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\(5x^2-19x-4\)
\(=\left(5x^2-20x\right)+\left(x-4\right)\)
\(=5x\left(x-4\right)+\left(x-4\right)\)
\(=\left(x-4\right)\left(5x+1\right)\)
\(5x^2-19x-4=5x^2+x-20x-4\)
\(=x\cdot\left(5x+1\right)-4\cdot\left(5x+1\right)\)
\(=\left(5x+1\right)\cdot\left(x-4\right)\)
a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)
\(=\left(x+y\right)\left(5y-5x+6\right)\)
b, \(12x^2+19x+7=12x^2+12x+7x+7\)
\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)
5x3 + 38x2 + 19x - 14
= ( 5x3 + 35x2 ) + ( 3x2 + 21x ) - ( 2x + 14 )
= 5x2 ( x + 7 ) + 3x ( x + 7 ) - 2 ( x + 7 )
= ( x + 7 ) ( 5x2 + 3x - 2 )
= ( x + 7 ) [ ( 5x2 - 2x ) + ( 5x - 2 ) ]
= ( x + 7 ) [ x ( 5x - 2 ) + ( 5x - 2 ) ]
= ( x + 7 ) ( x + 1 ) ( 5x - 2 )
\(5x^3+38x^2+19x-4\)
\(=\left(5x^3+35x^2\right)+\left(3x^2+21x\right)-\left(2x+14\right)\)
\(=5x^2\left(x+7\right)+3x\left(x+7\right)-2\left(x+7\right)\)
\(=\left(5x^2+3x-2\right)\left(x+7\right)\)
\(=\left(5x^2-2x+5x-2\right)\left(x+7\right)\)
\(=\left[x\left(5x-2\right)+\left(5x-2\right)\right]\left(x+7\right)\)
\(=\left(x+1\right)\left(5x-2\right)\left(x+7\right)\)
a)( x3 -19x-30=(x-5)(x+2)(x+3)
b) 2x3 -5x2+8x-3=(2x-1)(x2-2x+3)
\(6x^2-19x+15=6x^2-9x-10x+15\)
\(=3x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(3x-5\right)\left(2x-3\right)\)
\(3x^3-19x^2+44x-32=3x^3-4x^2-15x^2+20x+24x-32\)
\(=x^2\left(3x-4\right)-5x\left(3x-4\right)+8\left(3x-4\right)\)
\(=\left(3x-4\right)\left(x^2-5x+8\right)\)
Ta có: \(x^2-19x-30=\frac{4x^2-76x-120}{4}\)
\(=\frac{1}{4}.\left[\left(4x^2-76x+361\right)-481\right]\)
\(=\frac{1}{4}.\left[\left(2x-19\right)^2-481\right]\)
\(=\frac{1}{4}.\left(2x-19-\sqrt{481}\right).\left(2x-19+\sqrt{481}\right)\)
Nghiệm xấu nên phân tích khó :) Sửa thành x3 - 19x - 30 cho dễ
x3 - 19x - 30
= x3 + 3x2 - 3x2 - 9x - 10x - 30
= ( x3 + 3x2 ) - ( 3x2 + 9x ) - ( 10x + 30 )
= x2( x + 3 ) - 3x( x + 3 ) - 10( x + 3 )
= ( x + 3 )( x2 - 3x - 10 )
= ( x + 3 )( x2 + 2x - 5x - 10 )
= ( x + 3 )[ x( x + 2 ) - 5( x + 2 ) ]
= ( x + 3 )( x + 2 )( x - 5 )
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(5x^2-19x-4=5x^2-20x+x-4\)
\(=\left(5x^2-20x\right)+\left(x-4\right)\)
\(=5x\left(x-4\right)+\left(x-4\right)\)
\(=\left(5x-1\right)\left(x-4\right)\)
= 5x^2 + x - 20x - 4
= (5x^2 + x) - (20x + 4)
= x(5x+1) - 4 (5x + 1)
= (5x+1) (x - 4)