Tim x biet
\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\frac{x-12}{4}=\frac{9-3x}{x}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
1 tháng 6 2020
\(\frac{x-1}{21}=\frac{3}{x+1}\)
=> \(\left(x-1\right)\left(x+1\right)=21\cdot3\)
=> \(x^2-1=63\)
=> \(x^2=64\)
=> \(\orbr{\begin{cases}x^2=8^2\\x^2=\left(-8\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=8\\x=-8\end{cases}}\)
\(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
=> \(\frac{12}{13}x=2\)
=> \(x=\frac{13}{6}\)
1 tháng 6 2020
d, \(\frac{x-1}{21}=\frac{3}{x+1}\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=63\)
\(\Leftrightarrow x^2-1=63\Leftrightarrow x^2=64\Leftrightarrow x=\pm8\)
e, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)
T
5 tháng 6 2019
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
T
5 tháng 6 2019
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
25 tháng 1 2017
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
20 tháng 7 2020
\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
a, \(2\frac{7}{9}-\frac{12}{13}x=\frac{7}{9}\)
\(\Leftrightarrow\frac{25}{9}-\frac{12}{13}x=\frac{7}{9}\Leftrightarrow\frac{12}{13}x=2\Leftrightarrow x=\frac{13}{6}\)
b, \(\frac{x-12}{4}=\frac{9-3x}{x}\)
\(\Leftrightarrow x^2-12x=36-12x\Leftrightarrow x^2-12x-36+12x=0\)
\(\Leftrightarrow x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)