đk 1 - x\(\ge\)0

=> x \(\le\)1

Khi đó |x - 2| = -(x - 2) 

|x - 3| = -(x - 3) 

....

|x - 9|  = -(x - 9) 

Khi đó |x - 2| + |x - 3| +... + |x - 9| = 1-x                    (8 cặp số ở VT)

<=> -(x - 2) + -(x - 3) + .... + -(x - 9) = 1 - x                  

=> -x + 2 - x + 3  -  .... - x + 9 = 1 - x

=> -(x + x + ... x) + (2 + 3 + ... + 9) = 1 - x

     8 hạng tử x         8 hạng tử 

=> -8x + 44 = 1 - x

=> 7x = 43 

=> x = 43/7

3: Ta có: \(\sqrt{4x+1}=x+1\)

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

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

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

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

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

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

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

học tốt nha

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

các bn lm đến đâu cx dc miễn là lm hộ mk cái ạ, ai đang lm vào nhắn tin vs mk để mk bít nha

a; \(-\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{71}{15}< x< -\dfrac{13}{7}+\dfrac{19}{14}-\dfrac{7}{2}\)

              -\(\dfrac{19}{15}\) - \(\dfrac{71}{15}\) < \(x\) < -\(\dfrac{1}{2}\) - \(\dfrac{7}{2}\)

              -6 < \(x\) < -4

             vì \(x\) \(\in\) Z nên \(x\) = -5

- \(\dfrac{19}{6}+\dfrac{-15}{2}+\dfrac{11}{3}\) < \(x\) < - \(\dfrac{5}{4}\) + \(\dfrac{19}{12}\) - \(\dfrac{10}{3}\)

- \(\dfrac{32}{3}+\dfrac{11}{3}< x< \) \(\dfrac{1}{3}\) - \(\dfrac{10}{3}\)

-7 < \(x\) < - 3

Vì \(x\in\) Z nên \(x\in\) {-6;-5;-4}

ĐKXĐ: x khác + -2

 A=( 1/(x-2) + 2x/(x-2)(x+2) +1/(x+2)) . (x-1)/2

   =((x+2+2x+x-2)/(x-2)(x+2)).((x-1)/2)

   =(4x/(x-2)(x+2)).(x-1)/2 =2x/ (x-1)(x-2)(x+2)
 

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999+2004}\).

Có sai đề không vậy???

Sửa đề một chút :v

\(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{1999\cdot2004}\)

\(=\frac{1}{5}\left[\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{1999\cdot2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{1999}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\left[\frac{1}{4}-\frac{1}{2004}\right]\)

\(=\frac{1}{5}\cdot\frac{125}{501}=\frac{25}{501}\)

Đặt A =1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004. Ta có:

        A= 1/4 x 9 + 1/9 x 14 + 1/14 x 19 +...+ 1/1999 + 2004

      5A= 5/4 x 9 + 5/9 x 14 + 5/14 x 19 +...+ 5/1999 + 2004

      5A= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 +...+ 1/1999 - 1/2004

      5A= 1/4 - 1/2004

        A= (1/4 - 1/2004)/5

13/276

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)