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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. b: \(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x-y\right)^3+3xy\left(x-y\right)\right]\) \(=3\left(1-2xy\right)-2\left(1+3xy\right)\) \(=3-6xy-2-6xy=-12xy+1\) c: \(=\left(x+y\right)^3-3\left(x^2+y^2+2xy\right)+3\left(x+y\right)+2012\) \(=101^2-3\cdot101^2+3\cdot101+2012\) =1002013 \(a.\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{-2\left(y-x\right)}{3\left(y-x\right)}=\dfrac{-2}{3}\) \(b.\dfrac{x-2}{-x}=\dfrac{2-x}{x}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{8-x^3}{x\left(x^2+2x+4\right)}\) \(\dfrac{3x}{x+y}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{-3x\left(x-y\right)}{\left(x+y\right)\left(y-x\right)}=\dfrac{-3x\left(x-y\right)}{y^2-x^2}\) c: \(\dfrac{-3x\left(x-y\right)}{y^2-x^2}=\dfrac{3x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}=\dfrac{3x}{x+y}\) a: \(\dfrac{2\left(x-y\right)}{3y-3x}=\dfrac{2\left(x-y\right)}{-3\left(x-y\right)}=\dfrac{-2}{3}\) b: \(\dfrac{8-x^3}{x\left(x^2+2x+4\right)}=\dfrac{\left(2-x\right)\left(x^2+2x+4\right)}{x\left(x^2+2x+4\right)}=\dfrac{2-x}{x}\) 1.a (3x-2y)2= (3x)2 - 2. 3x . 2y - (2y)2 = 9x2 - 12xy - 4y2 2.b (2x - 1/2)2 = (2x)2 - 2.2x.1/2 - (1/2)2= 4x2 - 2 - 1/4 3.c (x/2 - y) (x/2+y)= (x/2)2 - (y)2 = x/4 - y2 Bài 1 : \(\left(3x-2y\right)^2=9x^2-12xy+4y^2\) \(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\) \(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\) \(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\) \(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\) a)2(x-y)/(-3)(x-y)=-2/3 b)8-x^3=(2-x)(x^2+2x+4) => Vế phải =(2-x)/x=(x-2)/-x c)y^2-x^2=(y+x)(y-x) bạn đổi dấu rồi rút gọn là được,cũng tương tự như trên ý Bài 1. Rút gọn: \(a,
x\left(1-x\right)+6\left(x+3\right)\left(x+3\right)\) \(=x-x^2+6\left(x^2+6x+9\right)\) \(=x-x^2+6x^2+36x+54\) \(=5x^2+37x+54\) \(b,
\left(2-3x\right)\left(2+3x\right)-\left(x+5\right)\left(x-5\right)\) \(=\left(4-9x^2\right)-\left(x^2-25\right)\) \(=-10x^2+29\) \(c,
\left(3x+1\right)\left(x+5\right)-\left(x-1\right)\left(x+1\right)\) \(=3x^2+15x+x+5-x^2+1\) \(=2x^2+16x+6\) \(d,\left(2-3x\right)\left(2x+3\right)+6\left(x-1\right)^2\) \(=\left(4x+6-6x^2-9x\right)+6\left(x^2-2x+1\right)\) \(=4x+6-6x^2-9x+6x^2-12x+6\) \(=-17x+12\) \(e,
x\left(5-x\right)-\left(2x+2\right)\left(3x+2\right)-\left(x-2\right)\left(x+2\right)\) \(=5x-x^2-\left(6x^2+4x+6x+4\right)-\left(x^2-4\right)\) \(=5x-x^2-6x^2-4x-6x-4-x^2+4\) \(=-8x^2-5x\) Bài 2: a: VT\(=x^3-xy+x^2y^2-y^3-x^3+y^3-x^2y^2\) =-xy b: \(VT=x^2+6xy+9y^2-x^2+9y^2-6xy=18y^2=VP\) 1. a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\) b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\) 2. a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\) b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ 3. \(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -) 4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\) \(A\ge\frac{7}{4}\) Vậy GTNN của A là 7/4
