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Bài 21: 

Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+\dfrac{x+3}{97}+1+\dfrac{x+4}{96}+1=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

mà \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}>0\)

nên x+100=0

hay x=-100

Vậy: x=-100

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

Em cảm ơn ạ.

\(=\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)x \(\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\) x\(\left(\dfrac{4}{4}+\dfrac{1}{4}\right)\) x \(\left(\dfrac{5}{5}+\dfrac{1}{5}\right)\)

\(=\dfrac{3}{2}\) x \(\dfrac{4}{3}\) x \(\dfrac{5}{4}\) x \(\dfrac{6}{5}\) 

\(=3\)

= 2

3

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

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

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$

`5/9+4/9:x=1/3`

`=>4/9:x=1/3-5/9`

`=>4/9:x=3/9-5/9`

`=>4/9:x=-2/9`

`=>x=4/9:(-2/9)`

`=>x=4/9.(-9/2)`

`=>x=-4/2`

`=>x=-2`

`5/9 + 4/9 : x= 1/3`

`=> 4/9 : x= 1/3-5/9`

`=> 4/9 : x= 3/9-5/9`

`=> 4/9 : x= -2/9`

`=> x= 4/9 :(-2/9)`

`=>x= 4/9 xx (-9/2)`

`=>x= -36/18`

`=>x=-2`

5/9+4/9:x=1/3 

          1:x=1/3

             x= 1: 1/3

             x= 3 

Theo tui là thế vì có mỗi phép cộng làm cộng trước xong đó tính x là ra nếu thấy cách làm thế nào thì mn bình luộn nha 

a:=>0,75x-x+1,25x=0,2

=>x=0,2

b: =>1/3-x=-3/6+4/6=1/6

=>x=1/3-1/6=1/6

c: =>(x-1)/45=-6/30=-1/5

=>x-1=-9

=>x=-8

d: =>(2/5x-1)=-5/7

=>2/5x=2/7

=>x=5/7

1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)

2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)

3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)

4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)

5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)

6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\)
\(\Rightarrow-6< x< -4\)
\(\Rightarrow x=-5\)

`(-1)/4 < x/24 < (-1)/6`

`(-6)/24 < x/24 < (-4)/24`

`->x=-5`