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

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

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

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

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

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

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

\(\Rightarrow2009.99=33x\)

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

\(\Rightarrow2009.3=x\)

\(\Rightarrow6027=x\)

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

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

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

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

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

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:

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

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

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

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

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

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

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

=> \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{97.100}=\frac{0.99x}{2009}\)

=> \(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{97.100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{1}{3}\left(1-\frac{1}{100}\right)=\frac{0,33x}{2009}\)

=> \(\frac{33}{100}=\frac{0,33x}{2009}\Rightarrow33.2009=100.0,33x\)

=> 33.2009 = 33x

=> x = 2009

Thanks bn nhìu nha, mình sẽ K cho bn ngay. Bn kb với mình nha.

\(3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\)

\(3M=\frac{4-1}{1.4}+\frac{7-4}{4.7}+...+\frac{100-97}{97.100}\)

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

\(3M=1-\frac{1}{100}\)

\(3M=\frac{99}{100}\)

\(M=\frac{33}{100}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\dfrac{3}{2}xA=\dfrac{3}{1x4}+\dfrac{3}{4x7}+\dfrac{3}{7x10}+...+\dfrac{3}{97x100}=\)

\(=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+\dfrac{10-7}{7x10}+...+\dfrac{100-97}{97x100}=\)

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

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\Rightarrow A=\dfrac{99}{100}x\dfrac{2}{3}=\dfrac{33}{50}\)

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

=> 3/1.4+3/4.7+.....+3/97.100 = 0,99x/2014

=> 1-1/4+1/4-1/7+....+1/97-1/100 = 0,99x/2014

=> 0,99x/2014 = 1-1/100 = 99/100

=> x = 99/100 : 0,99/2014 = 2014

Vậy x = 2014

Tk mk nha