đây là hằng đẳng thức số 7 nhá bn

(x - 2y) (4x² + 2xy + y²) = x³ - 8y³

\((x-2y)(4x^2+2xy+y^2 \\ =(x-2y)[(2x^2)+2x.y+y^2] \\ =x^3-(2y)^3 \\ =x^3-8y^3\)

\(\dfrac{6x25x42}{5x7x15}\)

=  \(\dfrac{2x3x5x5x6x7}{5x7x3x5}\)

= 12.

1. \(\dfrac{3x2x14}{4x7x19}\)

= \(\dfrac{3x2x7x2}{2x2x7x19}\)

= 3/19.

Lưu ý:

Dấu x kia là nhân nhé!

Mua 5m vải hết 80.000 đồng. Hỏi mua 7m vải loại đó hết bao nhiêu tiền?

\(A=\dfrac{\left(1-2x\right)\left(1+2x\right)}{2\left(1+2x\right)}:\dfrac{2\left(1-2x\right)}{3}\)

\(=\dfrac{1-2x}{2}\cdot\dfrac{3}{2\left(1-2x\right)}=\dfrac{3}{4}\)

ai giúp mik với ak mik cần gấp

\(\dfrac{1}{1.4}+\dfrac{1}{4.2}+\dfrac{1}{2.8}+\dfrac{1}{8.4}+\dfrac{1}{4.16}+\dfrac{1}{16.8}\)

= \(\dfrac{1}{4}.\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)\)

= \(\dfrac{1}{4}.\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}\right)\)

= \(\dfrac{1}{4}.\dfrac{3}{2}\left(1+\dfrac{1}{4}+\dfrac{1}{16}\right)\)

= \(\dfrac{1}{4}.\dfrac{3}{2}\left(\dfrac{16}{16}+\dfrac{4}{16}+\dfrac{1}{16}\right)\)

= \(\dfrac{3}{8}.\dfrac{21}{16}\)

= \(\dfrac{63}{128}\)

Bài 1: 

a: \(49-4x^2=\left(7-2x\right)\left(7+2x\right)\)

b: \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

c: \(x^2+18xy+81y^2=\left(x+9y\right)^2\)

mik ko cần bài 1 ạ

Bài 2: 

b: =>4x=20

hay x=5

a: \(x^2=2\)

=>\(x^2=\left(\sqrt{2}\right)^2\)

=>\(x=\pm\sqrt{2}\)

b: \(x^2=9\)

=>\(x^2=3^2\)

=>\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

c: \(\left(x-\sqrt{2}\right)^2=2\)

=>\(\left[{}\begin{matrix}x-\sqrt{2}=\sqrt{2}\\x-\sqrt{2}=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}\\x=0\end{matrix}\right.\)

d: \(4x^2-1=0\)

=>\(4x^2=1\)

=>\(x^2=\dfrac{1}{4}\)

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(x^3-4x^2+8x-8\)

\(=x^3-2x^2+4x-2x^2+4x-8\)

\(=\left(x^3-2x^2+4x\right)-\left(2x^2-4x+8\right)\)

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

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