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\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
2x -8 + 3(x-2)= 50
=> 2x -8 + 3x -6 = 50
=> 5x - 14 = 50
=> 5x = 64
=> x = 64/5
\(2x-8+3\left(x-2\right)=50\)
\(\Rightarrow2x-8+3x-6=50\)
\(\Rightarrow2x+3x=50+6+8\)
\(\Rightarrow5x=64\)
\(\Rightarrow x=\frac{64}{5}\)
-x-20-(8-2x)=(-12-3)
-x-20-8+2x=-15
x-28=-15
x=13
Vậy x=13
\(A=\frac{3x+2}{5x+3}\)
Gọi d là ƯC(3x+2 ; 5x+3)
\(\Rightarrow\hept{\begin{cases}3x+2⋮d\\5x+3⋮d\end{cases}\Rightarrow}\hept{\begin{cases}5\left(3x+2\right)⋮d\\3\left(5x+3\right)⋮d\end{cases}\Rightarrow}\hept{\begin{cases}15x+10⋮d\\15x+9⋮d\end{cases}}\)
\(\Rightarrow\left(15x+10\right)-\left(15x+9\right)⋮d\)
\(\Rightarrow15x+10-15x-9⋮d\)
\(\Rightarrow1⋮d\Leftrightarrow d=1\)
=> ƯCLN( 3x+2 ; 5x+3 ) = 1
=> \(A=\frac{3x+2}{5x+3}\)tối giản ( đpcm )
\(B=\frac{2x+3}{4x+8}\)
Gọi d là ƯC( 2x+3;4x+8 }
\(\Rightarrow\hept{\begin{cases}2x+3⋮d\\4x+8⋮d\end{cases}}\Rightarrow\hept{\begin{cases}2\left(2x+3\right)⋮d\\4x+8⋮d\end{cases}}\Rightarrow\hept{\begin{cases}4x+6⋮d\\4x+8⋮d\end{cases}}\)
\(\Rightarrow\left(4x+8\right)-\left(4x+6\right)⋮d\)
\(\Rightarrow4x+8-4x-6⋮d\)
\(\Rightarrow2⋮d\Leftrightarrow d=\left\{1;2\right\}\)
Với d = 1 => \(2x+3⋮d\)
Với d = 2 => \(2x+3⋮̸d\)vì \(3⋮̸2\)
=> d = 1
=> ƯCLN( 2x+3 ; 4x+8 ) = 1
=> \(B=\frac{2x+3}{4x+8}\)tối giản ( đpcm )
a) 3(x + 5) - 8 = x + 17
=> 3x + 15 - 8 = x + 17
=> 3x + 7 = x + 17
=> 3x - x = 17 - 7
=> 2x = 10
=> x = 10 : 2
=> x = 5
b) 5(x + 3) - 10 = 2x - 19
=> 5x + 15 - 10 = 2x - 19
=> 5x + 5 = 2x - 19
=> 5x - 2x = -19 - 5
=> 3x = -24
=> x = -24 : 3
=> x = -8
a. 3x+15-8=x+17
3x-x=17-15+8
2x=10
x=5
b.5x+15-10=2x-19
5x+5=2x-19
5x-2x=-19-5
3x=-24
x=-8
thank
2x-8=-6
2x =-6+8
2x =2
x =2:2
x=1
=>2x=-6+8
=>2x=2
=>x=2:2
=>x=1