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\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)(1)
Đặt \(x^2+5x=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-3=t^2+2t-3\)
\(=t^2+3t-t-3=t\left(t+3\right)-\left(t+3\right)\)
\(=\left(t-1\right)\left(t+3\right)\)(2)
Mà \(x^2+5x=t\)nên \(\left(2\right)=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)
hay \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)\(=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)
a ) \(2x-1-x^2\)
\(=\left(x-1\right)-\left(x^2-x\right)\)
\(=\left(x-1\right)\left(1-x\right)\)
\(=-\left(x-1\right)^2\)
b) \(8x^3+y^6\)
\(=\left(2x+y^2\right)\left(4x^2-2xy^2+y^4\right)\)
c) \(x^2-16+4xy+4y^2\)
\(=\left(x^2+4xy+4y^2\right)-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
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)
x2 + 5x + 4
= x2 + x + 4x + 4
(Tách 5x = x + 4x)
= x(x + 1) + 4(x + 1)
(có x + 1 là nhân tử chung)
= (x + 1)(x + 4)
b) x3 – x2 – 5x + 125
= (x3 + 125) - (x2 + 5x)
= (x + 5)(x2 - 5x + 25) - x(x + 5)
= (x + 5)(x2 - 5x + 25 - x)
= (x + 5)(x2 - 6x + 25)
mày không phân tích được ak,tự làm đi
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\cdot\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\cdot\left(x-3\right)\)
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right).\left(x-3\right)\)