Phân tích các đa thức sau thành nhân tử:
b ) 2 x y + z + 2 x + y z
nhờ giải giupws em với a
1. Phân tích các đa thức sau thành nhân tử:
a) 5x2 – 10xy
b) 3x(x – y) – 6(x – y)
c) 2x(x – y) – 4y(y – x)
d) 9x2 – 9y2
e) x2 – xy – x + y
f) xy – xz – y + z
2. Phân tích các đa thức sau thành nhân tử:
a)a2 – 4b2 b) x2 – y2 + 6y - 9
c) (2a + b)2 – a2 d) 16(x – 1)2 – 25(x + y)2
e)x2 + 10x + 25 f) 25x2 – 20xy + 4y2
g)9x4 + 24x2 + 16 h) x3 – 125
i)x6 – 1 k) x3 + 15x2 + 75x + 125
3. Tìm x biết :
a) 3x2 + 8x = 0 b) 9x2 – 25 = 0 c) x3 – 16x = 0 d) x3 + x = 0.
4. Chứng minh rằng với mọi số nguyên a thì: a3 – a chia hết cho 6
Bài `1`
\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)
Bài `3`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
Phân tích các đa thức sau thành nhân tử
a) (x+y+z)^3 - x^3 - y^3 - z^3
b) x^4 + 2012x^2 + 2011x + 2012
= x3 + y3 + z3 + 3x2yz + 3xy2z + 3xyz2 - x3 -y3 - z3
=3x2yz + 3xy2z + 3xyz2
= 3xyz( x + y + z)
b.
x^4+2012x^2+2012x-x+2012=
(x^4-x)+2012(x^2+x+1)=
x(x-1)(x^2+x+1)+2012(x^2+x+1)=
(x+2012)(x^2+x+1)
Phân tích các đa thức sau thành nhân tử : a) (x-y)z^3+(z-x)y^3+(y-z)x^3
t chỉ cho kết quả thôi nhá, còn nhóm nhân tử you tự xử nhá !
=(x-y)(z-x)(z-y)(x+y+z)
\(\left(x-y\right)z^3+\left(z-z\right)y^3+\left(y-z\right)x^3\)
\(=z^3\left(x-y\right)+y^3\left(z-x\right)+x^3\left(y-z\right)\)
\(=xz^3-yz^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+\left(y-z\right)x^3\)
\(=xz^3-yz^3+y^3z-xy^3+y^3z-xy^3+x^3y-x^3z\)
Mk ko chắc
\(\left(x-y\right)z^3+\left(z-x\right)y^3+\left(y-z\right)x^3\)
\(=\left(x-y\right)z^3-\left[\left(x-y\right)+\left(y-z\right)\right]y^3+\left(y-z\right)x^3\)
\(=\left(x-y\right)z^3-\left(x-y\right)y^3-\left(y-z\right)y^3+\left(y-z\right)x^3\)
\(=\left(x-y\right)\left(z^3-y^3\right)+\left(y-z\right)\left(x^3-y^3\right)\)
\(=\left(x-y\right)\left(z-y\right)\left(z^2+zy+y^2\right)+\left(y-z\right)\left(x-y\right)\left(x^2+y^2+xy\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x^2+y^2+xy-z^2-y^2-zy\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\left(x+y+z\right)\)
Phân tích các đa thức sau thành nhân tử: x 2 y + x y 2 + x 2 z + x z 2 + y 2 z + y z 2 + 3xyz.
x 2 y + x y 2 + x 2 z + x z 2 + y 2 z + y z 2 + 3xyz.
= ( x 2 y + x 2 z + xyz) + (x y 2 + y 2 z + xyz) + (x z 2 + y z 2 + xyz)
= x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)
= (x + y + z)(xy + xz + yz).
