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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)
Bài 2:
a) \(x^2+y^2-9-2xy\)
\(=\left(x^2-2xy+y^2\right)-3^2\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
b) \(4x^2-5x-9\)
\(=4x^2+4x-9x-9\)
\(=4x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x+1\right)\left(4x-9\right)\)
\(\left(2x-3\right)^2-\left(4x-1\right)\left(x+2\right)=4x^2-12x+9-4x^2-7x+2=-19x+11\)
\(\left(3x+2\right)\left(3x-2\right)-\left(3x-1\right)^2=9x^2-4-9x^2+6x-1=6x-5\)
\(x^2+y^2-9-2xy=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)
\(4x^2-5x-9=\left(4x-9\right)\left(x+1\right)\)
\(\left(x-3\right)^2-\left(x-1\right)\left(x-2\right)=5\Leftrightarrow x^2-6x+9-x^2+3x-2=5\)
\(\Leftrightarrow-3x=-2\Leftrightarrow x=x=\frac{2}{3}\)
\(3x^2+5x-8=0\Leftrightarrow\left(x-1\right)\left(3x+8\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{8}{3}\end{cases}}\)
\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9 \)
\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)
\(A=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)
Đặt \(x^2+6x+4=t\)
Ta được: \(A=\left(t-4\right)\left(t+4\right)-9\)
\(A=t^2-25\)
\(A=\left(t+5\right)\left(t-5\right)\)
\(A=\left(x^2+6x+9\right)\left(x^2+6x-1\right)\)
\(A=\left(x+3\right)^2\left(x^2+6x-1\right)\)
\(B=\left(x^2-3x\right)^2+5x^2-15x+6\)
\(B=\left(x^2-3x\right)^2+5\left(x^2-3x\right)+6\)
\(B=\left(x^2-3x\right)\left(x^2-3x+5\right)+6\)
Đặt \(x^2-3x=a\)
Ta được: \(B=a\left(a+5\right)+6\)
\(B=a^2+5a+6\)
\(B=a^2+2a+3a+6\)
\(B=a\left(a+2\right)+3\left(a+2\right)\)
\(B=\left(a+2\right)\left(a+3\right)\)
\(B=\left(x^2-3x+2\right)\left(x^2-3x+3\right)\)
\(B=\left(x^2-x-2x+2\right)\left(x^2+3x+3\right)\)
\(B=\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x^2+3x+3\right)\)
\(B=\left(x-1\right)\left(x-2\right)\left(x^2+3x+3\right)\)
\(3x^2-5x+2=3x^2-3x-2x+2=3x\left(x-1\right)-2\left(x-1\right)=\left(3x-2\right)\left(x-1\right)\)
b.)x^4+5x^3+15x-9
=x^4-9+5x^3+15x
=(x^2-3)(x^2+3)+5x(x^2+3)
=(x^2+3)(x^2-3+5x)