\(\dfrac{2012:0,125:0,25:0,5}{1006\times8\times8\times8}\)

=  \(\dfrac{1006\times2\times8\times4\times2}{1006\times8\times4\times2\times4\times2}\)

= \(\dfrac{1}{4}\)

umm..đề bài đâu bn?

Đề bài pay mất r bn<3

\(f,=\left(5^2+3\right):7=28:7=4\\ g,=7^2-9+8\cdot25=49-9+200=240\\ h,=600+72+18=690\\ i,=5^2+5-20=10\\ j,=45-28+83=100\)

\(2A=\frac{4}{1.5}+\frac{6}{5.11}+\frac{8}{11.19}+\frac{10}{19.29}+\frac{12}{29.41}\)

\(=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{19}+...+\frac{1}{29}-\frac{1}{41}=1-\frac{1}{41}=\frac{40}{41}\)

\(\Rightarrow A=\frac{20}{21}\)

\(3B=\frac{3}{1.4}+\frac{6}{4.10}+\frac{9}{10.19}+\frac{12}{19.31}=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{10}+\frac{1}{10}-\frac{1}{19}+\frac{1}{19}-\frac{1}{31}\)

\(=1-\frac{1}{31}=\frac{30}{31}\)

\(\Rightarrow B=\frac{10}{31}=\frac{20}{62}<\frac{20}{41}\)

Do đó $A>B$

Ta có: \(A=\dfrac{2}{1.5}+\dfrac{3}{5.11}+\dfrac{4}{11.19}+\dfrac{5}{19.29}+\dfrac{6}{29.41}\)

\(2A=1-\dfrac{1}{5}+\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{41}\)

\(2A=1-\dfrac{1}{41}=\dfrac{40}{41}\)

\(A=\dfrac{20}{41}\)

Lại có: \(B=\dfrac{1}{1.4}+\dfrac{2}{4.10}+\dfrac{3}{10.19}+\dfrac{4}{19.31}\)

\(3B=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}\)

\(3B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{31}\)

\(3B=1-\dfrac{1}{31}=\dfrac{30}{31}\)

\(B=\dfrac{10}{31}\)

Vì \(\dfrac{20}{41}>\dfrac{10}{31}\) nên...

đk : \(n\ne-\dfrac{1}{3}\)

gọi d là ƯCLN(18n+3,21n+7)

ta có 18n+3chia hết cho d

          21n+7 chia hết cho d

⇔21n+7-18n-3 chia hết cho d

⇔126n+42-126n-21 chia hết cho d 

21 chia hết cho d

⇒d∈Ư(21)=1;3;7;21

n ≠ 3k-1;3k-3;3k-7;3k-21

Bài 7:

Gọi số bị chia và số chia lần lượt là $a$ và $b$. Theo bài ra ta có:

$a+b=275$

$a=8b+5$

$\Rightarrow 8b+5+b=275$

$9b+5=275$

$9b=270$

$b=30$

Vậy số chia là $30$

Bài 8:

a. $(5-x)y=6=1.6=6.1=2.3=3.2$. Do đó có các trường hợp sau xảy ra:

TH1: $5-x=1; y=6$

$\Rightarrow x=4; y=6$

TH2: $5-x=6; y=1$

$\Rightarrow x=-1$ (loại vì $x$ là số tự nhiên)

TH3: $5-x=2; y=3$

$\Rightarrow x=3; y=3$

TH4: $5-x=3; y=2$

$\Rightarrow x=2; y=2$

b.

$1+3+5+....+x=2500$

$[\frac{x-1}{2}+1].\frac{x+1}{2}=2500$

$\frac{(x+1)(x+1)}{4}=2500$

$(x+1)(x+1)=10000=100.100$

$\Rightarrow x+1=100$

$\Rightarrow x=99$

\(8,\\ a,\Rightarrow\left(5-x\right)y=6=2\cdot3=\left(-2\right)\left(-3\right)=1\cdot6=\left(-1\right)\left(-6\right)\)

Vậy \(\left(x;y\right)=\left\{\left(3;3\right);\left(2;2\right);\left(7;-3\right);\left(8;-2\right);\left(4;6\right);\left(-1;1\right);\left(6;-6\right);\left(11;-1\right)\right\}\)