bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35

 Tích các thừa số là 0 chứng tỏ có ít nhất một tổng có kết quả là 0 

 Xét 1/7x - 2/7 = 0 

=> 1/7 . x = 2/7

     x = 2 

 Xét -1/5x + 3/5 = 0 

=> -1/5 . x = -3/5

     x = 3

Xét 1/3x + 4/3 = 0

=> 1/3x = -4/3

      x = -4

Bài 1:

a) (2x-3). (x+1) < 0

=>2x-3 và x+1 ngược dấu

Mà 2x-3<x+1 với mọi x

\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)

b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu

Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)

Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

=>....

Bài 2:

\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\cdot\frac{998}{3003}\)

\(=\frac{499}{3003}\)

tự làm nhé. bài cô Kiều cho dễ mừ :)

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)

<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)

<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)

\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)

<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)

<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)

<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)

câu c tương tự nha

học tốt

\(3\left(2x-\frac{5}{4}\right)=\left(3-1\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)

\(\Leftrightarrow6x-\frac{15}{4}=\frac{3}{2}x+\frac{1}{12}\)

\(\Leftrightarrow\frac{9}{2}x+\frac{3}{4}=\frac{15}{4}\)

\(\Leftrightarrow\frac{9}{2}x=3\)

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

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{6}{11}\)

\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=\frac{2}{5}\)

\(\Leftrightarrow x=\frac{2}{5}\div\frac{11}{15}=\frac{2.15}{5.11}=\frac{6}{11}\)

Vậy x = 6/11 

a) \(\frac{1}{3}.x+\frac{2}{5}.\left(x-1\right)=0\)

\(\frac{1}{3}.x+\frac{2}{5}.x-\frac{2}{5}=0\)

\(x.\left(\frac{1}{3}+\frac{2}{5}\right)-\frac{2}{5}=0\)

\(x.\frac{11}{15}-\frac{2}{5}=0\)

\(x.\frac{11}{15}=\frac{2}{5}\)

\(x=\frac{2}{5}:\frac{11}{15}\)

\(x=\frac{6}{11}\)

b) \(3.\left(x-\frac{1}{2}\right)-5.\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)

\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)

\(3x-5x-\left(\frac{3}{2}+3\right)=x+\frac{1}{5}\)

\(-2x-\frac{9}{2}=x+\frac{1}{5}\)

\(\Rightarrow-2x-x=\frac{1}{5}+\frac{9}{2}\)

\(-3x=\frac{47}{10}\)

\(x=\frac{47}{10}:\left(-3\right)\)

\(x=\frac{-47}{30}\)