\(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3x+5}\right)=\frac{4}{25}\)

\(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)

\(\frac{1}{3x+5}=\frac{1}{50}\)

=> 3x+5 = 50

3x = 45

x = 15

=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x²+4x+30x+6=20x²+25x+8x+10

=>20x²-20x²+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

B)=>(3x-1).(5x-34)=(40-5x).(25-3x)

=>15x²-102x-5x+34=1000-120x-125x+15x²

=>15x²-107x+34=1000-245x+15x²

=>15x²-15x²-107x+245x=1000-34

=>0-107x+245x=966

=>138x=966

=>x=7

A,=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x2+4x+30x+6=20x2+25x+8x+10

=>20x2-20x2+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

\(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{\left(x+2\right)\left(x+5\right)}=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{1}{3}\cdot\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{\left(x+2\right)\left(x+5\right)}\right)=\dfrac{3}{20}\)

\(\Rightarrow\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+..+\dfrac{3}{\left(x+2\right)\left(x+5\right)}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{x+2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+5}=\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{2}-\dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{x+5}=\dfrac{1}{20}\)

\(\Rightarrow x+5=20\)

\(\Rightarrow x=20-5\)

\(\Rightarrow x=15\)

\(\dfrac{1}{8.11}\) chứ ko phải \(\dfrac{1}{8.13}\) nhé

a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1

<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5

<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3

<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5

<=> |x - 5/3|/6 = 9/5

<=> |x - 5/3| = 9/5.6

<=> |x - 5/3| = 54/5

<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5

       x = 54/5 + 5/3         x = -54/5 - 5/3

       x = 187/15              x = -137/15

b) 2.|3x + 1| = 1/3.|3x + 1| + 5

<=> 2.|3x + 1| - 1/3.|3x + 1| = 5

<=> 5/3.|3x + 1| = 5

<=> 5.|3x + 1| = 5.3

<=> 5.|3x + 1| = 15

<=> |3x + 1| = 15 : 5

<=> |3x + 1| = 3

       3x + 1 = 3 hoặc 3x + 1 = -3 

       3x = 3 - 1           3x = -3 - 1

       3x = 2                3x = -4

       x = 2/3               x = -4/3

=> x = 2/3 hoặc x = -4/3

c) làm tương tự câu a) mình hơi lời

Làm câu c) cho

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

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

\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)

\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)

Giải tiếp nha

         Bài 1:

- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1

         2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

Bài 3:

a: \(S=1+5^2+5^4+\cdots+5^{200}\)

=>25S=\(5^2+5^4+5^6+\cdots+5^{202}\)

=>25S-S=\(5^2+5^4+\cdots+5^{202}-1-5^2-\cdots-5^{200}\)

=>24S=\(5^{202}-1\)

=>\(S=\frac{5^{202}-1}{24}\)

b: \(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}\cdot2^{30}=8^{10}\cdot4^{15}\)

\(3\cdot24^{10}=3\cdot3^{10}\cdot8^{10}=8^{10}\cdot3^{11}\)

mà \(4^{15}>3^{11}\)

nên \(4^{30}>3\cdot24^{10}\)

=>\(2^{30}+3^{30}+4^{30}>3\cdot24^{10}\)

Bài 2:

a: |2x-3|>5

=>\(\left[\begin{array}{l}2x-3>5\\ 2x-3<-5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x>8\\ 2x<-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ x<-1\end{array}\right.\)

c: |3x-1|<=7

=>-7<=3x-1<=7

=>-6<=3x<=8

=>\(-2\le x\le\frac83\)

d: \(\left|3x-5\right|+\left|2x+3\right|=7\) (1)

TH1: \(x<-\frac32\)

=>2x+3<0; 3x-5<0

(1) sẽ trở thành: -2x-3-3x+5=7

=>-5x+2=7

=>-5x=5

=>x=-1(loại)

TH2: -3/2<=x<5/3

=>2x+3>=0; 3x-5<0

(1) sẽ trở thành: 2x+3-3x+5=7

=>-x+8=7

=>-x=-1

=>x=-1(nhận)

TH3: x>=5/3

=>2x+3>0; 3x-5>=0

(1) sẽ trở thành: 2x+3+3x-5=7

=>5x-2=7

=>5x=9

=>x=9/5(nhận)