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a) Ta có: \(4\left(2-x\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left[4\left(x-2\right)+y\right]\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(3a^2x-3a^2y+abx-aby\)
\(=3a^2\left(x-y\right)+ab\left(x-y\right)\)
\(=\left(x-y\right)\left(3a^2+ab\right)\)
\(=a\left(x-y\right)\left(3a+b\right)\)
c) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
\(=\left(x-y\right)\left[x\left(x^2-2xy+y^2\right)-yx+y^2-y^2\right]\)
\(=\left(x-y\right)\left(x^3-2x^2y+xy^2-xy\right)\)
d) Ta có: \(2ax^3+6ax^2+6ax+18a\)
\(=2ax^2\left(x+3\right)+6a\left(x+3\right)\)
\(=\left(x+3\right)\left(2ax^3+6a\right)\)
\(=2a\left(x+3\right)\left(x^3+3\right)\)
e) Ta có: \(x^2y-xy^2-3x+3y\)
\(=xy\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-3\right)\)
Bài 4 :
a) \(x^3+x^2y-xy^2-y^3=x^2\left(x+y\right)-y^2\left(x+y\right)=\left(x^2-y^2\right)\left(x+y\right)=\left(x-y\right)\left(x+y\right)^2\)
b)\(x^2y^2+1-x^2-y^2=\left(x^2y^2-x^2\right)-\left(y^2-1\right)=x^2\left(y^2-1\right)-\left(y^2-1\right)=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(x+1\right)\left(y-1\right)\left(y+1\right)\)
c) \(x^2-y^2-4x+4y=\left(x^2-y^2\right)-\left(4x-4y\right)=\left(x-y\right)\left(x+y\right)-4\left(x-y\right)=\left(x-y\right)\left(x+y-4\right)\)
d)
\(x^2-y^2-2x-2y=\)\(\left(x^2-y^2\right)-\left(2x+2y\right)=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(x-y-2\right)\)
e) Trùng câu d
f) \(x^3-y^3-3x+3y=\left(x-y\right)\left(x^2-xy+y^2\right)-3\left(x-y\right)=\left(x-y\right)\left(x^2-xy+y^2-3\right)\)
Bài 5:
a) \(x^3-x^2-x+1=0\)
\(\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
b) Sửa đề : \(\left(2x-3\right)^2-\left(4x^2-9\right)=0\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(-6\right)=0\)\
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow x=\frac{9}{2}\)
vậy........
c) \(x^4+2x^3-6x-9=0\)
\(\Leftrightarrow\left(x^4-9\right)+\left(2x^3-6x\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow x^2-3=0\Leftrightarrow x^2=3\Leftrightarrow x=\pm\sqrt{3}\)
Vậy
d) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
Vậy ........
a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
a, -x - y2 + x2 - y = (x2 - y2) - (x + y)
= (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
b, x( x + y ) - 5x - 5y = x(x + y) - 5(x + y)
= (x - 5)(x + y)
c, x2 - 5x + 5y - y2 = (x - y)(x + y) - 5(x - y)
= (x - y)(x + y - 5)
d, 5x3 - 5x2y - 10x2 + 10xy = 5x2(x - y) - 10x(x - y)
= 5x(x - y)(x - 2)
e, 27x3 - 8y3 = (3x - 2y)(9x2 + 6xy + 4y2)
f, x2 - y2 - x - y = (x - y)(x + y) - (x + y)
= (x + y)(x - y - 1)
g, x2 - y2 - 2xy + y2 = (x2 - 2xy + y2) - y2
= (x - y)2 - y2
= (x - y - y)(x - y + y) = x(x - 2y)
h, x2 - y2 + 4 - 4x = (x2 - 4x + 4) - y2
= (x - 2)2 - y2
= (x - y - 2)(x + y - 2)
i, x3 + 3x2 + 3x + 1 - 27z3 = (x + 1)3 - 27z3
= (x+1-3z)(x2+2x+1+3xz+3z+9z2)
k, 4x2 + 4x - 9y2 + 1 = (2x + 1)2 - 9y2
= (2x - 3y + 1)(2x + 3y + 1)
m, x2 - 3x + xy - 3y = x(x - 3) + y(x - 3)
= (x - 3)(x + y)
a, ( x2 + x )2 - 14 ( x2 + x ) + 24
= (x2 + x)2 - 2(x2 + x) -12(x2 + x) + 24
= (x2 + x).(x2 + x -2) - 12(x2 + x -2)
= (x2 + x -2).(x2 + x -12)
= (x2 + 2x - x - 2).(x2 + 4x - 3x - 12)
=[x.(x+2)-(x+2)].[x.(x+4)-3(x+4)]
= (x+2).(x-1).(x+4).(x-3)
= x4 + 2x3 - 13x2 - 14x + 24
b, ( x2 + x )2 + 4x2 + 4x - 12
= x4 + 2x3 + x2 + 4x2 + 4x -12
= x4 + 2x3 + 5x2 + 4x -12
c, x4 + 2x3 + 5x2 + 4x - 12
= x4 - x3 + 3x3 - 3x2 + 8x2 - 8x +12x -12
= x3(x-1) + 3x2(x-1) + 8x(x-1) + 12(x-1)
= (x-1) . (x3 + 3x2 + 8x +12)
= (x-1) . ( x3 +2x2 + x2 + 2x + 6x +12)
= (x-1). [x2(x+2) + x(x+2) + 6(x+2)]
= (x-1).(x+2).(x2 + x+ 6)
bài 1:
a) 2m(x-y) + x-y = 2m(x-y) + (x-y) = (2m+1)(x-y)
b) x(y-2) + y\(^2\) - 2y = x(y-2) + y(y-2) = (x+y)(y-2)
c) x\(^2\) +xy - 2x - 2y = x(x+y) - 2(x+y) = (x-2)(x+y)
d) x + x\(^2\) - x\(^3\) - x\(^4\) = x(1 + x - x\(^2\) - x\(^3\))
e) 2+2x-xy-y = 2(1+x) - y(x+1) = (2-y)(x+1)
f) x\(^2\) + 2y - 1 -2x + 1 - y\(^2\) = (x\(^2\) -2x+1) - (y\(^2\)-2y+1) = (x-1)\(^2\) - (y-1)\(^2\)
g) (x+1)\(^2\) -x-1 = (x+1)\(^2\) -(x+1) =(x+1)(x+1-1) = (x+1)x
Bài 3:
\(P=x^2-4x+4+5=\left(x-2\right)^2+5>=5\)
Dấu = xảy ra khi x=2
Bài 4:
a: \(=-\left(x^2-4x-5\right)\)
\(=-\left(x^2-4x+4-9\right)\)
\(=-\left(x-2\right)^2+9< =9\)
Dấu = xảy ra khi x=2
b: \(=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}< =\dfrac{1}{4}\)
Dấu = xảy ra khi x=1/2
c: \(=x^2-6x+9+3=\left(x-3\right)^2+3>=3\)
Dấu '=' xảy ra khi x=3
a) ktra lại đề
b) \(5x\left(x-y\right)-\left(y-x\right)=\left(x-y\right)\left(5x+1\right)\)
c) \(x\left(x+3\right)+\left(3+x\right)=\left(x+3\right)\left(x+1\right)\)
f) \(4x\left(x-2\right)-\left(2x\right)^2=4x^2-8x-4x^2=-8x\)
g) \(\left(x-2\right)^2-\left(2-x\right)^3=\left(x-2\right)^2+\left(x-2\right)^3=\left(x-2\right)^2\left(x-1\right)\)