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a)
\(\frac{x-3}{10}=\frac{4}{x-3}\)
=> ( x - 3 )2 = 4 . 10.
( x - 3 )2 = 40
Mà x - 3 thuộc Z ( vì x thuộc Z ) nên ( x - 3 )2 là số chính phương.
Do 40 không là số chính phương.
=> Ko tìm được x thuộc Z thỏa mãn đề bài.
b)
\(\frac{x+5}{9}=\frac{4}{x+5}\)
=> ( x + 5 )2 = 4 . 9
( x + 5 )2 = 36
=> x + 5 = 6 hoặc x + 5 = -6.
+) x + 5 = 6
x = 1.
+) x + 5 = -6
x = -11.
Vậy x = 1; x = -11.
a) \(\frac{2}{3x}=\frac{5}{2}\Leftrightarrow3x=\frac{4}{5}\Rightarrow x=\frac{4}{15}\)
b) \(\frac{x-3}{4}=\frac{1}{2}\Rightarrow x-3=2\Rightarrow x=5\)
c) \(\frac{5}{24+x}=\frac{7}{12}\Rightarrow24+x=\frac{60}{7}\Rightarrow x=\frac{-108}{7}\)
d) \(-6x=18\Rightarrow x=-3\)
a) \(\frac{x-5}{7}=\frac{5}{3}\)
\(\Rightarrow\left(x-5\right)\cdot3=7\cdot5\)
\(\Rightarrow3x-15=35\)
\(\Rightarrow3x=15+35\)
\(\Rightarrow3x=50\)
\(\Rightarrow x=\frac{50}{3}\)
b) \(\frac{x-3}{3}=\frac{12}{x-3}\)
\(\Rightarrow\left(x-3\right)\cdot\left(x-3\right)=3\cdot12\)
\(\Rightarrow\left(x-3\right)^2=36\)
\(\Rightarrow\left(x-3\right)^2=6^2\)hoặc \(\left(x-3\right)^2=\left(-6\right)^2\)
\(\Rightarrow x-3=6\) \(x-3=-6\)
\(\Rightarrow x=6+3\) \(x=-6+3\)
\(\Rightarrow x=9\) hoặc \(x=-3\)
c) \(\frac{5-2x}{4}=\frac{7}{3}\)
\(\Rightarrow\left(5-2x\right)\cdot3=4\cdot7\)
\(\Rightarrow15-6x=28\)
\(\Rightarrow6x=15-28\)
\(\Rightarrow6x=-13\)
\(\Rightarrow x=-\frac{13}{6}\)
d) \(\frac{x+1}{4}=\frac{7}{3}\)
\(\Rightarrow\left(x+1\right)\cdot3=2\cdot7\)
\(\Rightarrow3x+3=28\)
\(\Rightarrow3x=28-3\)
\(\Rightarrow3x=25\)
\(\Rightarrow x=\frac{25}{3}\)
Chúc bạn học tốt !!!
\(\frac{3}{x}\)- 5= \(\frac{-4}{x}\)+2
\(\frac{3}{x}-\frac{-4}{x}=-5+2\)
\(\frac{7}{x}=-3\)
x=-3*7
x=-21
\(\frac{1}{5.8}\)+\(\frac{1}{8.11}\)+\(\frac{1}{11.14}\)+........+\(\frac{1}{x.\left(x+3\right)}\)=\(\frac{101}{1540}\)
3(.\(\frac{1}{5.8}+\frac{1}{8.11}\)+\(\frac{1}{11.14}+.......+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}.3=\frac{303}{1540}\)
\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+.....+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)
=>\(x+3=308\)
\(x=308-3=305\)
Vậy \(x=305\)
\(\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1540}\)
=> \(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)
=> \(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
=> \(\frac{1}{x+3}=\frac{1}{308}\)
=> x + 3 = 308
x = 308 - 5
x = 303
\(x+\frac{3}{4}=5-\frac{x}{3}\)
\(x+\frac{x}{3}=5-\frac{3}{4}\)
\(\frac{4}{3}x=\frac{17}{4}\)
\(x=\frac{17}{4}:\frac{4}{3}\)
x = 51/16
\(\frac{x+3}{4}=\frac{5-x}{3}\)
=> 3(x+3)=4(x-5)
=> 3x+9=4x-20
=> 9=4x-20-3x
=> 9=x-20
=> x=9+20
=> x=29