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\(\text{e) 2x^4 + 5x^3+13x^2+25x+15 }\)
\(\text{=2x^3(x+1)+3x^2(x+1)+10x(x+1)+15(x+1) }\)
\(\text{=(x+1)[x^2(2x+3)+5(2x+3)]}\)
\(\text{=(x+1)(2x+3)(x^2+5)}\)
\(2x^4+5x^3+13x^2+25x+15\)
\(=2x^4+2x^3+3x^3+3x^2+10x^2+10x+15x+15\)
\(=2x^3\left(x+1\right)+3x^2\left(x+1\right)+10x\left(x+1\right)+15\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^3+3x^2+10x+15\right)\)
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\(=2x^4-6x^3-x^3+3x^2-5x^2+15x-2x+6\)
\(=2x^3\left(x-3\right)-x^2\left(x-3\right)-5x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(2x^3-x^2-5x-2\right)\)
\(=\left(x-3\right)\left(2x^3-4x^2+3x^2-6x+x-2\right)\)
\(=\left(x-3\right)\left[2x^2\left(x-2\right)+3x\left(x-2\right)+\left(x-2\right)\right]\)
\(=\left(x-3\right)\left(x-2\right)\left(2x^2+3x+1\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(2x+1\right)\)
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=6x^2 - 4x - 9x +6
=(6x^2 -4x) - (9x-6)
=2x(3x -2) - 3(3x-2)
=(3x-2) (2x - 3)
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x3+13x2+7x-5
= x3+x2+12x2+12x-5x-5
=(x3+12x2-5x)+(x2+12x-5)
=x(x2+12x-5)+1(x2+12x-5)
=(x+1)(x2+12x-5)
phân tích đa thức thành nhân tử:
x3+13x2+7x-5
= x3+x2+12x2+12x-5x-5
=(x3+12x2-5x)+(x2+12x-5)
=x(x2+12x-5)+1(x2+12x-5)
=(x+1)(x2+12x-5)
Ai trên 10 điểm hỏi đáp thì mình nha mình đang cần gấp chỉ còn 99 điểm là tròn rồi mong các bạn hỗ trợ mình sẽ đền bù xứng đáng
tích nha
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\(3x^3-4x^2+13x-4\)
\(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right) \)
\(=\left(3x-1\right)\left(x^2-x+4\right)\)
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\(6x^2+13x-15\)
\(=6x^2+18x-5x-15\)
\(=6x.\left(x+3\right)-5.\left(x+3\right)\)
\(=\left(x+3\right).\left(6x-5\right)\)
\(6x^2+13x-15\)
\(=6x^2+18x-5x-15\)
\(=6x\left(x+3\right)-5\left(x+3\right)\)
\(=\left(x+3\right)\left(6x-5\right)\)
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Bài làm
a) x3 - 12 + 3x2 - 4x
= ( x3 + 3x2 ) + ( -12 - 4x )
= x2( x + 3 ) - 4( x + 3 )
= ( x2 - 4 )( x + 3 )
b) 2x2 + 13x - 7
= 2x2 + 14x - 1x - 7
= ( 2x2 + 14x ) - ( x + 7 )
= 2x( x + 7 ) - ( x + 7 )
= ( 2x - 1 )( x + 7 )
# Học tốt #
\(-2x^2+13x-23=0\)
\(2x^2-13x+23=0\)
\(x^2-\frac{13}{2}x+\frac{23}{2}=0\)
\(x^2-2.\frac{13}{4}x+\frac{169}{16}-\frac{169}{16}+\frac{23}{2}=0\)
\(\left(x-\frac{13}{4}\right)^2+\frac{15}{16}=0\)
\(\Rightarrow x\in\varnothing\)