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Cái này chưa học bt làm mấy câu
b. x^2 + 2x - 3
= x^2 + 3x - x - 3
= x ( x - 1 ) + 3 ( x - 1 )
= ( x + 3 ) ( x - 1 )
\(4x^2-3x-4\)
\(=\left(2x\right)^2-2.2x.\frac{3}{4}+\frac{9}{16}-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\frac{73}{16}\)
\(=\left(2x-\frac{3}{4}\right)^2-\left(\frac{\sqrt{73}}{4}\right)^2\)
\(=\left(2x-\frac{3}{4}-\frac{\sqrt{73}}{4}\right)\left(2x-\frac{3}{4}+\frac{\sqrt{73}}{4}\right)\)
\(=\left(2x-\frac{3+\sqrt{73}}{4}\right)\left(2x+\frac{-3+\sqrt{73}}{4}\right)\)
\(x^2+2x-3\)
\(=x^2-x+3x-3\)
\(=x\left(x-1\right)+3\left(x-1\right)\)
\(=\)\(\left(x+3\right)\left(x-1\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\) \(\left(1\right)\)
đặt \(x^2+5x+5=t\)
\(\left(1\right)\)\(=\) \(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
hay \(\left(1\right)=\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
học tốt
6, (x^2 +1) -4x^2 = x^2 + 1 - 4x^2 = 1 - (4x^2 - x^2) = 1 - 3x^2 = (1-\(\sqrt{3}\)x)(1+\(\sqrt{3}\)x)
7, x^2 - 4x -5 = x^2 - 2.x.2 + 4 - 9 = (x^2 - 2.x.2 +4) - 3^2 = (x-2)^2 - 3^2 = (x-2-3)(x-2+3) = (x-5)(x+1)
8, x^5 - 3x^4 + 3x^3 - x^2 = x^2(x^3 -3x^2 + 3x -1) = x^2(x-1)^3
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Ta có: ( 4x + 1)(12x - 1)(3x + 2)(x+1) - 4
= [(4x+1)(3x+2)]. [(12x-1)(x+1)] - 4 = (12x2 +11x + 2)(12x2 + 11x - 1) - 4
Đặt a = 12x2 + 11x - 1. Thay vào biểu thức ta có:
(a+3).a - 4 = a2 + 3a - 4 =a2 + 4a - a - 4 = a(a+4) - (a+4)
= (a+4)(a-1)
=> (4x+1)(12x-1)(3x+2)(x+1) - 4 = (12x2 + 11x + 3)(12x2+11x - 2)
d) mk chỉnh lại đề
\(8xy^2-5xyz-24y+15z\)
\(=xy\left(8y-5z\right)-3\left(8y-5z\right)\)
\(=\left(8y-5z\right)\left(xy-3\right)\)
e) \(x^4-x^3-x+1=\left(x-1\right)^2\left(x^2+x+1\right)\)
f) \(x^4+x^2y^2+y^4=\left(x^2-xy+y^2\right)\left(x^2+xy-y^2\right)\)
g) \(x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)\)
h) \(x^3-3x^2+2=\left(x-1\right)\left(x^2-2x-2\right)\)
i) \(2x^3+x^2-4x-12=\left(x-2\right)\left(2x^2+5x+6\right)\)
k) \(25x^2\left(x-5\right)-x+y=\left(1-5x\right)\left(1+5x\right)\left(y-x\right)\)
Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
1.\(5\left(x^2-9y^2-6y-1\right)=5.\left[x^2-\left(9y^2+6y+1\right)\right]=5\left[x^2-\left(3y+1\right)^2\right]=5\left(x+3y+1\right)\left(x-3y-1\right)\)
2.kiểm tra lại đề nha bạn
3.\(4x^2+10x-2x-5=2x\left(2x-1\right)+5\left(2x-1\right)=\left(2x-1\right)\left(2x+5\right)\)