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a: Sửa đề: x^3-x^2+5x-5
=x^2(x-1)+5(x-1)
=(x-1)(x^2+5)
b: x^3+4x^2+x-6
=x^3-x^2+5x^2-5x+6x-6
=(x-1)(x^2+5x+6)
=(x-1)(x+2)(x+3)
c: \(=\left(x+2\right)^3+y^3\)
\(=\left(x+2+y\right)\left(x^2+4x+4-xy-2y+y^2\right)\)
\(1.x^3+2x+x^2=x\left(x^2+x+2\right)\)
\(2.2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
\(3.-3x^3-5x^2+8x=-3x^3+3x^2-8x^2+8x\)
\(=-3x^2\left(x-1\right)-8x\left(x-1\right)=\left(3x^2+8x\right)\left(1-x\right)\)
\(=x\left(3x+8\right)\left(1-x\right)\)
\(4.x^2+4x-5=x^2-x+5x-5=\left(x-1\right)\left(x+5\right)\)
\(5.6x^2-3x-3=6x^2-6x+3x-3=3\left(x-1\right)\left(2x+1\right)\)
\(6.3x^2-2x-5=3x^2+3x-5x-5=\left(x+1\right)\left(3x-5\right)\)
\(8.x^2-2x-4y^2-4y=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)\(=\left(x+2y\right)\left(x-y-2\right)\)
\(9.x^3+2x^2y+xy^2-9x=x\left(x^2+2xy+y^2-9\right)\)
\(=x\left(x+y-3\right)\left(x+y+3\right)\)
\(10.x^2-y^2+6x+9=\left(x+3-y\right)\left(x+3+y\right)\)
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
a, \(x^3-x+3x^2y+3xy^2+y^3-y\)
= \((x^3+3x^2y+3xy^2+y^3)-x-y\)
= \(\left(x+y\right)^3-\left(x+y\right)\)
= \(\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a, = (x + y)5 - (x5 + y5)
= (x + y)5 - (x + y)(x4 - x3y + x2y2 - xy3 + y4)
= (x + y) [(x + y)4 - x4 + x3y - x2y2 + xy3 - y4]
= (x + y) (5x3y + 5x2y2 + 5xy3)
= 5xy(x + y)(x2 + xy + y2)
b, = x(x2 - 5xy - 14y2)
= x(x2 - 7xy + 2xy - 14y2)
= x(x + 2y)(x - 7y)
chắc bn nảy hỏi lun cả bài tâp về nhà quá, làm km 1 câu
a) = a+a+a + a +a +1 -a -a -a = a(a+a+1) +(a+a+1) - a(a+a+1)= (a+a+1)(a-a+1)
tự bn thêm mũ 4;3;2 vào được là bn làm dc cac câu sau
Phân tích đa thức thành nhân tử:(em làm luôn đấy,ko ghi lại đề)
\(\left(x^3+y^3\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)+3xy\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)\(=\left(x+y\right)\left[\left(x+y\right)^2-1^2\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3=0\)
\(\Leftrightarrow\left(2x+1\right)^3=0\)
\(\Leftrightarrow2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(2x^2+5x-3=0\Leftrightarrow\left(2x^2+6x\right)+\left(-x-3\right)=0\)
\(\Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)
\(x^2-2x-3=0\Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}.}\)
\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
\(=5x-1+2\left(4+5x-20x-25x^2\right)+25x^2+40x+16\)
\(=25x^2+45x+15+8+10x-40x-50x^2\)
\(=-25x^2+15x+23\)
\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(=\left(x-y\right)^3-\left(x-y\right)^3+\left(x+y\right)^3-3x^2y-3xy^2\)
\(=\left(x+y\right)^3-3x^2y-3xy^2\)
\(=x^3+3x^2y+3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)