7\)
3) Ta có: \(\left(x-5\right)\left(x+5\right)-\left(x+5\right)^2\)
\(=x^2-25-x^2-10x-25\)
=-10x-50
4) Ta có: \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)\)
\(=x^2-6x+9-x^2+16\)
=-6x+25
5) Ta có: \(\left(y^2-6y+9\right)-\left(y-3\right)^2\)
\(=y^2-6y+9-y^2+6y-9\)
=0
6) Ta có: \(\left(2x+3\right)^2-\left(2x-3\right)\left(2x+3\right)\)
\(=4x^2+12x+9-4x^2+9\)
=12x+18
\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)
Bài 1:
\(P=2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2+4xy-7y^2\)
Đáp án D.
- Cách 1:
- Cách 2: sử dụng hằng đẳng thức
Ta có:
x 3 + 8 = x 3 + 2 3 x + 2 x 2 - 2 x + 4 ⇒ x 3 + 8 : x + 2 = x 2 - 2 x + 4
⇒ Chọn D
a , x(3-x)-y(y-2x)+y(y-2)-x(2y-x)
=3x-x2-y2+2xy+y2-2y-2xy+x2
=3x-(-x2+x2)+(2xy-2xy)-2y
=3x-2y
b,(x2-7)(x+2)-(2x-1)(x-14)+x(x2-2x-22)+35
=x3+2x2-7x-14-2x2+29x-14+x3-2x2-22x+35
=(x3+x3)+(2x2-2x2-2x2)-(-7x+29x-22x)-14-14+35
=2x3-2x2+7
c,(2x+y)(4x2-2xy+y2)-8x3-y3
=2xy2-4x2y+8x3+y3-2xy2+4x2y-8x3-y3
=(2xy2-2xy2)-(-4x2y+4x2y)+(8x3-8x3)+(y3-y3)
=0
\(B=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)-x^4+y^4\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4-x^4+y^4\)
\(=0\)
\(C=\left(2-x\right)\left(1+2x\right)+\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)
\(=2+3x-2x^2+2x^2-29x+14+x^3-2x^2-22x+35\)
\(=x^3-2x^2-48x+51\)
Đề bài sai
\(1,\\ a,=x^2+2xy+y^2\\ b,=x^2-4xy+4y^2\\ c,=x^2y^4-1\\ d,=\left[\left(x-y\right)\left(x+y\right)\right]^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\\ 2,\\ a,=\left(x+2\right)^2\\ b,=\left(3x-2\right)^2\\ c,=\left(\dfrac{x}{2}+1\right)^2\\ d,=\left(x+y-2\right)^2\)
Ta có:
Chọn đáp án A.