2x=5x và x3 +y3=133
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10: \(x\left(x-y\right)+x^2-y^2\)
\(=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+x+y\right)\)
\(=\left(x-y\right)\left(2x+y\right)\)
11: \(x^2-y^2+10x-10y\)
\(=\left(x^2-y^2\right)+\left(10x-10y\right)\)
\(=\left(x-y\right)\left(x+y\right)+10\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+10\right)\)
12: \(x^2-y^2+20x+20y\)
\(=\left(x^2-y^2\right)+\left(20x+20y\right)\)
\(=\left(x-y\right)\left(x+y\right)+20\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+20\right)\)
13: \(4x^2-9y^2-4x-6y\)
\(=\left(4x^2-9y^2\right)-\left(4x+6y\right)\)
\(=\left(2x-3y\right)\left(2x+3y\right)-2\left(2x+3y\right)\)
\(=\left(2x+3y\right)\left(2x-3y-2\right)\)
14: \(x^3-y^3+7x^2-7y^2\)
\(=\left(x^3-y^3\right)+\left(7x^2-7y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\cdot\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+7\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+7x+7y\right)\)
15: \(x^3+4x-\left(y^3+4y\right)\)
\(=x^3-y^3+4x-4y\)
\(=\left(x^3-y^3\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+4\right)\)
16: \(x^3+y^3+2x+2y\)
\(=\left(x^3+y^3\right)+\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+2\right)\)
17: \(x^3-y^3-2x^2y+2xy^2\)
\(=\left(x^3-y^3\right)-\left(2x^2y-2xy^2\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-2xy\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-2xy\right)\)
\(=\left(x-y\right)\left(x^2-xy+y^2\right)\)
18: \(x^3-4x^2+4x-xy^2\)
\(=x\left(x^2-4x+4-y^2\right)\)
\(=x\left[\left(x^2-4x+4\right)-y^2\right]\)
\(=x\left[\left(x-2\right)^2-y^2\right]\)
\(=x\left(x-2-y\right)\left(x-2+y\right)\)
1: \(=\dfrac{x-1}{x^2+x+1}+\dfrac{x+1}{x-1}\)
\(=\dfrac{x^2-2x+1+x^3+x^2+x^2+x+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3+3x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
2: \(=\dfrac{\left(x^2-y^2\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)^2}{x^2+xy+y^2}\)
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
1: =(2x+y-2y)(2x+y+2y)
=(2x-y)(2x+3y)
2: =(4-5x)(16+20x+25x^2)
3: =x(x^2-2xy+y^2-4)
=x[(x-y)^2-4]
=x(x-y-2)(x-y+2)
4: =(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
Nếu hai đại lượng y và x tỉ lệ nghịch và y = 5 x nên hệ số tỉ lệ a = 5, do đó
x 1 y 1 = x 2 y 2 = x 3 y 3 = ... = 5
Đáp án cần chọn là C
11: \(2x^2-12xy+18y^2\)
\(=2\left(x^2-6xy+9y^2\right)\)
\(=2\left(x-3y\right)^2\)
12: \(\left(x^2+x\right)^2+3\left(x^2+x\right)+2\)
\(=\left(x^2+x+2\right)\left(x^2+x+1\right)\)
a) Ta thấy \(xy=\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}=\dfrac{3^2-5}{2}=2\)
\(\Rightarrow x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\) \(=3\left(5-2\right)=9\)
b) Ta thấy \(xy=\dfrac{-\left(x-y\right)^2+\left(x^2+y^2\right)}{2}=\dfrac{15-5^2}{2}=-5\)
\(\Rightarrow x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)\) \(=5\left(15-5\right)=50\)
xem lại đề đi sai đề rồi