Tìm x
n) 8x - 16( x - 5)=0
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\(\dfrac{3}{5}x-\dfrac{3}{5}+x=5\)
\(\Rightarrow\dfrac{3}{5}x+x-\dfrac{3}{5}=5\)
\(\Rightarrow\dfrac{8}{5}x=5+\dfrac{3}{5}\)
\(\Rightarrow\dfrac{8}{5}x=\dfrac{28}{5}\)
\(\Rightarrow x=\dfrac{28}{5}:\dfrac{8}{5}\)
\(\Rightarrow x=\dfrac{28}{5}.\dfrac{5}{8}\)
\(\Rightarrow x=\dfrac{7}{2}\)
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


\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(.....\)
\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)
\(\Rightarrow B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}< 1\)
\(\Rightarrow B< 1\left(dpcm\right)\)
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(B< \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\)
\(B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(B< 1-\dfrac{1}{10}\)
\(B< \dfrac{9}{10}< 1\)
Vậy \(B< 1\)

\(53.51+4+53.49+91+53\)
\(=53.\left(51+49+1\right)+4+91\)
\(=53.101+4+91\)
\(=5353+4+91\)
\(=5448\)

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) \(xy-2x+2y=10\left(x;y\inℕ\right)\)
\(\Rightarrow2xy-4x+4y=20\)
\(\Rightarrow2x\left(y-2\right)+4y-8+8=20\)
\(\Rightarrow2x\left(y-2\right)+4\left(y-2\right)=12\)
\(\Rightarrow\left(2x+4\right)\left(y-2\right)=12\)
\(\Rightarrow\left(2x+4\right);\left(y-2\right)\in\left\{1;2;3;4;6;12\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(-\dfrac{3}{2};14\right);\left(-1;8\right);\left(-\dfrac{1}{3};6\right);\left(0;5\right);\left(1;3\right);\left(4;3\right)\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(0;5\right);\left(1;3\right);\left(4;3\right)\right\}\left(x;y\inℕ\right)\)

\(\text{13.58.4+32.26.2+52.10}\)
\(\text{=(13.4).58 + (26.2)32 + 52.10}\)
\(\text{=52.58+52.32+52.10}\)
\(\text{=52(58+32+10)}\)
\(\text{=52.100}\)
\(\text{= 5200}\)

\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(\Rightarrow x=2\)
8x -16(x-5)=0
8x-16x+80=0
8x - 16x=0-80
x.(8-16)=-80
x.(-8)=-80
x=-80:(-8)
x=10
Vậy x=10
8x - 16x + 80 = 0
-8x = -80
x = 10