$\dfrac14+\dfrac1{28}+\dfrac1{70}+\cdots+\dfrac1{9700}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\cdots+\dfrac1{97\cdot100}$

$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac{50+25-20-1}{100}\right)$

$=\dfrac13\cdot\dfrac{54}{100}$

$=\dfrac{9}{50}$

Theo đề bài:

$\dfrac{9}{50}=\dfrac{0,33x}{2009}$

$\Leftrightarrow\dfrac{9}{50}=\dfrac{33x}{100\cdot2009}$

$\Leftrightarrow 9\cdot100\cdot2009=50\cdot33x$

$\Leftrightarrow x=\dfrac{9\cdot100\cdot2009}{50\cdot33}$

$\Leftrightarrow x=\dfrac{6\cdot2009}{11}$

$\Leftrightarrowx=\dfrac{12054}{11}$.

\(A=\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)

\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\)

\(A=\frac{3}{3}\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{97.100}\right)\)

\(A=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{97.100}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(A=\frac{1}{3}\left(1-\frac{1}{100}\right)\)

\(A=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+...+\frac{1}{9700}\)

\(=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}\)

\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{100}{100}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+...+\frac{1}{9700}=\frac{0,33x}{2009}\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{97.100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{1}{1}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{100}{100}-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{2009}\)

\(\Rightarrow2009.99=100.0,33x\)

\(\Rightarrow2009.99=33x\)

\(\Rightarrow2009.99:33=x\)

\(\Rightarrow2009.3=x\)

\(\Rightarrow6027=x\)

Vậy \(x=6027\)(MK KO CHẮC NÓ ĐÚNG NHÉ )

A = 1/4 + 1/28 + 1/70 +...+ 1/9700

A = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/97.100

3A = 3/1.4 + 3/4.7 + 3/7.10 +...+ 3/97.100

3A = 1 - 1/100

3A = 99/100

A=99/100:3=33/100

\(=\frac{1}{1.4}+\frac{1}{4.7}+..+\frac{1}{97.100}\)

\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{1}{3}\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(=\frac{1}{3}.\frac{99}{100}=\frac{33}{100}\)

1.

Gọi chiều rộng của căn phòng là $x$ (m), $(x>0)$.

Khi đó chiều dài là: $x+2$ (m).

Theo đề bài:

$2(x+x+2)=36$

$\Leftrightarrow 4x+4=36$

$\Leftrightarrow 4x=32$

$\Leftrightarrow x=8$.

Vậy chiều rộng là $8$ m, chiều dài là $10$ m.

a)

Diện tích căn phòng là:

$S=8\cdot10=80\text{ m}^2$.

Vậy: $S=80\text{ m}^2$.

b)

Cạnh viên gạch là:

$40\text{ cm}=0,4\text{ m}$.

Diện tích một viên gạch là:

$0,4\cdot0,4=0,16\text{ m}^2$.

Số viên gạch cần dùng là:

$\dfrac{80}{0,16}=500$ (viên).

Vậy: 500 viên gạch.

2.

$A=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots+\dfrac1{9700}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{97\cdot100}$

$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\dfrac13\left(\dfrac1{10}-\dfrac1{13}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\dfrac1{10}-\dfrac1{13}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac{50+25-20-1}{100}\right)$

$=\dfrac13\cdot\dfrac{54}{100}$

$=\dfrac{9}{50}$.

Vậy: $A=\dfrac{9}{50}$.

\(\frac{3}{1.4}+\frac{3}{4.7}+..+\frac{3}{97.100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{4}+\frac{1}{4}-...-\frac{1}{100}=\frac{0,33x}{2009}\)

\(1-\frac{1}{100}=\frac{0,33x}{2009}\)

\(\frac{99}{100}=\frac{0,33x}{20009}\Rightarrow2009.99=100.0,33x\)

x=6027

$\dfrac14+\dfrac1{28}+\dfrac1{70}+\cdots+\dfrac1{9700}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\cdots+\dfrac1{97\cdot100}$

$=\dfrac13\left(\dfrac12-\dfrac15\right)+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12-\dfrac15+\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\cdots+\dfrac1{97}-\dfrac1{100}\right)$

$=\dfrac13\left(\dfrac12+\dfrac14-\dfrac15-\dfrac1{100}\right)$

$=\dfrac13\cdot\dfrac{54}{100}$

$=\dfrac9{50}$.

Theo đề bài:

$\dfrac9{50}=\dfrac{0,33x}{2009}$

$\Leftrightarrow\dfrac9{50}=\dfrac{33x}{100\cdot2009}$

$\Leftrightarrow 9\cdot100\cdot2009=50\cdot33x$

$\Leftrightarrow x=\dfrac{9\cdot100\cdot2009}{50\cdot33}$

$\Leftrightarrow x=\dfrac{6\cdot2009}{11}$

$\Leftrightarrowx=\dfrac{12054}{11}$.

M=1/4+1/4.7+1/7.10+1/10.13+.............+1/88.91

3M=3/4+3/4.7+3/7.10+3/10.13+........+3/88.91

3M= 3/4+1/4-1/7+1/7-1/10+1/10-1/13+......+1/88-1/91

3M=3/4+1/4-1/91=1-1/91=90/91 

----->M= 30/91

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots+\dfrac1{8008}$

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\dfrac13\left(\dfrac1{10}-\dfrac1{13}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17+\dfrac17-\dfrac1{10}+\dfrac1{10}-\dfrac1{13}+\cdots+\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.

Vào câu hỏi tương tự nha bạn !

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots$ (có $30$ số hạng)

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

Với mọi số hạng từ số hạng thứ hai:

$\dfrac1{a(a+3)}=\dfrac13\left(\dfrac1a-\dfrac1{a+3}\right)$

Do đó:

$M=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.

$M=\dfrac14+\dfrac1{28}+\dfrac1{70}+\dfrac1{130}+\cdots$ (có $30$ số hạng)

$=\dfrac1{2\cdot2}+\dfrac1{4\cdot7}+\dfrac1{7\cdot10}+\dfrac1{10\cdot13}+\cdots+\dfrac1{88\cdot91}$

Với mọi số hạng từ số hạng thứ hai:

$\dfrac1{a(a+3)}=\dfrac13\left(\dfrac1a-\dfrac1{a+3}\right)$

Do đó:

$M=\dfrac14+\dfrac13\left(\dfrac14-\dfrac17\right)+\dfrac13\left(\dfrac17-\dfrac1{10}\right)+\cdots+\dfrac13\left(\dfrac1{88}-\dfrac1{91}\right)$

$=\dfrac14+\dfrac13\left(\dfrac14-\dfrac1{91}\right)$

$=\dfrac14+\dfrac{29}{364}$

$=\dfrac{91+29}{364}$

$=\dfrac{120}{364}$

$=\dfrac{30}{91}$.