Chương III : Phân số

đề là j vậy e?

(1-2/2.3).(1-2/3.4).....(1-2/n.(n-1))=2021/6062

= ( \(\dfrac{6}{2.3}\) -\(\dfrac{2}{2.3}\) ). (\(\dfrac{12}{3.4}\) - \(\dfrac{2}{3.4}\) )......( \(\dfrac{9900}{99.100}\) - \(\dfrac{2}{99.100}\) )

= 4/2.3 .10/3.4..... 9898/99.100

= 1.4/2.3 . 2.5/3.4 ....  98.101/99.100

=\(\dfrac{1.2.3.4...98}{2.3...99}\) . \(\dfrac{4.5.6...101}{3.4.5...100}\) 

= 1/99.101/3

= 101/297

(1-2/2.3).(1-2/3.4).....(1-2/n.(n-1))=2021/6062

= ( 22.322.3 ). (23.423.4 )......( 299.100299.100 )

= 4/2.3 .10/3.4..... 9898/99.100

= 1.4/2.3 . 2.5/3.4 ....  98.101/99.100

=4.5.6...1013.4.5...1004.5.6...1013.4.5...100 

= 1/99.101/3

= 101/297

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)

\(3A=\dfrac{24}{49}\)

\(A=\dfrac{24}{49}:3\)

\(A=\dfrac{8}{49}\)

`-1,5 \times (7/3-5/3 \times 4)`

`= -1,5 \times (7/3-20/3)`

`= -1,5 \times (-13/3)`

`= 13/2`

`@` `\text {Ans}`

`\downarrow`

`j)`

`-3/4 + 2/7 + (-1)/4 + 3/5 + 5/7`

`= (-3/4 - 1/4) + (2/7 + 5/7) + 3/5`

`= -1 + 1 + 3/5`

`= 3/5`

`k)`

`-2/17 + 15/23 + (-15)/17 + 4/19 + 8/23`

`= (-2/17 - 15/17) + (15/23 + 8/23) + 4/19`

`= -1 + 1 + 4/19`

`= 4/19`

$#KDN040510$

j: =-3/4-1/4+2/7+5/7+3/5

=-1+1+3/5

=3/5

k: =-2/17-15/17+15/23+8/23+4/19

=-1+1+4/19

=4/19

\(j)\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{-1}{4}+\dfrac{3}{5}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{-1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)+\dfrac{3}{5}\\ =\dfrac{-4}{4}+\dfrac{7}{7}+\dfrac{3}{5}\\ =-1+1+\dfrac{3}{5}\\ =0+\dfrac{3}{5}\\ =\dfrac{3}{5}\\ k)\dfrac{-2}{17}+\dfrac{15}{23}+\dfrac{-15}{17}+\dfrac{4}{19}+\dfrac{8}{23}\\ =\left(\dfrac{-2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{15}{23}+\dfrac{8}{23}\right)+\dfrac{4}{19}\\ =\dfrac{-17}{17}+\dfrac{23}{23}+\dfrac{4}{19}\\ =-1+1+\dfrac{4}{9}\\ =0+\dfrac{4}{19}\\ =\dfrac{4}{19}\)

ĐKXĐ: x<>-2

\(B=\dfrac{3x-5}{x+2}=\dfrac{3x+6-11}{x+2}\)

\(=3-\dfrac{11}{x+2}\)

Để B là số nguyên âm lớn nhất thì \(\dfrac{11}{x+2}\) min

=>x+2=1

=>x=-1

ừm .....

bài này mình cũng ko làm đc

mà bạn hỏi giờ này ko ai trả lời đâu

bạn ngủ đi mai hỏi tiếp 

giờ này mn ngủ hết rồi 

Lời giải:
Gọi $d=ƯCLN(3n+2, 7n-1)$
$\Rightarrow 3n+2\vdots d; 7n-1\vdots d$

$\Rightarrow 7(3n+2)-3(7n-1)\vdots d$

$\Rightarrow 17\vdots d\Rightarrow d\in\left\{1; 17\right\}$

Để phân số rút gọn được thì $d=17$

Điều này xảy ra khi $3n+2\vdots 17$

$\Rightarrow 3n+2-17\vdots 17$

$\Rightarrow 3n-15\vdots 17$

$\Rightarrow 3(n-5)\vdots 17$

$\Rightarrow n-5\vdots 17$

$\Rightarrow n=17k+5$ với $k$ tự nhiên bất kỳ.

Vì $100\leq n\leq 150\Rightarrow 100\leq 17k+5\leq 150$

$\Rightarrow 5,6\leq k\leq 8,5$

Vì $k$ tự nhiên nên $k\in\left\{6;7;8\right\}$

$\Rightarrow n\in \left\{107; 124; 141\right\}$

Không có đề bạn nhé, hình như bạn quên đăng ảnh

\(-\dfrac{5}{6}-x=\dfrac{7}{12}+\left(-\dfrac{1}{3}\right)\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}-\dfrac{1}{3}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{7}{12}-\dfrac{4}{12}\)

\(\Leftrightarrow-\dfrac{5}{6}-x=\dfrac{3}{12}\)

\(\Leftrightarrow-x=\dfrac{3}{12}+\dfrac{5}{6}\)

\(\Leftrightarrow-x=\dfrac{3}{12}+\dfrac{10}{12}\)

\(\Leftrightarrow-x=\dfrac{13}{12}\)

\(\Leftrightarrow x=-\dfrac{13}{12}\)