NHÂN ĐA THỨC VS ĐƠN THỨC :(x-1)(x^5+x^4+x^3+x^2+x+1)
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a) (x-1)*(x+2)-(x-3)*(-x+4)=19
\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)
\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)
\(\Leftrightarrow2x^2-3x+7-19=0\)
\(\Leftrightarrow2x^2-3x-12=0\)
Đề sai??
b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)
\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)
\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)
\(\Leftrightarrow-30x^2+7x-4+17=0\)
\(\Leftrightarrow-30x^2+7x+13=0\)
???
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(x3 - 2x2 + x - 1)(5 - x) = (x3 - 2x2 + x - 1).5 - [ (x3 - 2x2 + x - 1).x ] = 5x3 - 10x2 + 5x - 5 - (x4 - 2x3 + x2 - x)
= 5x3 - 10x2 + 5x - 5 - x4 + 2x3 - x2 + x = x4 + (5x3 + 2x3) + (-10x2 - x2) + (5x + x) - 5 = x4 + 7x3 - 11x2+ 6x - 5
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Làm này mới đúng chứ:
(3x-5)(x+1) - (x-1)(x+2)= (2x-3)(x+2) + 1
=> (3x-5)(x+1) = (2x-3)(x+2) + (x-1)(x+2) + 1
=> (3x - 4 - 1)(x + 1) = (x+2) [(2x-3) + (x-1)] + 1
=> (3x-4)(x+1) - x - 1 = (x+2).(3x-4) + 1
=> (3x-4)(x+1) - (x+2).(3x-4) = x + 1 + 1
=> (3x-4).[(x+1) - (x+2)] = x + 2
=> (3x-4).(-1) = x + 2
=> - 3x + 4 = x + 2
=> 3x + x = 4 + 2
=> 4x = 6
=> x = 6 : 4
=> x = 3/2
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a)\(\left(x+1\right)\left(x+3\right)-x\left(x-2\right)=7\)
\(x\left(x+3\right)+x+3-x^2+2x=7\)
\(x^2+3x+x+3-x^2+2x=7\)
\(6x+3=7\)
\(6x=4\)
\(x=\frac{4}{6}=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b) \(2x\left(3x+5\right)-x\left(6x-1\right)=33\)
\(6x^2+10x-6x^2-x=33\)
\(9x=33\)
\(x=\frac{33}{9}\)
Vậy \(x=\frac{33}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x+1\right)\left(x+2\right)\left(x-3\right)=\left(x^2+2x+x+2\right)\left(x-3\right)\)
\(=\left(x^2+3x+2\right)\left(x-3\right)\)
\(=x\left(x^2+3x+2\right)-3\left(x^2+3x+2\right)\)
\(=x^3+3x^2+2x-3x^2-9x-6\)
\(=x^3-7x-6\)
b) \(\left(2x-1\right)\left(x+2\right)\left(x+3\right)=\left(2x-1\right)\left(x^2+3x+2x+6\right)\)
\(=\left(2x-1\right)\left(x^2+5x+6\right)\)
\(=2x\left(x^2+5x+6\right)-\left(x^2+5x+6\right)\)
\(=2x^3+10x^2+12x-x^2-5x-6\)
\(=2x^3+9x^2+7x-6\)
(x-1)(x\(^5\)+x\(^4\)+x\(^3\)+x\(^2\)+x+1)
=x(\(x^5+x^4+x^3+x^2+x+1\))-1(\(x^5+x^4+x^3+x^2+x+1\))
= x.\(x^5+x\cdot x^4+x\cdot x^3+x\cdot x^2+x\cdot x+x\cdot1\)-1.\(x^5-1\cdot x^4-1\cdot x^3-1\cdot x^2-1\cdot x-1\cdot1\)
=\(x^6\)+\(x^5\)\(+x^4\)+\(x^3\)+\(x^2\)+1x -1\(x^5\)-1\(x^4\)-1\(x^3\)-1\(x^2\)-1x -1
=\(x^6\)+(\(x^5\)-1\(x^5\))+(x\(^4\)-1\(x^4\))+(\(x^3\)-1\(x^3\))+(x\(^2\)-1\(x^2\))+(1x-1x)-1
=x\(^6\)-1
thanks