Tìm x biết
1/1*4+1/4*7+...+1/x(x+3) = 125/376
Giải giúp nhé . Đang cần gấp . Cảm ơn
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)
Câu a) xem lại đề giùm nhé em
b) \(\left(x-1\right)^3=9^3\)
\(x-1=9\)
\(x=10\)
Vậy \(x=10\)
c) \(\left(x-1\right)^2=25\)
\(x-1=5\) hoặc \(x-1=-5\)
* \(x-1=5\)
\(x=6\)
* \(x-1=-5\)
\(x=-4\)
Vậy \(x=-4\); \(x=6\)
d) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy \(x=2\)
e) Sửa đề: \(\left(2x+4\right)^3=64\)
\(\left(2x+4\right)^3=4^3\)
\(2x+4=4\)
\(2x=0\)
\(x=0\)
Vậy \(x=0\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
\(3\dfrac{2}{5}\cdot1\dfrac{4}{7}=\dfrac{17}{5}\cdot\dfrac{11}{7}=\dfrac{187}{35}\)
\(a)3\left(x-1\right)^2=75\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=\left(-5\right)^2\\\left(x-1\right)^2=5^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
\(b)170+\left(84-5x\right):2^2=186\)
\(\Leftrightarrow\frac{84-5x}{4}=16\)
\(\Leftrightarrow84-5x=64\)
\(\Leftrightarrow5x=20\)
\(\Leftrightarrow x=4\)
\(c)125-5\left(x+4\right)=38\)
\(\Leftrightarrow5\left(x+4\right)=87\)
\(\Leftrightarrow x+4=\frac{87}{5}\)
\(\Leftrightarrow x=\frac{87}{5}-4\)
\(\Leftrightarrow x=\frac{67}{5}\)
ta có: \(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{x}.\)
\(A=1+\frac{1}{2}+\frac{1}{2.2}+\frac{1}{2.2.2}+...+\frac{1}{x}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2.2}+...+\frac{1}{x:2}\)
\(\Rightarrow2A-A=2-\frac{1}{x}\)
\(A=2-\frac{1}{x}=\frac{4095}{2048}\)
=> 1/x = 1/2048
=> x = 2048 ( 2048 = 211 )
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{x(x+3)}=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\left[\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{x(x+3)}\right]=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\left[1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right]=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\left[1-\frac{1}{x+3}\right]=\frac{125}{376}\)
\(\Leftrightarrow1-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}\)
\(\Leftrightarrow1-\frac{1}{x+3}=\frac{125}{376}\cdot3\)
\(\Leftrightarrow1-\frac{1}{x+3}=\frac{375}{376}\)
\(\Leftrightarrow\frac{1}{x+3}=1-\frac{375}{376}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
\(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\cdot\left(1-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{x+2}{x+3}=\frac{125}{376}\)
\(\Leftrightarrow\frac{x+2}{x+3}=\frac{125}{376}\div\frac{1}{3}\)
\(\Leftrightarrow\frac{x+2}{x+3}=\frac{375}{376}\)
\(\Leftrightarrow\left(x+2\right).376=\left(x+3\right).375\)
\(\Leftrightarrow376\text{x}+752=375\text{x}+1125\)
\(\Leftrightarrow376\text{x}-375\text{x}=1125-752\)
\(\Leftrightarrow x=373\)
Vậy x = 373