@Thùy Nguyễn Thị Bích

xin lỗi mik mới lớp 6

-4;-3;-2;-1 nha bạn.

\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5\cdot1\frac{31}{45}\right):\left(-21\frac{1}{2}\right)\)

\(\Leftrightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}\cdot\frac{76}{45}\right):\left(-\frac{43}{2}\right)\)

\(\Leftrightarrow\frac{41}{9}\cdot\frac{18}{41}-7< x< \frac{43}{5}:\left(-\frac{43}{2}\right)\)

\(\Leftrightarrow2-7< x< -\frac{2}{5}\)

\(\Leftrightarrow-5< x< -0,4\)

\(\Leftrightarrow x\in\left\{-4;-3;-2;-1\right\}\)

tao ko biết tao mới lên lớp 5 thôi mày 

Mình ko biết thông cảm nha .Năm nay mình mới lên lớp 5 thui à

            THẬT LÒNG XIN LỖI VÌ KO GIÚP ĐƯỢC GÌ

S=311​+321​+331​+...+601​

=(131+132+...+140)+(141+...+150)+(151+...+160)=(311​+321​+...+401​)+(411​+...+501​)+(511​+...+601​)

<(130+130+...+130)+(140+...+140)+(150+...+150)<(301​+301​+...+301​)+(401​+...+401​)+(501​+...+501​)

=1030+1040+1050=13+14+15=4760<4860=45=3010​+4010​+5010​=31​+

S có 30 số hạng . Nhóm thành 3 nhóm , mỗi nhóm 10 số hạng.

\(S=\left[\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right]+\left[\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right]+\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right]\)

\(S< \left[\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right]+\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]\)

\(S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}\)

\(S< \frac{37}{60}< \frac{48}{60}=\frac{4}{5}(1)\)

Lại có : \(S>\left[\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right]+\left[\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right]+\left[\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right]\)

\(S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}\)

\(S>\frac{37}{60}>\frac{36}{60}=\frac{3}{5}(2)\)

Từ 1 và 2 suy ra \(\frac{3}{5}< S< \frac{4}{5}\)

Lời giải:

$A=(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})+(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50})+(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60})$

$> \frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}> \frac{36}{60}=\frac{3}{5}(1)$

Lại có:

$A=(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40})+(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50})+(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60})$

$< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}(2)$

Từ $(1); (2)\Rightarrow$ ta có đpcm.

Bài 1 :

\(\left(-2\right)\left(x+1\right)-3\left(1-x\right)=4\)

\(\Leftrightarrow-2x-2-3+3x=4\)

\(\Leftrightarrow x=4+2+3=9\)

Bài 2 :

Cho \(S=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\)

\(\Leftrightarrow S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Rightarrow S< \left(\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)

\(+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(\Leftrightarrow S< \frac{10}{30}+\frac{10}{40}+\frac{10}{50}=\frac{47}{60}< \frac{48}{60}=\frac{4}{5}\)⁽¹⁾

Lại có :

\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\right)\)

\(+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)

\(\Leftrightarrow S>\frac{10}{40}+\frac{10}{50}+\frac{10}{60}=\frac{37}{60}>\frac{36}{60}=\frac{3}{5}\)⁽²⁾

Từ ⁽¹⁾ và ⁽²⁾ , ta có :

\(\frac{3}{5}< S< \frac{4}{5}hay\frac{3}{5}< \frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}< \frac{4}{5}\)

Nguyen Ribi Nkok Ngok Khôi Bùi nguyễn ngọc dinh Phùng Tuệ Minh Akai Haruma buithianhtho ?Amanda? Nguyễn Thành Trương Nguyễn Ngô Minh Trí

Bài 1 : x=9