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26 tháng 11 2023

a: \(70a+84b-20ab-24b^2\)

\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)

\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)

\(=\left(5a+6b\right)\left(14-4b\right)\)

\(=2\left(7-2b\right)\left(5a+6b\right)\)

b: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2\right)+\left(y^2z+yz^2\right)+3xyz\)

\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+3xyz\)

\(=x^2\left(y+z\right)+x\left(y^2+z^2\right)+yz\left(y+z\right)+2xyz+xyz\)

\(=x^2\left(y+z\right)+x\left(y^2+z^2+2yz\right)+yz\left(y+z+x\right)\)

\(=x^2\left(y+z\right)+x\left(y+z\right)^2+yz\left(y+z+x\right)\)

\(=\left(y+z\right)\cdot x\left(x+y+z\right)+yz\left(y+z+x\right)\)

\(=\left(y+z+x\right)\cdot\left(xy+xz+yz\right)\)

c: \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)

\(=\left(x^2y+x^2z\right)+\left(xy^2+xz^2+2xyz\right)+\left(y^2z+yz^2\right)\)

\(=x^2\left(y+z\right)+x\left(y^2+z^2+2xz\right)+yz\left(y+z\right)\)

\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)

\(=\left(y+z\right)\left(x^2+yz+xy+xz\right)\)

\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)

26 tháng 11 2023

a) \(70a+84b-20ab-24b^2\)

\(=\left(70a+84b\right)-\left(20ab+24b^2\right)\)

\(=14\left(5a+6b\right)-4b\left(5a+6b\right)\)

\(=\left(5a+6b\right)\left(14-4b\right)\)

\(=2\left(5a+6b\right)\left(7-2b\right)\)

b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xyz+xz^2\right)+\left(xyz+y^2z+yz^2\right)\)

\(=xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)

c) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)

\(=\left(x^2y+xy^2\right)+\left(xz^2+yz^2\right)+\left(x^2z+2xyz+y^2z\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x^2+2xy+y^2\right)\)

\(=xy\left(x+y\right)+z^2\left(x+y\right)+z\left(x+y\right)^2\)

\(=\left(x+y\right)\left[xy+z^2+z\left(x+y\right)\right]\)

\(=\left(x+y\right)\left(xy+z^2+xz+yz\right)\)

\(=\left(x+y\right)\left[\left(xy+yz\right)+\left(xz+z^2\right)\right]\)

\(=\left(x+y\right)\left[y\left(x+z\right)+z\left(x+z\right)\right]\)

\(=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

26 tháng 11 2023

a, 70a + 84b - 20ab - 24b2

 = 14.(5a + 6b) - 4b(5a + 6b)

= (5a + 6b).(14 - 4b) 

24 tháng 9 2023

\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)

\(=\left(x^2y+xy^2+xyz\right)+\left(y^2z+yz^2+xyz\right)+\left(x^2z+xz^2+xyz\right)\)

\(=xy\left(x+y+z\right)+yz\left(y+z+x\right)+xz\left(x+z+y\right)\)

\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)

27 tháng 9 2019

\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)

27 tháng 9 2019

\(49\left(y-4\right)^2-9y^2-36y-36\)

\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)

\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)

\(=\left(10y-22\right)\left(4y-34\right)\)

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

\(=\left(x^3+y^3\right)-\left(x^2y+xy^2\right)\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x+y\right)\left(x-y\right)^2\)

14 tháng 9 2023

3) \(x^2\left(x+2y\right)-x-2y\)

\(=x^2\left(x+2y\right)-\left(x+2y\right)\)

\(=\left(x^2-1\right)\left(x+2y\right)\)

\(=\left(x+1\right)\left(x-1\right)\left(x+2y\right)\)

4) \(x^3-4x^2-9x+36\)

\(=\left(x^3-4x^2\right)-\left(9x-36\right)\)

\(=x^2\cdot\left(x-4\right)-9\left(x-4\right)\)

\(=\left(x-4\right)\left(x^2-9\right)\)

\(=\left(x-4\right)\left(x+3\right)\left(x-3\right)\)

 

 

15 tháng 9 2023

\(x^2\left(x+2y\right)-x-2y\\ =x^2\left(x+2y\right)-\left(x+2y\right)\\ =\left(x^2-1\right)\left(x+2y\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+2y\right)\\ ---\\ x^3-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\ =\left(x^2-9\right)\left(x-4\right)\\ =\left(x-3\right)\left(x+3\right)\left(x-4\right)\)

22 tháng 10 2021

\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)

\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)

\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)

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 –...
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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

 

1
19 tháng 12 2023

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`

\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)

\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1\)

Ta có ĐT tương đương

\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

Thay \(x=9\) ; \(y=10\) ; \(z=11\) vào BT có :

\(\left(9-1\right)\left(10-1\right)\left(11-1\right)=720\)

Vậy .........

17 tháng 7 2018

C = xyz - xy - yz - xz + x + y +z- 1

= xy(z-1) - y(z-1) - x(z-1) + 1(z-1)

(xy-y-x+1)(z-1)