giúp minh với ai nhanh minh k

anh nhanh k

( 3x/7 + 1 ) : (-4 ) = -1/28

 3x/7 + 1           = -1/28 x (-4 ) 

(3x/7 + 1           = 1/7

3x/7                  = 1/7  - 1

3x/7                  = -6/7

Suy ra 3x = -6

            x = -6 : 3

            x = -2

=>3x/7+1=-1/28x(-4)

=>3x/7+1=1/7

=>3x/7=1/7-1

=>3x/7=-6/7

=>3x=6

=>x=6:3=2

k mình đi

Bài 1: 

\(=\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}+\dfrac{1}{\left(x+13\right)\left(x+16\right)}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+13\right)}+\dfrac{3}{\left(x+13\right)\cdot\left(x+16\right)}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+7}+\dfrac{1}{x+7}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+13}+\dfrac{1}{x+13}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{x+1}-\dfrac{1}{x+16}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{x+16-x-1}{\left(x+1\right)\left(x+16\right)}=\dfrac{5}{\left(x+1\right)\left(x+16\right)}\)

Bài 2: 

\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b+4\right)^2+\left(2c-1\right)^2=0\)

Dấu '=' xảy ra khi a=1; b=-4; c=1/2

6) \(2\left(x-8\right)=2^2\)

\(\Rightarrow x-8=2^2:2\)

\(\Rightarrow x-8=2\)

\(\Rightarrow x=2+8\)

\(\Rightarrow x=10\)

tíc mình nha

5) 14 chia hết cho 2x+3

=>2x+3 thuộc ước 14

mà Ư(14)={1,2,7,14}

ta có

vậy x=0

a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)

\(\dfrac{1}{2}-x=\dfrac{6}{5}\)

\(x=\dfrac{1}{2}-\dfrac{6}{5}\)

\(x=-\dfrac{7}{10}\)

b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)

\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)

\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)

\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)

\(x=\dfrac{60}{13}:\dfrac{12}{13}\)

\(x=5\)

Dễ thấy dãy số có công thức tổng quát: 1/[(3n-2)(3n+1)] 
Xét: 1/[(3n-2)(3n+1)] = 1/3.[1/(3n-2) - 1/(3n+1)] 
Với bài này n = 34 
Ta có: 
1/4 = 1/3( 1-1/4) 
1/28 = 1/3( 1/4 - 1/7) 
1/70 = 1/3( 1/7 - 1/10) 
.............................. 
1/10300 = 1/3( 1/100 - 1/103) 
Cộng vế với vế ta có: 
S = 1/4+1/28+1/70+1/130+...+1/10300 = 1/3( 1-1/103) 
S = 34/103

I don't know

Ta có:

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

Dựa vào quy luật trên=>Số hạng thứ 30 là:\(\frac{1}{98.101}\)

\(\Rightarrow3M=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{98}-\frac{1}{101}\)

\(\Rightarrow M=\left(1-\frac{1}{101}\right):3=\frac{100}{101}.\frac{1}{3}=\frac{100}{303}\)

Mình viết hơi tắt mong bạn thông cảm.

3m là gì vậy

\(\orbr{\begin{cases}x-1=4\\x-2=4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4+1\\x=4+2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=6\end{cases}}\)

Vậy x\(\in\){5;6}

\(\text{Th1: }\hept{\begin{cases}x-1< 0\\x-2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x\ge2\end{cases}\Rightarrow}x\in\varnothing}\)

\(\text{Th2: }\hept{\begin{cases}x-1>0\\x-2\le0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x\le2\end{cases}\Rightarrow1< x\le}2}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow x-1+2-x=4\)

\(\Leftrightarrow x-x=4-2+1\)

\(\Leftrightarrow0x=3\left(\text{vô lí}\right)\)

Vậy \(x\in\varnothing\)

\(\text{TH3: }\hept{\begin{cases}x-1\ge0\\x-2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge1\\x\ge2\end{cases}\Rightarrow}x\ge1}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow x-1+x-2=4\)

\(\Leftrightarrow2x=7\)

\(\Leftrightarrow x=\frac{7}{2}\left(\text{nhận}\right)\)

\(\text{TH4: }\hept{\begin{cases}x-1\le0\\x-2\le0\end{cases}\Rightarrow\hept{\begin{cases}x\le1\\x\le2\end{cases}\Rightarrow}x\le2}\)

\(\text{Khi đó: }\left|x-1\right|+\left|x-2\right|=4\)

\(\Leftrightarrow1-x+2-x=4\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=-\frac{1}{2}\left(\text{nhận}\right)\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{7}{2}\right\}\)