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\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{2}-\frac{1}{50}\)
\(=\frac{12}{25}\)Mà 12< 25 nên \(\frac{12}{25}< 1\)
Vậy \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}< 1\)
đặt A=\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)=\(\frac{1}{2}-\frac{1}{50}\)<\(\frac{1}{2}\)<1
suy ra A<1
Vậy \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)<1
\(x^{x+1}-x^{x+1}\cdot x^3=0\)
\(x^{x+1}\left(1-x^3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x+1=0\Rightarrow x=-1\\1-x^3=0\Rightarrow x=1\end{cases}}\)
Vậy,..........
x : xx+1 = xx+4
=> x = xx+4 . xx+1
=> x = xx+4+x+1
=> x = x2x+5
=> x \(\in\left\{-1;0;1\right\}\)
a,\(\left(3x-4\right)\left(x-1\right)^3=0\)
\(=>\orbr{\begin{cases}3x-4=0\\x-1=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=\frac{4}{3}\\x=1\end{cases}}\)
b,\(\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=7450\)
\(=>x.100+\left(1+2+3+...+100\right)=7450\)
\(=>100x+5050=7450\)
\(=>100x=2400\)
\(=>x=24\)
Cold Wind , Đinh Tuấn Việt, Kiệt ღ ๖ۣۜLý๖ۣۜ , Ngọc Mai , Nhật Minh,......bn có thể xem trong bảng xếp hạng nha
bn nên vào trang toán để hỏi nha, có nhìu bn giỏi mon toán lém
\(3x-8⋮x-4\Rightarrow3\left(x-4\right)+4⋮x-4\)
\(\Rightarrow4⋮x-4\Rightarrow x-4\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow n\in\left\{5;3;6;2;8;0\right\}\)
Vậy...................
\(2-1+x+5-x+2=-14+x\)
\(\Leftrightarrow-1+x+5-x+2+14-x=0\)
\(\Leftrightarrow20-x=0\)
\(\Leftrightarrow x=20\)
1, ( - 12 ) - ( 13 - x ) = - 15 - ( - 17 )
=> - 12 - 13 + x = - 15 + 17
=> - 25 + x = 2
=> x = 2 + 25
=> x = 27
2, 305 - x + 14 = 48 + ( x - 23 )
=> 305 - x + 14 = 48 + x - 23
=> x + x = 305 + 14 - 48 + 23
=> 2x = 294
=> x = 147
3, - ( x - 6 + 85 ) = ( x + 51 ) - 54
=> - x + 6 - 85 = x + 51 - 54
=> x + x = 6 - 85 - 51 + 54
=> 2x = - 76
=> x = - 38
4, - ( 35 - x ) - ( 37 - x ) = 33 - x
=> - 35 + x - 37 + x = 33 - x
=> x + x + x = 33 + 35 + 37
=> 3x = 105
=> x = 35
\(\dfrac{x}{2.3}+\dfrac{x}{3.4}+\dfrac{x}{4.5}+...+\dfrac{x}{49.50}=1\\ \Rightarrow x\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x\left(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{50-49}{49.50}\right)=1\\ \Rightarrow x\left(\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+\dfrac{5}{4.5}-\dfrac{4}{4.5}+...+\dfrac{50}{49.50}-\dfrac{49}{49.50}\right)=1\)
\(\Rightarrow x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ \Rightarrow x\left(\dfrac{25}{50}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{24}{50}=1\\ \Rightarrow x=1:\dfrac{24}{50}=\dfrac{50}{24}\)