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AH
Akai Haruma
Giáo viên
6 tháng 1

Lời giải:
a. $5x^2-10xy=5x(x-2y)$

b. $3x(x-y)-6(x-y)=(x-y)(3x-6)=3(x-y)(x-2)$
c. $2x(x-y)-4y(y-x)=2x(x-y)+4y(x-y)=(x-y)(2x+4y)=2(x-y)(x+2y)$

d. $9x^2-9y^2=9(x^2-y^2)=9(x-y)(x+y)$

e. $x^2-xy-x+y=(x^2-xy)-(x-y)=x(x-y)-(x-y)=(x-y)(x-1)$

f. $xy-xz-y+z=(xy-y)-(xz-z)=y(x-1)-z(x-1)=(x-1)(y-z)$

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 –...
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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`

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

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

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

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

a: \(81x^5-x^3\)

\(=x^3\left(81x^2-1\right)\)

\(=x^3\left(9x-1\right)\left(9x+1\right)\)

b: \(9x^2y-12xy+4y\)

\(=y\left(9x^2-12x+4\right)\)

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

c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)

\(=\left(x-5\right)^2-\left(4x-8\right)^2\)

\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)

\(=-3\left(x-1\right)\left(5x-13\right)\)

d: Ta có: \(9x^2-y^2-21x-7y\)

\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)

\(=\left(3x+y\right)\left(3x-y-7\right)\)

e: Ta có: \(-y^2+8y-16+9x^2\)

\(=-\left(y^2-8y+16-9x^2\right)\)

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

f: Ta có: \(5x^2-4x-1\)

\(=5x^2-5x+x-1\)

\(=5x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x-1\right)\left(5x+1\right)\)

\(a,x\left(x+6\right)\\ b,\left(9x-1\right)\left(9x+1\right)\\ c,\left(x+y\right)-3^2\\ =\left(x+y-3\right)\left(x+y+3\right)\\ d,\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x+y-1\right)\)

22 tháng 8 2021

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

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

\(=xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+z-1\)

\(=\left(xy-y-x+1\right)\left(z-1\right)\)

\(=[\left(x-1\right)y-\left(x-1\right)]\left(z-1\right)\)

\(=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)

27 tháng 8 2023

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

\(=\left(2x+y\right).3y\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3\)

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

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

c) \(9x^2-3x+2y-4y^2\)

\(=9x^2-4y^2-3x+2y\)

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

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

d) \(4x^2-4xy+2x-y+y^2\)

\(=4x^2-4xy+y^2+2x-y\)

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

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

e) \(x^3+3x^2+3x+1-y^3\)

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

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

g) \(x^3-2x^2y+xy^2-4x\)

\(=x\left(x^2-2xy+y^2\right)-4x\)

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

\(=x\left[\left(x-y\right)^2-4\right]\)

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

27 tháng 8 2023

a) (x + 2y)² - (x - y)²

= (x + 2y - x + y)(x + 2y + x - y)

= 3y(2x + y)

b) (x + 1)³ + (x - 1)³

= (x + 1 + x - 1)[(x + 1)² - (x + 1)(x - 1) + (x - 1)²]

= 2x(x² + 2x + 1 - x² + 1 + x² - 2x + 1)

= 2x(x² + 3)

c) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

= (3x - 2y)(3x + 2y) - (3x - 2y)

= (3x - 2y)(3x + 2y - 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) x³ + 3x² + 3x + 1 - y³

= (x³ + 3x² + 3x + 1) - y³

= (x + 1)³ - y³

= (x + 1 - y)[(x + 1)² + (x + 1)y + y²]

= (x - y + 1)(x² + 2x + 1 + xy + y + y²)

g) x³ - 2x²y + xy² - 4x

= x(x² - 2xy + y² - 4)

= x[(x² - 2xy + y²) - 4]

= x[(x - y)² - 2²]

= x(x - y - 2)(x - y + 2)

14 tháng 7 2021

a) \(=x^2+7x-12x-84-2x+14\)

\(=x^2-7x-70\)

b)\(=x^2-4x-2x+8\)

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

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

c) \(=9x\left(x+y\right)-\left(x+y\right)\)

\(=\left(9x-1\right)\left(x+y\right)\)

d)\(=\left(x-y\right)^2-9^2\)

\(=\left(x-y+9\right)\left(x-y-9\right)\)

e)\(=x^2+8x+16-60+15x\)

\(=x^2+23x-44\)

b) Ta có: \(x^3-x^2y-xy^2+y^3\)

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

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

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

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

22 tháng 10 2021

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

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

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