\(\dfrac{2}{3}\)x\(\dfrac{7}{4}\):\(\dfrac{a}{12}\)=\(\dfrac{2}{3}\)

=>\(\dfrac{7}{6}\):\(\dfrac{a}{12}\)=\(\dfrac{2}{3}\)

=>\(\dfrac{a}{12}\)=\(\dfrac{7}{6}\):\(\dfrac{2}{3}\)=\(\dfrac{21}{12}\)

=>a=21.

cô Nguyễn Thị Thương Hoài ơi, chỉ e bài này đi

\(\dfrac{2}{3}x\dfrac{7}{4}:\dfrac{a}{12}=\dfrac{2}{3}\)\(=>\dfrac{7}{4}:\dfrac{a}{12}=\dfrac{2}{3}\)\(=>\dfrac{a}{12}=\dfrac{7}{6}:\dfrac{2}{3}=\dfrac{21}{12}\)

A= 66/67

B= 11.a/67

Các bạn tìm a giúp mình với

Quang Anh ơi cho mình hỏi đề có sai chỗ nào không vậy ?

Không sai đâu

xin lỗi em mới học lớp 4

\(\frac{3}{2\times5}+\frac{2}{5\times8}+\frac{3}{8\times11}+...+\frac{3}{602\times605}\)

\(=\frac{5-2}{2\times5}+\frac{8-5}{5\times8}+\frac{11-8}{8\times11}+...+\frac{605-602}{602\times605}\)

\(=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{602}-\frac{1}{605}\)

\(=\frac{1}{2}-\frac{1}{605}=\frac{603}{1210}\)

\(\frac{4}{3\times7}+\frac{5}{7\times12}+\frac{1}{12\times13}+\frac{2}{13\times15}\)

\(=\frac{7-4}{3\times7}+\frac{12-7}{7\times12}+\frac{13-12}{12\times13}+\frac{15-13}{13\times15}\)

\(=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\)

\(=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)

Lời giải:

Gọi tổng trên là $A$
$A=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{25.27}\right)$

$=2\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{27-25}{25.27}\right)$

$=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{25}-\frac{1}{27}\right)$

$=2\left(1-\frac{1}{27})=\frac{52}{27}$

a = 1/5-1/8 +1/8-1/11+....+1/602-1/605 = 1/5-(1/8-1/8) -(1/11-1/11)-......-(1/602-1/602)-1/605

1/5-1/605=120/605=24/121 

câu b bạn làm tương tự nhé 

`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ

`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`

`=1/1-1/101`

`=101/101-1/101`

`=100/101`

(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)

Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\times\left(1-\dfrac{1}{101}\right)\)

\(=2\times\dfrac{100}{101}\)

\(=\dfrac{200}{101}\)

Sửa bài ( dòng 3 đến hết bài )

... = \(2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)