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\(\left(x+2y\right)^2-\left(x-2y\right)^2\\ =\left[\left(x+2y\right)-\left(x-2y\right)\right]\left[\left(x+2y\right)+\left(x-2y\right)\right]\\ =\left(x+2y-x+2y\right)\left(x+2y+x-2y\right)\\ =4y.\left(2x\right)\\ =8xy\)
\(\left(3x+y\right)^2+\left(x-y\right)^2\\ =\left[\left(3x\right)^2+2.3x.y+y^2\right]+\left(x^2-2xy+y^2\right)\\ =6x^2+6xy+y^2+x^2-2xy-y^2\\ =7x^2+4xy\)
\(-\left(x+5\right)^2-\left(x-3\right)^2\\ =-\left(x^2+10x+25\right)-\left(x^2-6x+9\right)\\ =-x^2-10x-25-x^2+6x-9\\ =-2x^2-4x-34\)
\(\left(3x-2\right)^2-\left(3x-1\right)^2\\ =\left[\left(3x-2\right)-\left(3x-1\right)\right]\left[\left(3x-2\right)+\left(3x-1\right)\right]\\ =\left(3x-2-3x+1\right)\left(3x-2+3x-1\right)\\ =-1.\left(6x-3\right)\\ =-6x+3\)
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)
\(=4x\cdot2y=8xy\)
b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Câu A) là \(\left(2x+y\right)^2-\left(y-2x\right)^2\)
Chứ ko phải là\(\left(2x+y\right)^2-\left(2x-y\right)^2\)
Nhưng dù sao thì cũng cảm ơn
\(=\dfrac{\left(x-1\right)^3}{xy\left(x-1\right)+\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{xy+1}\)
\(\dfrac{x^3-3x^2+3x-1}{1-x+x^2y-xy}=\dfrac{\left(x-1\right)^3}{\left(xy-1\right)\left(x-a\right)}=\dfrac{\left(x-1\right)^2}{xy-1}\)
a) \(=x^2-5-x^2+49=44\)
b) Nhân tử cuối cùng bạn ghi gì vậy?
a: =x^3+8-1+27x^3=28x^3+7
b: Sửa đề: (2+y)(y^2-2y+4)+(5-y)(25+5y+y^2)
=8+y^3+125-y^3
=133
\(B=\left(3x+2y\right)\left(3x-2y\right)-\left(3x-1\right)^2+\left(2y+1\right)^2\)
\(B=9x^2-4y^2-9x^2+6x-1+4y^2+4y+1\)
B = 6x + 4y