If x = 3 , y= -1 then x^5 + 5x^4y + 10x^3y^2 + 5xy^4 + y^2 = .....
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A) \(2x\left(x+1\right)=2x^2+2x\)
B) \(\left(5x-6y\right)\left(x-2\right)=5x^2-10x-6xy+12y\)
C) \(\dfrac{10x^4y^2-5x^3y^2+15xy^4}{5xy}=\dfrac{5xy\left(2x^3y-x^2y+3y^3\right)}{5xy}\)
\(=2x^3y-x^2y+3y^3\)
D) \(2x\left(x+y\right)-3\left(x+y\right)=2x^2+2xy-3x-3y\)
\(E)4x^2-49\)
F) \(x\left(x+2\right)-2x-4=0\)
\(x\left(x+2\right)-2\left(x+2\right)=0\)
\(\left(x-2\right)\left(x+2\right)=0\)
\(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
A, 2x.(x+1) = 2x\(^2\)+2x B,(5x-6y)(x-2) = 5x\(^2\)-10x-6xy+12y C,(10x\(^4\)y\(^2\)-5x\(^3\)y\(^2\)+15xy\(^4\)) : 5xy = 2x\(^3\)y - x\(^2\)y + 3y\(^3\)
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a) \(2x\left(x-7\right)-5y\left(x-7\right)=\left(x-7\right)\left(2x-5y\right)\)
b) \(5x^3y+10x^2y+5xy=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)
c) \(4y^2-4y-x^2+1=\left(2y-1\right)^2-x^2=\left(2y-1-x\right)\left(2y-1+x\right)\)
d) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1=\left(x^2+3x+1\right)^2\)
a: \(=\left(x-7\right)\left(2x-5y\right)\)
b: \(=5xy\left(x^2+2x+1\right)=5xy\left(x+1\right)^2\)
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Bài 1:
a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)
\(=\dfrac{15x^2y^2z}{3xyz}\)
\(=5xy\)
b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)
\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)
\(=15x^4-12x^3+9x^2\)
c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)
\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)
\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)
\(=2x+5+\dfrac{20}{x-4}\)
d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)
\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)
\(=-15x^3y^2+25x^2y^2-5xy^3\)
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1, a, \(2x\left(x+1\right)=2x^2+2x\)
b, \(\left(5x-6\right)\left(x-2\right)=5x^2-16x+12\)
c, \(\left(10x^4y^2-5x^3y^2+15xy^2\right):5xy=2x^3y-x^2y+3y\)
2, a, \(2x\left(x+y\right)-3\left(x+y\right)=\left(x+y\right)\left(2x-3\right)\)
b, \(4x^2-49=\left(2x\right)^2-7^2=\left(2x+7\right)\left(2x-7\right)\)
3, \(x\left(x+2\right)-2x-4=0\)
\(\Leftrightarrow x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
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a) (10x3y - 5x2y2 - 25x4y3) : (- 5xy)
= - 2x2 + xy + 5x3y2
b) (27x3 - y3) : (3x - y)
= (3x - y)(9x2 - 3xy + y2) : (3x - y)
= 9x2 - 3xy + y2
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
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3: 10x=6y=5z
\(\Leftrightarrow\dfrac{10x}{30}=\dfrac{6y}{30}=\dfrac{5z}{30}\)
hay x/3=y/5=z/6
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{24}{2}=12\)
Do đó: x=36; y=60; z=72
4: Ta có: 9x=3y=2z
nên \(\dfrac{9x}{18}=\dfrac{3y}{18}=\dfrac{2z}{18}\)
hay x/2=y/6=z/9
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{6}=\dfrac{z}{9}=\dfrac{x-y+z}{2-6+9}=\dfrac{50}{5}=10\)
Do đó: x=20; y=60; z=90