chuyển về dạng nguyên thể rồi tính thể chất khối lượng sau đó quay về đang tìm mũ của nhiều số làm ra rồi thì dễ lắm bạn ạ k minh nha

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

b)\(\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)

c)\(-2\left(x-4\right)\left(2x+1\right)\)

d)\(\left(x-5\right)\left(4x+1\right)\)

e)\(3\left(x-2\right)\left(3x-2\right)\)

g)\(2\left(a-b\right)^2\)

h)\(\left(xy-3\right)\left(5y^2-2z\right)\)

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

l)\(abc^2\left(b-a\right)\left(b+c\right)\)

m)\(\left(x-y\right)\left(y-z\right)\left(x-z\right)\)

x² - x - y² - y

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

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

***

9x² + y² - 16z² + 6xy

= (3x + y)² - (4z)²

= (3x + y - 4z)(3x + y + 4z)

***

a³ - a²x - ay + xy

= a²(a - x) - y(a - x)

= (a - x)(a² - y)

***

2x² - 8y² + 3x + 6y

= 2(x² - 4y²) + 3(x + 2y)

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

= (x + 2y)(2x - 4y + 3)

***

xy(x + y) + yz(y + z) + xz(x + z) + 2xyz

= xy(x + y + z) + yz(x + y + z) + xz(x + z)

= y(x + y + z)(x + z) + xz(x + z)

= (x + z)(xy + y² + yz + xz)

= (x + z)[y(x + y) + z(x + y)]

= (x + z)(x + y)(y + z) 

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

\(=\left(2x+1\right)\left(3-2x+5\right)\)

\(=\left(2x+1\right)\left(8-2x\right)\)

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

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

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)\(a^4+a^3+a^3b+a^2b=\left(a^4+a^3b\right)+\left(a^3+a^2b\right)\)

\(=a^3\left(a+b\right)+a^2\left(a+b\right)\)

\(=\left(a^3+a^2\right)\left(a+b\right)\)

\(=a^2\left(a+1\right)\left(a+b\right)\)

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

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

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

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

c)\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)

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

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

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

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

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

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

a) \(x^3y^3+125=\left(xy\right)^3+5^3=\left(xy+5\right)\left(x^2y^2-5xy+25\right)\)

b) \(8x^3+y^3-6xy\left(2x+y\right)=\left(8x^3+y^3\right)-6xy\left(2x+y\right)=[\left(2x\right)^3+y^3]-6xy\left(2x+y\right)\)

\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-6xy\left(2x+y\right)=\left(2x+y\right)\left(4x^2-2xy+y^2-6xy\right)\)

\(=\left(2x+y\right)\left(4x^2-8xy+y^2\right)\)

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

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