d: ta có: \(\left(x-3\right)^2-x+3\)

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

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

a: \(A=\dfrac{3n^2+3n}{12n}=\dfrac{3n\left(n+1\right)}{3n\cdot4}=\dfrac{n+1}{4}\)

Vì 4=2^2 ko có thừa số nguyên tố nào khác 2 và 5

nên \(A=\dfrac{n+1}{4}\) viết được dưới dạng số thập phân hữu hạn

c: \(37-2\cdot1^{11}+3\cdot5\cdot5^2\)

\(=37-2+15\cdot5^2\)

=35+375

=410

\(c,=37-2.1+3.5^3=37-2+3.125=35-375=-340\\ d,=2^5+2\left[12+2.\left(3.3^2+1\right)\right]=2^5+2\left(12+2.10\right)=2^5+2.32=32+64=96\)tik mik nha

a: \(230+75-125=180\)

b: \(2\cdot3\cdot4\cdot5:15=8\)

Bài 5: 

Ta có: \(\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}=10\)

\(\Leftrightarrow\dfrac{158-x}{31}+\dfrac{185-x}{29}+\dfrac{208-x}{27}+\dfrac{227-x}{25}-10=0\)

\(\Leftrightarrow\dfrac{158-x}{31}-1+\dfrac{185-x}{29}-2+\dfrac{208-x}{27}-3+\dfrac{227-x}{25}-4=0\)

\(\Leftrightarrow\dfrac{127-x}{31}+\dfrac{127-x}{29}+\dfrac{127-x}{27}+\dfrac{127-x}{25}=0\)

\(\Leftrightarrow\left(127-x\right)\left(\dfrac{1}{31}+\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}\right)=0\)

mà \(\dfrac{1}{31}+\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}>0\)

nên 127-x=0

hay x=127

Vậy: S={127}