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\(3x^2-14x-5=3x\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(3x+1\right)\)
\(3x^2+x-4=3x^2-3x+4x-4=3x\left(x-1\right)+4\left(x-1\right)=\left(3x+4\right)\left(x-1\right)\)
\(\left(x^2-5x\right)^2-3x^2+15x-18\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+3\right)\)
\(=\left(x^2-5x+3\right)\left(x-6\right)\left(x+1\right)\)
\(=\left(x^2-5x\right)^2-3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)^2-6\left(x^2-5x\right)+3\left(x^2-5x\right)-18\\ =\left(x^2-5x\right)\left(x^2-5x-6\right)+3\left(x^2-5x-6\right)\\ =\left(x^2-5x+3\right)\left(x^2-5x-6\right)\\ =\left(x-6\right)\left(x+1\right)\left(x^2-5x+3\right)\)
\(4\left(x^2+15x+50\right)\left(x^2+18x+72\right)-3x^2\\ =4\left(x+5\right)\left(x+10\right)\left(x+6\right)\left(x+12\right)-3x^2\\ =4\left(x^2+16x+60\right)\left(x^2+17x+60\right)-3x^2\)
Đặt \(x^2+16x+60=a\)
\(=4a\left(a+x\right)-3x^2\\ =4a^2+4ax-3x^2\\ =\left(2a-x\right)\left(2a+3x\right)\\ =\left[2\left(x^2+16x+60\right)-x\right]\left[2\left(x^2+16x+60\right)+3x\right]\\ =\left(2x^2+31x+120\right)\left(2x^2+35x+120\right)\)
(x2+15x+50)(x2+18x+72)−3x2=4(x+5)(x+10)(x+6)(x+12)−3x2=4(x2+16x+60)(x2+17x+60)−3x24(�2+15�+50)(�2+18�+72)−3�2=4(�+5)(�+10)(�+6)(�+12)−3�2=4(�2+16�+60)(�2+17�+60)−3�2
Đặt x2+16x+60=a�2+16�+60=�
=4a(a+x)−3x2=4a2+4ax−3x2=(2a−x)(2a+3x)=[2(x2+16x+60)−x][2(x2+16x+60)+3x]=(2x2+31x+120)(2x2+35x+120)
Đa thức này không phân tích được thành nhân tử.
Bạn coi lại đề.
Ta có: \(4\left(x+5\right)\left(x+6\right)\left(x+10\right)\left(x+12\right)+3x^2\)
\(=4\left(x^2+60+17x\right)\left(x^2+60x+16x\right)+3x^2\)
\(=4\left[\left(x^2+60\right)^2+33x\left(x^2+60\right)+272x^2\right]+3x^2\)
\(=4\left(x^2+60\right)^2+132x\left(x^2+60\right)+1091x^2\)
\(2x^2+x-6\)
\(=2x^2-3x+4x-6\)
\(=x\left(2x-3\right)+2\left(2x-3\right)\)
\(=\left(2x-3\right)\left(x+2\right)\)
\(6x^2-13x+6\)
\(=6x^2-9x-4x+6\)
\(=\left(2x-3\right)\left(3x-2\right)\)
=3x(x+2y)
=3x(x+2y)