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\(B=x\left(x-2\right)\left(x+2\right)-\left(x+3\right)\left(x^2-3x+9\right)\)
\(B=x\left(x^2-4\right)-\left(x^3-3x^2+9x+3x^2-9x+27\right)\)
\(B=x^3-4x-\left(x^3+27\right)\)
\(B=-4x-27\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)
x(1 - 3x)(4 - 3x) - (x - 4)(3x + 5)
= (x - 3x2)(4 - 3x) - 3x2 - 5x + 12x + 20
= 4x - 3x2 - 12x2 + 9x3 - 3x2 - 5x + 12x + 20
= 9x3 - 18x2 + 11x + 20
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(x+1\right)\left(x^2-x+1\right)-x^3+3x=15\)
\(\Leftrightarrow4x^3+4-x^3+3x-15=0\)
\(\Leftrightarrow3x^3+3x-11=0\)
Đây là phương trình vô tỉ nhé bạn, nên nghiệm rất xấu!
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Bài 2:
a: Để \(\dfrac{4}{x+2}>0\) thì x+2>0
hay x>-2
b: Để \(\dfrac{3x+2}{-4}>0\) thì 3x+2<0
hay x<-2/3
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\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)
\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)
\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)
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\(=\dfrac{2x^2y^2}{3xy^2}-\dfrac{2ax+3x}{3a}=\dfrac{2x}{3}-\dfrac{2ax+3x}{3a}\)
\(=\dfrac{2xa-2xa-3x}{3a}=\dfrac{-3x}{3a}=-\dfrac{x}{a}\)
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\(=\dfrac{5}{3}-\dfrac{5a-6}{3a}=\dfrac{5a-5a+6}{3a}=\dfrac{6}{3a}=\dfrac{2}{a}\)
\(x^2-3x+xy-3x\)
\(=x^2+xy-6x\)
\(=x.\left(x+y-6\right)\)