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xz^2+yz^2+xyz\right)+\left(xy^2+y^2z+xyz\right)\)
\(=x\left(xy+xz+yz\right)+z\left(xz+yz+xy\right)+y\left(xy+yz+xz\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
phân tích các đa thức sau thành nhân tử :
a , 49 * ( y - 4 ) ^2 - 9 *y^2 -36*y - 36
b, x*y*z - ( xy+yz+xz) + ( x+y+z) -1
\(a,49.\left(y-4\right)^2-9y^2-36y-36=49\left(y-4\right)^2-9\left(y^2+4y+4\right)\)
\(=49\left(y-4\right)^2-9\left(y+4\right)^2=\left(7y-28\right)^2-\left(3y+12\right)^2\)
\(=\left(7y-28+3y+12\right)\left(7y-28-3y-12\right)\)
\(=\left(10y-16\right)\left(4y-40\right)=8\left(5y-8\right)\left(y-10\right)\)
\(b,xyz-\left(xy+yz+xz\right)+\left(x+y+z\right)-1\)
\(=xyz-xy-yz-xz+x+y+z-1\)
\(=\left(xyz-xy\right)-\left(xz-x\right)-\left(yz-y\right)+\left(z-1\right)\)
\(=xy\left(z-1\right)-x\left(z-1\right)-y\left(z-1\right)+\left(z-1\right)\)
\(=\left(z-1\right)\left(xy-x-y+1\right)\)
\(=\left(z-1\right)\text{[}x\left(y-1\right)-\left(y-1\right)\text{]}\)
\(=\left(z-1\right)\left(y-1\right)\left(x-1\right)\)
Bài 1 : Phân tích các đa thức sau thành nhân tử :
1) 15x + 15y 2) 8x - 12y
3) xy - x 4) 4x^2- 6x
Bài 2 : Phân tích các đa thức sau thành nhân tử :
1) 2(x + y) - 5a(x + y) 2) a^2(x - 5) - 3(x - 5)
3) 4x(a - b) + 6xy(a - b) 4) 3x(x - 1) + 5(x -1)
Bài 3 : Tính giá trị của biểu thức :
1) A = 13.87 + 13.12 + 13
2) B = (x - 3).2x + (x - 3).y tại x = 13 và y = 4
Bài 4 : Tìm x :
1) x(x - 5) - 2(x - 5) = 0 2) 3x(x - 4) - x + 4 = 0
3) x(x - 7) - 2(7 - x) = 0 4) 2x(2x + 3) - 2x - 3 = 0
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
phân tích đa thức sau thành nhân tử x^2 y^2(y-x)+y^2 z^2(z-y)-z^2 x^2(z-x)
phân tích đa thức đa thức sau thành nhân tử x+y+z = x^3-y^3-z^3
phân tích đa thức sau thành nhân tử x^2 y^2 ( y-x) + y^2z^2 (z-y)- x^2 z^2 ( z-x)
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left[\left(z-y\right)+\left(y-x\right)\right]\)
\(=x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-x^2z^2\left(z-y\right)-x^2z^2\left(y-x\right)\)
\(=\left(y-x\right)\left(x^2y^2-x^2z^2\right)+\left(z-y\right)\left(y^2z^2-x^2z^2\right)\)
\(=x^2\left(y-x\right)\left(y-z\right)\left(y+z\right)+z^2\left(z-y\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left(-x^2y-x^2z+z^2y+z^2x\right)\)
\(=\left(y-x\right)\left(z-y\right)\left[xz\left(z-x\right)+y\left(z-x\right)\left(z+x\right)\right]\)
\(=\left(y-x\right)\left(z-y\right)\left(z-x\right)\left(xy+yz+xz\right)\)
Phân tích các đa thức sau thành nhân tử:
a) x 2 ( x - 3 ) 2 - ( x - 3 ) 2 - x 2 +1;
b) x 3 - 2 x 2 + 4x - 8;
c) ( x + y ) 3 - ( x - y ) 3 ;
d) 2 a 2 (x + y + z) - 4ab (x + y + z) + 2 b 2 (x + y + z).
a) (x - 1)(x + l)(x - 2)(x - 4). b) (x - 2)( x 2 + 4).
c) 2y(3 x 2 + y 2 ). d) 2(x + y + z) ( a - b ) 2 .
a. \(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left[\left(x-3\right)^2-1\right]\left(x^2-1\right)\)
\(=\left(x-3+1\right)\left(x-3-1\right)\left(x+1\right)\left(x-1\right)\)
\(=\left(x-2\right)\left(x-4\right)\left(x+1\right)\left(x-1\right)\)
b. \(x^3-2x^2+4x-8\)
\(=\left(x^3+4x\right)-\left(2x^2+8\right)\)
\(=x\left(x^2+4\right)-2\left(x^2+4\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
c. \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x^3-3x^2y+3xy^2-y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3\)
\(=2y\left(3x^2+y^2\right)\)
d. \(2a^2\left(x+y+z\right)-4ab\left(x+y+z\right)+2b^2\left(x+y+z\right)\)
\(=\left(2a^2-4ab+2b^2\right)\left(x+y+z\right)\)
\(=2\left(a^2-2ab+b^2\right)\left(x+y+z\right)\)
\(=2\left(a-b\right)^2\left(x+y+z\right)\)