Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9

\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)

\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)

b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)

<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)

<=>x=-43/30

Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại

a,A=1863/623:2/7

A=1863/178

b,

ta có:

1863/178.x+5/6=-3/4

1863/178.x=-19/12

=>x=-1691/11178

Bài 1 : Ta có:

\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)

= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)

= \(\frac{7}{9}\)

Bài 2 :

 \(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)

=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)

=> 50x = 10

=> x = 10 : 50

=> x = 1/5

Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3

                                         <=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy 

Câu 2 đây:

\(|x^2+|x-1||=x^2+2\)

\(\Rightarrow\orbr{\begin{cases}x^2+\left|x-1\right|=x^2+2\\x^2+\left|x-1\right|=-x^2-2\left(l\right)\end{cases}}\)

\(\Rightarrow\left|x-1\right|=2\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

a)    \(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+0,5}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{2}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2012}{2013}\)

\(=\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\right):\frac{2012}{2013}\)

\(=\left(\frac{2}{7}-\frac{2}{7}\right):\frac{2012}{2013}\)

\(=0\)

câu 2 đâu bn

a. \(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|\left(-3,2\right)+\frac{2}{5}\right|\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{16}{5}+\frac{2}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\left|-\frac{14}{5}\right|-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{14}{5}-\frac{4}{5}\)\(\Leftrightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=2\\x-\frac{1}{3}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{5}{3}\end{cases}.}\)

Vậy \(x\in\left\{-\frac{5}{3};\frac{7}{3}\right\}.\)

b. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\times\left(x-7\right)^{10}=0\)\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}.}\)Xét 2 trường hợp:

Vậy \(x\in\left\{6;7;8\right\}.\)

ban nguyen nhat minh giang lai cho mk dong 2 cau b cai

mk cam on

Dòng 2 câu b là đặt nhân tử chung \(\left(x-7\right)^{x+1}\)ra ngoài

a/ \(\frac{15}{x}-\frac{1}{3}=\frac{28}{51}\)

\(\frac{15}{x}=\frac{28}{51}+\frac{1}{3}\)

\(\frac{15}{x}=\frac{15}{17}\)

\(x=15:\frac{15}{17}\)

\(x=17\)

b) \(\frac{x}{20}-\frac{2}{5}=10\)

\(\frac{x}{20}=10+\frac{2}{5}\)

\(\frac{x}{20}=\frac{52}{5}\)

\(x=\frac{52}{5}\cdot20\)

\(x=208\)

c) \(x+\frac{18}{23}=2\frac{1}{3}\)

\(x+\frac{18}{23}=\frac{7}{3}\)

\(x=\frac{7}{3}-\frac{18}{23}\)

\(x=\frac{107}{69}\)

d) \(\frac{7}{11}< x-\frac{1}{7}< \frac{10}{13}\)

\(\Rightarrow\frac{7}{11}+\frac{1}{7}< x< \frac{10}{13}\)

\(\frac{60}{77}< x< \frac{60}{78}\)

Đến đây .....bí!

e) Tớ bỏ luôn đc ko.

D) 7/11<X-1/7<10/13

    <=> 7/11+1/7<x< 10/13+1/7

 <=> 60/77< x< 83/91

<=> 5460/1001 <x< 6391/1001

vậy X thuộc tập hợp các phÂN số lớn hơn 5460/1001 và bé hơn 913/1001

vd :  Y/1001 trong đó y là 5461;5462;5463...6389;6390

e) 7/9<13/11-x<15/16

<=> 7/9-13/11<-x< 15/16 -13/11

<=> -40/99< -x< -43/176

<=> 43/99< x < 40/176( nhân với -1 nên đổi dấu )

<=> 387/1584 < x< 640/1584

<=> vậy X là các phân số có dạng y/1584 , trong đó Y là 388;389;390 ........ 639

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

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

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)

Ta có:

\(\frac{x+7}{2010}+\frac{x+6}{2011}=\frac{x+5}{2012}+\frac{x+4}{2013}\)

\(\Leftrightarrow\left(\frac{x+7}{2010}+1\right)+\left(\frac{x+6}{2011}+1\right)=\left(\frac{x+5}{2012}+1\right)+\left(\frac{x+4}{2013}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2010}+\frac{x+2017}{2011}=\frac{x+2017}{2012}+\frac{x+2017}{2013}\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2010}+\frac{1}{2011}\right)=\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

Suy ra \(x+2017=0\)

Vậy \(x=-2017\)

b) Dễ tự làm nhé

thế mà cũng phải hỏi

Giỏi bạn làm hộ mk cái