Tìm x:
a. ( x + 2 ) + ( x + 5 ) + ( x + 8 ) + ( x + 11 ) + ( x + 14 ) = 75
b. ( x + 1 ) + ( x + 3 ) + ( x + 5 ) + ... + ( x + 101 ) = 5400
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a, (x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75
5x + (2 + 5 + 8 + 11 + 14) = 75
5x + 40 = 75
5x = 35
x = 7
Các câu dưới tương tự.
a)(x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75
x+x+x+x+x+2+5+8+11+14=75
5xx+40=75
5xx=75-40
5xx=35
x=35:5
x=7
b)(x+1)+(x+3)+(x+5)+....+(x+101)=5400
Nhận thấy dãy trên có (101-1):2+1=51(số) nên có 51 x
Tách như câu a
51xx+(1+3+5+...+101)=5400
51xx+[(101+1)x51:2]=5400
51xx+2601=5400
51xx=5400-2601
51xx=2799
x=2799:51
x=933/17
c)(x+1)+(x+4)+(x+7)+....+(x+94)=1840
Phần c này cũng giống phần b,em dựa vào mà làm nhé,giống,khác mỗi số thôi
Chúc em họ tốt^^
a)\(\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\Leftrightarrow x=305\)
b ko hiểu đề
\(a;343-4\times\left(x+1\right)=143\)
\(4\times\left(x+1\right)=343-143\)
\(4\times\left(x+1\right)=200\)
\(x+1=200:4\)
\(x+1=50\)
\(x=50-1=49\)
Mấy câu hỏi trẻ trâu thì tự tính nhé
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{308}\)
\(\Leftrightarrow x+3=308\)
\(\Leftrightarrow x=305\)
Vậy x=305
Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^
duyệt đi
\(\frac{1}{3}\left(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)
\(\frac{1}{5}+\frac{1}{8}-\frac{1}{8}+\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
\(\left(\frac{1}{5}-\frac{1}{x+3}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+...+\left(\frac{1}{x}-\frac{1}{x}\right)=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>x+3=308
x=308-3
x=305
Vậy x=305
a.\(\dfrac{1}{3}\) + x = \(\dfrac{5}{6}\)
x = \(\dfrac{5}{6}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{1}{2}\)
b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\)
| x-1| = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)
|x-1| = \(\dfrac{3}{2}\)
\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1
\(\dfrac{2}{3}\) (\(\dfrac{x}{2}\) + 3) = 1 - \(\dfrac{1}{3}\)
\(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = \(\dfrac{2}{3}\)
\(\dfrac{x}{2}\) + 3 = 1
\(\dfrac{x}{2}\) = 1 - 3
\(\dfrac{x}{2}\) = -2
\(x\) = -4
d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)
(x+2)2 = 27.3
(x+2) =92
\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)
a.(x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75
=>(x+x+x+x+x)+(2+5+8+11+14)=75
=>Xx5+40=75
=>Xx5=75-40
=>Xx5=35
=>x=35:5
=>x=7