Question

 lm giúp mik vs chiều nay kt r

Answer 1

Bài 4:

Ta có: \(\left(n+2\right)^2-\left(n-2\right)^2\)

\(=n^2+4n+4-n^2+4n-4\)

\(=8n⋮8\)

Answer 2

Bài 1: 

a: \(x^2-12x+36=\left(x-6\right)^2\)

b: \(4x^2+12x+9=\left(2x+3\right)^2\)

c: \(\dfrac{1}{4}x^2-5xy+25y^2=\left(\dfrac{1}{2}x-5y\right)^2\)

d: \(\left(x-5\right)^2-16=\left(x-5-4\right)\left(x-5+4\right)=\left(x-9\right)\left(x-1\right)\)

e: \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(x+2\right)\left(8-x\right)\)

g: \(\left(7x-4\right)^2-\left(2x+1\right)^2\)

\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)\)

\(=15\left(x-1\right)\left(3x-1\right)\)

f: \(8x^3+\dfrac{1}{27}=\left(2x+\dfrac{1}{3}\right)\left(4x^2-\dfrac{2}{3}x+\dfrac{1}{9}\right)\)

g: \(49\left(x-4\right)^2-9\left(x+2\right)^2\)

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

\(=\left(4x-34\right)\left(10x-24\right)\)

\(=4\left(2x-17\right)\left(5x-12\right)\)

Answer 3

Bài 2: 

a: Ta có: \(x^2-36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

b: Ta có: \(\left(x-4\right)^2-36=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(x+8\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=12\\x+8=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-20\end{matrix}\right.\)

d: Ta có: \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

\(\Leftrightarrow\left(2x-3\right)^2=0\)

\(\Leftrightarrow2x-3=0\)

hay \(x=\dfrac{3}{2}\)

Answer 4

Bài 3: 

a: \(75^2-25^2=5000\)

b: \(53^2-47^2=600\)

c: \(31.8^2-2\cdot31.8\cdot21.8+21.8^2=\left(31.8-21.8\right)^2=10^2=100\)

d: \(58.2^2+2\cdot58.2\cdot41.8+41.8^2=\left(58.2+41.8\right)^2=100^2=10000\)