A = \(\dfrac{7}{2.3}+\dfrac{7}{2.3.4}+\dfrac{7}{3.4.5}+...+\dfrac{7}{48.49.50}\)

A = \(7.\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)

\(\dfrac{A}{7}\) = \(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{48.49.50}\right)\)

Bạn để ý kĩ phần này:

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3}\)

\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{2}{2.3.4}\)

\(\dfrac{1}{3.4}-\dfrac{1}{4.5}=\dfrac{2}{3.4.5}\)

...

\(\dfrac{1}{48.49}-\dfrac{1}{49.50}=\dfrac{2}{48.49.50}\)

Qua đó, bạn có thể rút ra công thức tổng quát sau:

\(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}=\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)

Từ đó suy ra:

\(\dfrac{2A}{7}\) = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{48.49.50}\)

\(\dfrac{2A}{7}=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{48.49}-\dfrac{1}{49.50}\)

\(\dfrac{2A}{7}=\dfrac{1}{2}-\dfrac{1}{49.50}\)

\(\dfrac{2A}{7}=\dfrac{612}{1225}\)

\(2450A=4284\)

A = \(\dfrac{306}{175}\)

P = \(\left(m+1\right)\left(m+3\right)\left(m+5\right)\left(m+7\right)+15\)

P = \(\left(m^2+8m+7\right)\left(m^2+8m+15\right)+15\) (*)

Đặt \(m^2+8m+7=a\)

(*) \(\Leftrightarrow a.\left(a+8\right)+15\)

= \(a^2+8a+15\)

= \(\left(a+3\right)\left(a+5\right)\)

= \(\left(m^2+8m+7+3\right)\left(m^2+8m+7+5\right)\)

= \(\left(m^2+8m+10\right)\left(m^2+8m+12\right)\)

= \(\left(m^2+8m+10\right)\left(m+2\right)\left(m+6\right)⋮\left(m+6\right)\) ( đpcm )

Câu 1: Giải

Đặt CTHH: \(X_2\left(SO_4\right)_3\)

Theo đề : \(2X+3\left(SO_4\right)=342\)

\(2X+3.96=342\)

\(2X=54\Rightarrow X=27\)

Vậy X là Nhôm ( Al ).

Câu 2: Giải

Đặt CTHH: \(X_2CO_3\)

Theo đề: \(\dfrac{X_2CO_3}{H_2}=53\)

\(\dfrac{X_2CO_3}{2}=53\Rightarrow X_2CO_3=106\)

\(2X+40+16.3=106\)

\(2X=18\)

\(X=9\)

Vậy X là Beri ( Be )

a) \(x^2-x-30=0\)

\(\left(x-6\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

b) \(-2x^2-4x+2=0\)

\(\left(x+1-\sqrt{2}\right)\left(x+1+\sqrt{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{2}\\x=\sqrt{-1-\sqrt{2}}\end{matrix}\right.\)

b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)

\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)

\(\left|4-7x\right|=\dfrac{49}{30}\) (*)

+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)

PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)

\(-7x=-\dfrac{71}{30}\)

x = \(\dfrac{71}{210}\) (t/m)

+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)

Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)

x = \(\dfrac{169}{210}\) (t/m)

Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)

cái đó bằng 0 luôn còn phân tích gì chứ -_-

Tóm tắt: Khối trụ tròn

m = 5kg

p = 1250 Pa

d = ?

Giải:

Vì khối trụ đc đặt trên mặt bàn nên:

P = F = 10m = 10.5 = 50 N

Diện tích mặt bàn bị ép là:

S = \(\dfrac{p}{F}=\dfrac{1250}{50}=25\) ( m² )

Bán kính của đáy khối trụ là:

r = \(\sqrt{\dfrac{S}{3.14}}\) = \(\sqrt{\dfrac{25}{3,14}}\approx2,82\) ( m )

Đường kính của đáy khối trụ là:

d = 2r = 2.2,82 = 5,64 ( m )

Đs: ...

\(S_n=2-4+6-8+...+\left(-1\right)^{n-1}.n\)

\(S_n=2-4+6-8+...+\left(-1\right)^n:\left(-1\right).n\)

\(S_n=2-4+6-8+...+n\)

\(\Rightarrow S_{35}+S_{50}+S_{100}\)

= \(\left(2-4+6-8+...+35\right)+\left(2-4+6-8+...+50\right)+\left(2-4+6-8+...+100\right)\)

= \(17.\left(-2\right)+35+25.\left(-2\right)+50.\left(-2\right)\)

= -149

Bóc tem :v

a) \(\dfrac{63^2-47^2}{215^2-105^2}\)

= \(\dfrac{\left(63-47\right)\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

= \(\dfrac{16.110}{110.320}=\dfrac{16}{320}=\dfrac{1}{20}\)

b) \(\dfrac{437^2-363^2}{537^2-463^2}\)

= \(\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)

= \(\dfrac{74.800}{74.1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)

2)

A = \(26^2-24^2=\left(26-24\right)\left(26+24\right)=2.50=100\)

B = \(27^2-25^2=\left(27-25\right)\left(27+25\right)=2.52=104\)

Từ đó suy ra A < B