37+(13-/2x+7/)=6^30 :(9^14.4^15)
Tìm x
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b, xy-x+2y=3
<=> xy-x+2y-3=0
<=>x(y-1)+2(y-1)-1=0
<=> (y-1)(x+2) =1
Ta có bảng sau
x+2 | 1 | -1 |
y-1 | 1 | -1 |
x | -1 | -3 |
y | 2 | 0 |
Vậy...
a)x+1/15 + x+2/7 + x+4/4 + 6 = 0
\(\Rightarrow\left(\frac{x+1}{15}+1\right)+\left(\frac{x+2}{7}+2\right)+\left(\frac{x+4}{4}+3\right)+\left(\frac{x+16}{\frac{1}{6}\left(x+16\right)}\right)=0\)\(\Rightarrow\frac{x+16}{15}+\frac{x+16}{7}+\frac{x+16}{4}+\frac{x+16}{\frac{1}{6}\left(x+16\right)}=0\)
\(\Rightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}+\frac{1}{\frac{1}{6}\left(x+16\right)}\right)=0\)
\(\Rightarrow x+16=0\).Do \(\frac{1}{15}+\frac{1}{7}+\frac{1}{4}+\frac{1}{\frac{1}{6}\left(x+16\right)}\ne0\)
\(\Rightarrow x=-16\)
b)tương tự
\(\dfrac{7}{16}+\dfrac{-1}{8}+\dfrac{9}{32}=\dfrac{14}{32}-\dfrac{4}{32}+\dfrac{9}{32}=\dfrac{19}{32}\)
\(\dfrac{5}{4}-\dfrac{25}{30}\times\dfrac{37}{44}+\dfrac{-25}{30}\times\dfrac{13}{44}+\dfrac{-25}{30}\times\dfrac{-6}{44}\)
\(=-\dfrac{25}{30}\times\left(\dfrac{5}{4}+\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)\)
\(=-\dfrac{25}{30}\times\left(\dfrac{55}{44}+\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)\)
\(=-\dfrac{25}{30}\times\dfrac{99}{44}\)
\(=-\dfrac{5}{6}\times\dfrac{9}{4}\)
\(=-\dfrac{15}{8}\)
\(\left(\dfrac{7}{16}+\dfrac{-1}{8}+\dfrac{9}{32}\right):\dfrac{5}{4}\)
\(=\left(\dfrac{14}{32}+\dfrac{-4}{32}+\dfrac{9}{32}\right):\dfrac{5}{4}\)
\(=\dfrac{19}{32}:\dfrac{5}{4}\)
\(=\dfrac{19}{32}.\dfrac{4}{5}=\dfrac{19.4}{32.5}=\dfrac{19}{40}\)
\(\dfrac{-25}{30}.\dfrac{37}{44}+\dfrac{-25}{30}.\dfrac{13}{44}+\dfrac{-25}{30}.\dfrac{-6}{44}\)
\(=\dfrac{-25}{30}.\left(\dfrac{37}{44}+\dfrac{13}{44}+\dfrac{-6}{44}\right)\)
\(=\dfrac{-25}{30}.\dfrac{44}{44}=\dfrac{-5}{6}.1=-\dfrac{5}{6}\)
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