\(\frac{11}{12}-\frac{2}{3}|x|=\frac{3}{8}\)
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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
\(\frac{1}{3}x-1.2x-\frac{2}{23}.\left(\frac{-11}{12}+\frac{5}{6}-\frac{7}{8}\right)-\frac{1}{8}=-11\frac{2}{3}+12\frac{7}{8}\)
\(\Leftrightarrow\frac{1}{3}x-\frac{6}{5}x-\frac{2}{23}.\frac{-23}{24}-\frac{1}{8}=\frac{-35}{3}+\frac{103}{8}\)
\(\Leftrightarrow\frac{-13}{15}x+\frac{1}{12}-\frac{1}{8}=\frac{29}{24}\)
\(\Leftrightarrow\frac{-13}{15}x-\frac{1}{24}=\frac{29}{24}\)
\(\Leftrightarrow\frac{-13}{15}x=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{-75}{52}\)
Vậy \(x=\frac{-75}{52}\)
1/3x-1/2x-2/23.(-11/12+5/6-7/8)-1/8=-31/3+103/8
1/3x-1,2x-2/23.(-11/12+5/6-7/8)-1/8=61/24
1/3x-1,2x-2/23.-23/24-1/8=61/24
1/3x-1,2x-1/12-1/8=61/24
1/3x-1,2x+1/24=61/24
x.(1/3-1,2)+1/24=61/24
x.-13/15+1/24=61/24
x.-33/40=61/24
x=61/24:-33/40
x=61/24×-40/33
x=-305/99
vậy x=-305/99
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(M=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(M=\frac{3}{-5}=\frac{-3}{5}\)
#)Giải :
a) x + 2x + 3x + ... + 100x = - 213
=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213
=> 100x + 5049 = - 213
<=> 100x = - 5262
<=> x = - 52,62
#)Giải :
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)
\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{2}{3}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
a) =-5/7 +7/8-2/7+1/8- -1/12+ -13/12
=(-5/7-2/7)+(7/8+1/8)-(-1/12--13/12)
=-7/7+8/8 - 12/12
= -1+1+1
=1
b)= ( -3/8+11/8)-(12/11+ -1/11)+(-3/5- 2/5)
= 1- 1 + (-1)
=-1
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right)\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=-\frac{3}{5}\)
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(\frac{-5}{8}\right)-\left(\frac{-5}{10}\right)+\left(\frac{-5}{11}\right)+\left(\frac{-5}{12}\right)}\)
\(=\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{\left(-5\right).\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\)
\(=\frac{-3}{5}\)
\(\frac{11}{12}-\frac{2}{3}\left|x\right|=\frac{3}{8}\)
\(\Leftrightarrow\frac{2}{3}\left|x\right|=\frac{13}{24}\)
\(\Leftrightarrow\left|x\right|=\frac{13}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{16}\\x=-\frac{13}{16}\end{cases}}\)
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 \(x=\pm\frac{13}{16}\)