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

\(\Rightarrow\dfrac{3}{5}x+x-\dfrac{3}{5}=5\)

\(\Rightarrow\dfrac{8}{5}x=5+\dfrac{3}{5}\)

\(\Rightarrow\dfrac{8}{5}x=\dfrac{28}{5}\)

\(\Rightarrow x=\dfrac{28}{5}:\dfrac{8}{5}\)

\(\Rightarrow x=\dfrac{28}{5}.\dfrac{5}{8}\)

\(\Rightarrow x=\dfrac{7}{2}\)

 3/5 x ( X - 1 ) + X = 5 

 (X + 1) + X = 5 - 3/5 

  X + 1 + X = 22/5

 2X + 1 = 22/5 

= 2X = 22/5 - 1

 2X = 17/5 

 X = 17/5 : 2 

 X = 17/10

Trả lời

(5+5)x5+(5+5)x5=100

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(.....\)

\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)

\(\Rightarrow B< 1\left(dpcm\right)\)

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)

 \(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)

 \(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(B< 1-\dfrac{1}{10}\)

\(B< \dfrac{9}{10}< 1\)

Vậy \(B< 1\)

\(53.51+4+53.49+91+53\)

\(=53.\left(51+49+1\right)+4+91\)

\(=53.101+4+91\)

\(=5353+4+91\)

\(=5448\)

cảm ơn

Bài 1 : 

a) \(xy-2x+2y=10\)

\(\Leftrightarrow x\left(y-2\right)+2y=10\)

\(\Leftrightarrow x\left(y-2\right)+2y-4=6\)

\(\Leftrightarrow x\left(y-2\right)+2\left(y-2\right)=6\)

\(\Leftrightarrow\left(x+2\right)\left(y-2\right)=6\)

Ta có : \(x+2\ge2\) vì \(x\in N\)

Do đó : ta có bảng :

x+2 :  2       3      6

y-2 :    3       2       1

x    :     0        1       4

y    :      5        4        3

Vậy...........

a) \(xy-2x+2y=10\left(x;y\inℕ\right)\)

\(\Rightarrow2xy-4x+4y=20\)

\(\Rightarrow2x\left(y-2\right)+4y-8+8=20\)

\(\Rightarrow2x\left(y-2\right)+4\left(y-2\right)=12\)

\(\Rightarrow\left(2x+4\right)\left(y-2\right)=12\)

\(\Rightarrow\left(2x+4\right);\left(y-2\right)\in\left\{1;2;3;4;6;12\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-\dfrac{3}{2};14\right);\left(-1;8\right);\left(-\dfrac{1}{3};6\right);\left(0;5\right);\left(1;3\right);\left(4;3\right)\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;5\right);\left(1;3\right);\left(4;3\right)\right\}\left(x;y\inℕ\right)\)

\(\text{13.58.4+32.26.2+52.10}\)

\(\text{=(13.4).58 + (26.2)32 + 52.10}\)

\(\text{=52.58+52.32+52.10}\)

\(\text{=52(58+32+10)}\)

\(\text{=52.100}\)

\(\text{= 5200}\)

\(3^x+3^{x+1}+3^{x+2}=117\)

\(3^x+3^x.3+3^x.3^2=117\)

\(3^x\left(1+3+3^2\right)=117\)

\(3^x.13=117\)

\(3^x=9\)

\(\Rightarrow x=2\)

`3^{x}+3^{x+1}+3^{x+2}=117`

`3^{x}.(1+3+3^{2})=117`

`3^{x}.13=117`

`3^{x}=117:13=9`

`3^{x}=3^{2}`

`x=2`

\(8.9.14+6.17.12+19.4.18\)

\(=4.2.9.14+6.3.17.4+19.4.18\)

\(=4.18.14+18.4.17+19.4.18\)

\(=4.18.\left(14+17+19\right)\)

\(=4.18.50\)

\(=3600\)