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15x3y5z : 5x2y3
= (15:5).(x3:x2 ).(y5 : y3 ).z
= 3.x(3-2).y(5-3).z
= 3xy2z
1, x^3-2x^2+x
=x^3-x^2-x^2+x
=(x^3-x^2)-(x^2-x)
= x^2(x-1)-x(x-1)
=(x-1)(x^2-x)
=x(x-1)(x-1)
2, x^2-2x-15
=x^2-2x-9-6
= x^2-9-2x-6
=(x^2-9)-(2x+6)
=(x-3)(x+3)-2(x+3)
=(x+3)(x-3-2)=(x+3)(x-5)
3 , \(^{3x^3y^2-6x^2y^3+9x^2y^2}\)
= \(^{3x^2y^2\left(x-2y+3\right)}\)
4, \(^{5x^2y^3-25x^3y^4+10x^3y^3}\)
=\(^{5x^2y^2\left(y-5xy^2+2xy\right)}\)
5, \(^{12x^2y-18xy^2-30y^2}\)
=\(^{3y\left(4x^2-6xy-10y\right)}\)
Lời giải:
1. $x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$
2. $x^2-2x-15=(x^2+3x)-(5x+15)=x(x+3)-5(x+3)=(x+3)(x-5)$
3. $3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2(x-2y+3)$
4. $5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3(1-5xy+2x)$
5. $12x^2y-18xy^2-30y^2=6y(2x^2-3xy-5y)$
b: Ta có: \(xy-3x-2y+6\)
\(=x\left(y-3\right)-2\left(y-3\right)\)
\(=\left(y-3\right)\left(x-2\right)\)
\(a,=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ b,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ c,=5x^2y^3\left(1-5xy+2x\right)\\ d,=6y\left(2x^2-3xy-10y\right)\\ e,,=\left(x-y\right)\left(5-x\right)\\ f,=\left(2x+3\right)^2\)
1) PT \(\Leftrightarrow\left(\dfrac{x+1}{35}+1\right)+\left(\dfrac{x+3}{33}+1\right)=\left(\dfrac{x+5}{31}+1\right)+\left(\dfrac{x+7}{29}+1\right)\)
\(\Leftrightarrow\dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\)
\(\Leftrightarrow\left(x+36\right)\left(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}\right)=0\)
\(\Leftrightarrow x+36=0\) (Do \(\dfrac{1}{29}+\dfrac{1}{31}-\dfrac{1}{33}-\dfrac{1}{35}>0\))
\(\Leftrightarrow x=-36\).
Vậy nghiệm của pt là x = -36.
\(\left(20x^3y^5-5x^2y^3\right):5x^2y^3\\ =5x^2y^3\left(4xy^2-1\right):5x^2y^3\\ =4xy^2-1\)