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\(a)\) \(2x-5=21\)

\(\Leftrightarrow\) \(2x=21+5\)

\(\Leftrightarrow\) \(2x=26\)

\(\Leftrightarrow\) \(x=26:2\)

\(\Leftrightarrow\) \(=13\)

\(b)\) \(\frac{3}{4}+\frac{1}{4}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{5}{6}-\frac{3}{4}\)

\(\Leftrightarrow\) \(\frac{1}{4}x=\frac{1}{12}\)

\(\Leftrightarrow\) \(x=\frac{1}{3}\)

\(c)\) \(\frac{7}{8}-\frac{1}{2}x=\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{7}{8}-\frac{5}{6}\)

\(\Leftrightarrow\) \(\frac{1}{2}x=\frac{1}{24}\)

\(\Leftrightarrow\) \(x=\frac{1}{12}\)

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{90}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-...+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\left(1-\dfrac{1}{10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{9}{10}\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=1\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}\)

\(\Rightarrow x=\dfrac{11}{25}\)

b) \(x\cdot9,85+x\cdot0,15=0,1\)

\(\Rightarrow x\cdot\left(9,85+0,15\right)=0,1\)

\(\Rightarrow x\cdot10=0,1\)

\(\Rightarrow x=\dfrac{0,1}{10}\)

\(\Rightarrow x=0,01\)

c) \(\dfrac{2}{5}+2022x=\dfrac{4}{10}\)

\(\Rightarrow\dfrac{2}{5}+2022x=\dfrac{2}{5}\)

\(\Rightarrow2022x=\dfrac{2}{5}-\dfrac{2}{5}\)

\(\Rightarrow2022x=0\)

\(\Rightarrow x=\dfrac{0}{2022}\)

\(\Rightarrow x=0\)

a) \(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\left(1\right)\)

Ta có :

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

\(\left(1\right)\Rightarrow\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Rightarrow90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=89\)

\(\Rightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right].2=90-89\)

\(\Rightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{206}{100}=\dfrac{5}{2}:\dfrac{1}{2}\)

\(\Rightarrow x+\dfrac{103}{50}=\dfrac{5}{2}.\dfrac{2}{1}\)

\(\Rightarrow x+\dfrac{103}{50}=5\)

\(\Rightarrow x=5-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{250}{50}-\dfrac{103}{50}\)

\(\Rightarrow x=\dfrac{147}{50}\)

coi lại đề bài đi bạn

Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Leftrightarrow100\cdot\left(1-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Leftrightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)

\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=2\)

\(\Leftrightarrow x+\dfrac{103}{50}=\dfrac{5}{4}\)

hay \(x=\dfrac{331}{100}\)

\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-........-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)

\(\Leftrightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}\right]=89\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)

\(\Leftrightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right)=1.\frac{1}{2}=\frac{1}{2}\)

\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}\)

\(\Leftrightarrow x+\frac{103}{50}=\frac{5}{2}.2\)

\(\Leftrightarrow x+\frac{103}{50}=5\)

\(\Leftrightarrow x=5-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{250}{50}-\frac{103}{50}\)

\(\Leftrightarrow x=\frac{147}{50}\)

Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Leftrightarrow100\cdot\dfrac{9}{10}-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)

\(\Leftrightarrow\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]:\dfrac{1}{2}=1\)

\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=2\)

\(\Leftrightarrow x=-\dfrac{81}{100}\)

Bài 1 :

\(A=1+2+3+4+5+6+7+8+9\)

   \(=\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5\)

   \(=10+10+10+10+5\)      

   \(=45\)

Bài 2 bạn tự làm nha

Bài 2 : Tìm x :

a) ( x - 8 )⁵ = ( x - 8 )⁹

=> x - 8 thuộc { 0 ; 1 } vì :

Nếu x - 8 = 0 thì ( x - 8 )⁵ = 0⁵ = 0

Nếu x - 8 = 0 thì ( x - 8 )⁹ = 0⁹ = 0

Vậy x - 8 = 0 là thỏa mãn điều kiện.

Nếu x - 8 = 1 thì ( x - 8 )⁵ = 1⁵ = 1

Nếu x - 8 = 1 thì ( x - 8 )⁹ = 1⁹ = 1

Vậy x - 8 = 1 là thỏa mãn điều kiện.

=> x - 8 thuộc { 0 ; 1 }

=> x thuộc { 8 ; 9 }

Bài 2 : Tìm x :

b) ( x + 1 ) + ( x + 2 ) +.....+ ( x + 100 ) = 5750

    100x + ( 1 + 2 + 3 + 4 + .....+ 100 )   = 5750

    100x +                  5050                    = 5750

    100x                                                = 5750 - 5050

    100x                                                 =      700

         x                                                  = 700 : 100

         x                                                   =      7