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\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
a) A = 3 ( x − y ) 2 − 2 ( x + y ) 2 − ( x − y ) ( x + y ) 2 A = [ ( x − y ) − ( x + y ) ] 2 + 5 ( x − y ) 2 − 5 ( x + y ) 2 2 A = 4 y 2 + 5 [ ( x − y ) − ( x + y ) ] [ ( x − y ) + ( x + y ) ] 2 A = 4 y 2 + 5 [ − 2 y ] [ 2 x ] = 4 y 2 − 20 x y = 4 y ( y − 5 x ) A = 2 y ( y − 5 x )
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
\(a. 2x(3x^2-5x+3) = 6x^3-10x^2+6x \)
\(b. -2x(x^2+5x-3) = -2x^3-10x^2+6x\)
c. \(-\dfrac{1}{2}x^2\left(2x^3-4x+3\right)
=-x^5+2x^3-\dfrac{3}{2}x^2\)
\(d.\left(2x-1\right)\left(x^2+5-4\right)=\left(2x-1\right)\left(x^2+1\right)=2x^3+2x-x^2-1\)
e. \(-\left(5x-4\right)\left(2x+3\right)=10x^2+15x-8x-12=-10x^2+7x-12\)
f.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=\left(2x-y\right)\left(2x-y\right)^2=\left(2x-y\right)^3\)
g.\(\left(3x-4\right)\left(x+4\right)+\left(5-x\right)\left(2x^2+3x-1\right)=3x^2+12x-4x-16+10x^2+15x-5-2x^3-3x^2+x=-2x^3+10x^2+24x-21\)
e. \(7x\left(x-4\right)-\left(7x+3\right)\left(2x^2-x+4\right)=7x^2-28x-14x^3+7x^2-28x-6x^2+3x+-12=-14x^3+8x^2-53x-12\)
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
GIÚP MÌNH VỚI ĐỀ BÀI LÀ RÚT GỌN THÔI NHA THUỘC KIỂU HẰNG ĐẲNG THỨC 6 VÀ 7 GIÚP MÌNH VỚI MÌNH CẦN GẤP TRONG TỐI NAY GIÚP VỚI