
a)...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 1. a/ \(A=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\) \(=2\left(x+y\right)\left(x^2-xy+y^2\right)-3\left[\left(x+y\right)^2-2xy\right]\) \(=2\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]-3\left[\left(x+y\right)^2-2xy\right]\) \(=2.1.\left[1^2-3xy\right]-3\left[1^2-2xy\right]\) \(=2-6xy-3+6xy\) \(=-1\) Vậy... 2. a. \(127^2+146.127+73^2\) \(=127^2+2.73.127+73^2\) \(=\left(127+73\right)^2=200^2=40000\) b. \(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\) \(=18^8-18^8+1\) \(=1\) a) \(x^3-5x^2+8x-4\) \(=x^3-2x^2-3x^2+6x+2x-4\) \(=x^2\left(x-2\right)-3x\left(x-2\right)+2\left(x-2\right)\) \(=\left(x-2\right)\left(x^2-3x+2\right)\) \(=\left(x-2\right)\left(x^2-x-2x+2\right)\) \(=\left(x-2\right)\left[x\left(x-1\right)-2\left(x-1\right)\right]\) \(=\left(x-2\right)\left(x-1\right)\left(x-2\right)\) b) \(A=10x^2-15x+8x-12+7\) \(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\) \(A=\left(2x-3\right)\left(5x+4\right)+7\) Dễ thấy \(\left(2x-3\right)\left(5x+4\right)⋮\left(2x-3\right)=B\) Vậy để \(A⋮B\)thì \(7⋮\left(2x-3\right)\) \(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\) \(\Rightarrow x\in\left\{2;1;5;-2\right\}\) Vậy....... Bài 1: a) |x - 1,38| + |2y - 4,2| ≥ 0 ∀ x; y. Dấu '=' xảy ra khi x = 1,38; y = 2,1 Vậy .... b) |x - y| + |y + \(\frac{9}{25}\)| ≥ 0 ∀ x,y. Dấu "=" xảy ra khi x=y=\(-\frac{9}{25}\) Vậy ... Bài 2: a) 2|3x - 2| - 1 ≥ -1 ∀x. Vậy Amin= -1 khi x=\(\frac{2}{3}\) b) x2 + 3|y - 2| - 1 ≥ -1 ∀x,y. Vậy Bmin = -1 khi x = 0; y = 2 c) C = |x + 32| + |54 - x| ≥ |x + 32 + 54 - x| = 86. Cmin = 86 khi (x + 32)(54 - x) ≥ 0 ⇔ -32 ≤ x ≤ 54 d) |5x - 2| + |3y + 12| ≥ 0 ∀x,y. ⇒ 4 - |5x - 2| - |3y + 12| ≤ 4 Vậy Dmax = 4 khi x = \(\frac{2}{5}\); y = -4 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 : \(\left(3x-2y\right)^2=9x^2-12xy+4y^2\) \(\left(2x-\frac{1}{2}\right)^2=4x^2-4x+\frac{1}{4}\) \(\left(\frac{x}{2}-y\right)\left(\frac{x}{2}+y\right)=\frac{x^2}{4}-y^2\) \(\left(x+\frac{1}{3}\right)^3=x^3+x^2+\frac{1}{3}x+\frac{1}{27}\) \(\left(x-2\right)\left(x^2+2x+2^2\right)=x^3-8\)
