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M = (a + b + c)3 - a3 - b3 - c3
= (a + b)3 + c3 + 3(a + b)2c + 3(a + b)c2 - a3 - b3 - c3
= a3 + b3 + c3 + 3a2b + 3ab2 + 3(a + b)c(a + b + c) - a3 - b3 - c3
= 3ab (a + b) + 3c(a + b)(a + b + c)
= 3(a + b)[ab + c(a + b + c)]
= 3(a + b)(ab + bc + ac + c2)
= 3(a + b)[b(a + c) + c(a + c)]
= 3(a + b)(b + c)(c + a)
N = a3 + b3 + c3 - 3abc
= (a + b)3 + c3 - 3ab(a + b) - 3abc
= (a + b + c)3 - 3(a + b)c(a + b + c) - 3ab(a + b + c)
= (a + b + c)[(a + b + c)2 - 3(a + b)c - 3ab]
= (a + b + c)(a2 + b2 + c2 + 2ab + 2bc + 2ca - 2ac - 3bc - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
\(x^2\left(x-5\right)+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x^2+1\right)\)
x(x + 2) + x(x - 5) - 5(x + 2)
= [x(x + 2) - 5(x + 2)] + x(x - 5)
= (x - 5)(x + 2) + x(x - 5)
= (x - 5)(x + 2 + x)
= (x - 5)(2x + 2)
= 2(x - 5)(x + 1)
Biểu thức này không phân tích được thành nhân tử nữa bạn nhé.
Phân tích đa thức sau thành nhân tử:( x-5 )\(^2\)- 5 + x
Bài làm:
( x-5 )\(^2\)- 5 + x
= ( x - 5 )\(^2\)+ ( x - 5 ) Chỗ này có thể phân tích thành ( x - 5 ) ( x - 5 ) + ( x - 5 ) nha
= ( x - 5 ) \([\)( x - 5 ) + 1 \(]\)
= ( x- 5 ) ( x - 5 + 1 )
* Chúc bạn học tốt ^^
# Linh
\(=x^3\left(x^2+1\right)-\left(x^2+1\right)=\left(x^3-1\right)\left(x^2+1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+1\right)\)
không thể
M = x^2 - 5 = x^2 - (^5)^2= (x-^5)(x+^5)