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a: Ta có: \(x^2-4y^2-2x-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
c: Ta có: \(x^3+2x^2y-x-2y\)
\(=x^2\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
d: Ta có: \(3x^2-3y^2-2\cdot\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\cdot\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
e: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
f: Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
a) x2-2x-y2+2y
=(x2-y2)-(2x-2y)
=(x-y)(x+y)-2(x-y)
=(x-y)(x+y-2)
a) x2 ( x+ 2y) -x -2y
= x2 ( x+ 2y) -(x+2y)
= (x2-1)(x+2y)
= (x-1)(x+1)(x+2y)
b)3x2- 3y2 -2 (x-y)2
= 3(x2-y2) -2 (x-y)2
= 3(x-y)(x+y)-2(x-y)(x-y)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)
c) x2- 2x-4y2 - 4y
= (x2-4y2)-(2x+4y)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\)
d) x3 - 4x2 - 9x +36
= (x3+3x2)-(7x2+21x)+(12x+36)
= x2(x+3)-7x(x+3)+12(x+3)
=(x2-7x+12)(x+3)
\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
\(a,9x^2+y^2+2z^2-18x+4z-6y+20=0\\ \Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2+2\left(z+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)
\(b,5x^2+5y^2+8xy+2y-2x+2=0\\ \Leftrightarrow4\left(x+y\right)^2+\left(x-1\right)^2+\left(y+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(c,5x^2+2y^2+4xy-2x+4y+5=0\\ \Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=-y\\x=1\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
\(d,x^2+4y^2+z^2=2x+12y-4z-14\\ \Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+\left(z+2\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3}{2}\\z=-2\end{matrix}\right.\)
\(e,x^2+y^2-6x+4y+2=0\\ \Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Pt vô nghiệm do ko có 2 bình phương số nguyên có tổng là 11
e: Ta có: \(x^2-6x+y^2+4y+2=0\)
\(\Leftrightarrow x^2-6x+9+y^2+4y+4-11=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(y+2\right)^2=11\)
Dấu '=' xảy ra khi x=3 và y=-2
a) \(\left(x+2y\right)^2-\left(x-y\right)^2=\left(x+2y+x-y\right)\left(x+2y-x+y\right)\)
\(=\left(2x+y\right).3y\)
b) \(\left(x+1\right)^3+\left(x-1\right)^3\)
\(=\left(x+1+x-1\right)\left[\left(x+1\right)^2-\left(x+1\right)\left(x-1\right)+\left(x-1\right)^2\right]\)
\(=2x\left[\left(x+1\right)^2-\left(x^2-1\right)+\left(x-1\right)^2\right]\)
c) \(9x^2-3x+2y-4y^2\)
\(=9x^2-4y^2-3x+2y\)
\(=\left(3x-2y\right)\left(3x+2y\right)-\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left[3x+2y-1\right]\)
d) \(4x^2-4xy+2x-y+y^2\)
\(=4x^2-4xy+y^2+2x-y\)
\(=\left(2x-y\right)^2+2x-y\)
\(=\left(2x-y\right)\left(2x-y+1\right)\)
e) \(x^3+3x^2+3x+1-y^3\)
\(=\left(x+1\right)^3-y^3\)
\(=\left(x+1-y\right)\left[\left(x+1\right)^2+y\left(x+1\right)+y^2\right]\)
g) \(x^3-2x^2y+xy^2-4x\)
\(=x\left(x^2-2xy+y^2\right)-4x\)
\(=x\left(x-y\right)^2-4x\)
\(=x\left[\left(x-y\right)^2-4\right]\)
\(=x\left(x-y+2\right)\left(x-y-2\right)\)
a) (x + 2y)² - (x - y)²
= (x + 2y - x + y)(x + 2y + x - y)
= 3y(2x + y)
b) (x + 1)³ + (x - 1)³
= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]
= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)
= 2x(x² + 3)
c) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) x³ + 3x² + 3x + 1 - y³
= (x³ + 3x² + 3x + 1) - y³
= (x + 1)³ - y³
= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]
= (x - y + 1)(x² + 2x + 1 + xy + y + y²)
g) x³ - 2x²y + xy² - 4x
= x(x² - 2xy + y² - 4)
= x[(x² - 2xy + y²) - 4]
= x[(x - y)² - 2²]
= x(x - y - 2)(x - y + 2)
phân tích đa thức thành nhân tử bằng cách nhóm hạng tử
1) x2 - y2 - 2x - 2y
2) 3x2 - 3y2 - 2(x - y)2
1) \(x^2-y^2-2x-2y\)
\(=\left(x^2-y^2\right)-\left(2x+2y\right)\)
\(=\left(x+y\right)\left(x-y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
2) \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
1) x² - y² - 2x - 2y
= (x² - y²) - (2x + 2y)
= (x - y)(x + y) - 2(x + y)
= (x + y)(x - y - 2)
2) 3x² - 3y² - 2(x - y)²
= (3x² - 3y²) - 2(x - y)²
= 3(x² - y²) - 2(x - y)²
= 3(x - y)(x + y) - 2(x - y)²
= (x - y)[3(x + y) - 2(x - y)]
= (x - y)(3x + 3y - 2x + 2y)
= (x - y)(x + 5y)