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a, - \(\dfrac{1}{3}\).\(xy\).(3\(x^3\).y2 - 6\(x^2\) + y2)
= - \(x^4\).y3 + 2\(x^3\).y - \(\dfrac{1}{3}\).\(xy^3\)
b, (2\(x\) -3).(4\(x\)2 + 6\(x\) + 9)
= (2\(x\))3 - 33
= 8\(x^3\) - 27
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=-15x^3+10x^4+20x^2\\ b,=2x^3+2x^2+4x-x^2-x-2=2x^3+x^2+3x-2\)
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Thực hiện các phép tính sau:
\(\text{a) 3x². (2x³ – x + 5)=}6x^6-3x^3+15x^2\)
\(\text{ b) (4xy + 3y – 5x). x²y}=4x^3y^2+3x^2y^2-5x^3y\)
\(\text{ d) ( 3x – 5 )( x²– 5x + 7 )}=3x^3+21x-35\)
Học tốt
a)=6x^5-3x^3+15x^2
b)=4x^3y^2+3x^2y^2-5x^3y
c)=3x^3-5x^2-15x^2+25x+21x-35=3x^3-20x^2+46x-35
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Áp dụng hằng đẳng thức \(\left(a+b\right)\left(a-b\right)=a^2-b^2\)ta đc:
\(\left(x+2y+z\right)\left(x+2y-z\right)=\left[\left(x+2y\right)+z\right]\left[\left(x+2y\right)-z\right]\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^{ }\)
( x + 2y + z ) ( x + 2y - z )
<=> ( x + 2y)2 - z2
<=> x2 + 4xy + 4y2 - z2
^^ Học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
DK \(3+2x\ne0;3-2x\ne0\Leftrightarrow x\ne\left\{+-\frac{3}{2}\right\}\)
\(\left(\frac{3-2x+\left(3+2x\right)}{\left(3+2x\right)\left(3-2x\right)}\right).\frac{\left(3+2x\right)}{1}\)\(=\left(\frac{6}{\left(3+2x\right)\left(3-2x\right)}\right).\left(3+2x\right)\)\(=\frac{6}{\left(3-2x\right)}\)
\(=\frac{\left(3-2x\right)+\left(3+2x\right)}{\left(3+2x\right)\left(3-2x\right)}.\left(3+2x\right)\)
\(=\frac{6}{3-2x}\)
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\(a,\left(x-2\right).\left(x-3\right)-\left(x+3\right).\left(x-3\right)\)
\(=\left(x-3\right).\left(x-2-x+3\right)=x-3\)
\(b,\frac{\left(x^2+4x+4\right)}{x+2}-4x+5=\frac{\left(x+2\right)^2}{x+2}-4x+5\)
\(x+2-4x+5=-3x+7\)
a) \(\left(x-2\right)\left(x-3\right)-\left(x+3\right)\left(x-3\right)\)
\(=\left(x^2-5x+6\right)-\left(x^2-9\right)\)
\(=x^2-5x+6-x^2+9\)
\(=15-5x\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)-\left(4x-5\right)\)
\(=\left(x+2\right)^2:\left(x+2\right)-\left(4x-5\right)\)
\(=\left(x+2\right)-4x+5\)
\(=x+2-4x+5\)
\(=7-3x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(\left(5x+1\right)^2-\left(2xy-3\right)^2\)
\(=25x^2+10x+1-\left(2xy-3\right)^2\)
\(=25x^2+10x+1\left(4x^2y^2-12xy+9\right)\)
\(=25x^2+10x+1-4x^2y^2+12xy-9\)
\(=25x^2-4x^2y^2+10x+12xy-8\)
Bài 2:
\(\left(x-1\right)\left(x^2+x+1\right)=x^2\left(x-9\right)+2x+6\)
\(=x^3-1=x^3-9x^2+2x+6\)
\(=x^3-9x^2+2x+6=x^3-1\)
\(=x^3-9x^2+2x+6+1=x^3-1+1\)
\(=x^3-9x^2+2x+7=x^3\)
\(=x^3-9x^2+2x+7-x^3=x^3-x^3\)
\(=-9x^2+2x+7=0\)
\(\Rightarrow x=-\frac{7}{9};x=1\)