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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 \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)
x - ( 81 + 5x ) = 19
=> x - 81 - 5x = 19
=> ( x - 5x ) - 81 = 19
=> -4x = 19 + 81
=> -4x = 100
=> x = 100 : ( -4 )
=> x = -25
2x - ( 8 + 3x ) = 92
=> 2x - 8 - 3x = 92
=> ( 2x - 3x ) = 8 + 92
=> -x = 100
=> x = -100
\(\left(x+1\right)^4=64\)
kiểm tra lại đề xem bạn
thấy ra kq dài quá
thoạt tiên bn đã vt sai đề :(chỗ in đậm)
196:{64-196:{64-[18+2(25-21)2]}}
=196:{64-196:{64-[18+2.42]}}
=196:{64-196:{64-[18+32]}}
=196:{64-196:{64-50}}
=196:{64-196:14}}
=196:{64-14}
=196:50
=3.92
\(\Leftrightarrow\dfrac{1}{2x-4}=\dfrac{1}{14}\\ \Rightarrow2x-4=14\\ \Rightarrow2x=18\\ \Rightarrow x=9\)
\(2\) \(^{x+1}.\)18=64.9
=>2 \(^{x+1}.2.9=64.9\)
=>2\(^{x+1}.2\ \ =64\)
=>\(2^{\left(x+1\right)+1}=2^6\)
=>x+2 = 6
=> x =6 - 2
=> x =4
Vậy x=4
chúc bạn học tốt!
ai cho mình là đúng thì like nha