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Câu hỏi của Vũ Mai Linh - Toán lớp 7 - Học toán với OnlineMath
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
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ĐKXXD : \(x\ne20;8;3;1\)
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)
\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)
\(\frac{11}{12}-\frac{2}{3}\left|x\right|=\frac{3}{8}\)
\(\Leftrightarrow\frac{2}{3}\left|x\right|=\frac{13}{24}\)
\(\Leftrightarrow\left|x\right|=\frac{13}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{16}\\x=-\frac{13}{16}\end{cases}}\)
Ta có \(\frac{11}{12}-\frac{2}{3}.\left|x\right|=\frac{3}{8}\)
\(\Leftrightarrow\frac{2}{3}.\left|x\right|=\frac{11}{12}-\frac{3}{8}\)
\(\Leftrightarrow\frac{2}{3}\left|x\right|=\frac{22}{24}-\frac{9}{24}\)
\(\Leftrightarrow\frac{2}{3}\left|x\right|=\frac{13}{24}\)
\(\Leftrightarrow\left|x\right|=\frac{13}{16}\)
\(\Leftrightarrow x=\pm\frac{13}{16}\)
Vậy \(x=\pm\frac{13}{16}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(3^x=\frac{\left(2.3\right).\left(3.3\right).\left(4.3\right).\left(5.3\right).\left(9.2\right).\left(7.3\right).\left(2.12\right)\left(9.3\right)}{1.\left(2.3\right).\left(2.2\right).5.\left(3.2\right).7.\left(4.2\right)}=3^{10}\)
=> x=10
kb vs mk r mk giải cho nek
\(\frac{x}{12}=\frac{3}{8}\)
=> x.8 = 3.12
=> x.8 = 36
=> x = 36/8 = 9/2