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\(\frac{x+4}{x}=\frac{x}{x}+\frac{4}{x}\)
\(\Rightarrow x\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
Ta có :
\(x+4⋮x\)
=> \(x+4-x⋮x\)
=> \(4⋮x\)
=> \(x\inƯ_4\)
=> x\(\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\frac{x+1+3}{x+1}=\frac{x+1}{x+1}+\frac{3}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Rightarrow x\in\left\{0;-2;2;-4\right\}\)
Ta có :
\(x+1+3⋮x+1\)
=> \(x+1+3-\left(x+1\right)⋮x+1\)
=> \(3⋮x+1\)
=> \(x+1\inƯ_3\)
=> \(x+1\in\left\{1;3;-1;-3\right\}\)
=> \(x\in\left\{0;2;-2;-4\right\}\)
\(\frac{x+5}{x+1}=\frac{x+4+1}{x+1}=\frac{x+1}{x+1}+\frac{4}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(4\right)=\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{0;1;3;-2;-3;-5\right\}\)
x+5\(⋮\)x+1
x+1+4\(⋮\)x+1
Vì x+1\(⋮\)x+1
Buộc 4\(⋮\)x+1=>x+1ϵƯ(4)={1;2;4}
Với x+1=1=>x=0
x+1=2=>x=1
x+1=4=>x=3
Vậy xϵ{0;1;3}
Ta có :
\(9⋮2x+1\)
=> \(2x+1\inƯ_9\)
=> \(2x+1\in\left\{1;3;9\right\}\)
=> \(2x\in\left\{0;2;8\right\}\)
=> \(x\in\left\{0;1;4\right\}\)
Vậy \(x\in\left\{0;1;4\right\}\)
a, \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=72:3\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=24+1\)
\(\Rightarrow x=25\)
b, \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow-x.x=-9.4\)
\(\Rightarrow-\left(x^2\right)=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
c, \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.18\)
\(\Rightarrow x.x+x.1=72\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+x-8^2+8=0\)
\(\Rightarrow x=8\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
1) \(\frac{1}{7}=\frac{8}{-x}\Rightarrow-x=7\cdot8\)
\(-x=56\)
\(x=-56\)
5) \(\left(x-2\frac{1}{4}\right):\left(-\frac{5}{6}\right)=3\)
\(x-\frac{9}{4}=3\cdot\left(-\frac{5}{6}\right)\)
\(x-\frac{9}{4}=-\frac{5}{2}\)
\(x=-\frac{5}{2}+\frac{9}{4}\)
\(x=-\frac{1}{4}\)
6) \(-8\left(4\frac{1}{5}\cdot x+\frac{3}{10}\right)=4\frac{4}{9}\)
\(-8\left(\frac{21}{5}x+\frac{3}{10}\right)=\frac{40}{9}\)
\(\frac{21}{5}x+\frac{3}{10}=\frac{40}{9}:\left(-8\right)\)
\(\frac{21}{5}x+\frac{3}{10}=-\frac{5}{9}\)
\(\frac{21}{5}x=-\frac{5}{9}-\frac{3}{10}\)
\(\frac{21}{5}x=-\frac{77}{90}\)
\(x=-\frac{77}{90}:\frac{21}{5}\)
\(x=-\frac{11}{54}\)
c) pt <=> \(x-\frac{21}{5}=\frac{23}{7}< =>x=\frac{23}{7}+\frac{21}{5}=\frac{262}{35}\)
vậy x = \(\frac{262}{35}\)
d) \(x-\frac{3}{4}=\frac{51}{8}< =>x=\frac{51}{8}+\frac{3}{4}=\frac{57}{8}\)
vậy x = \(\frac{57}{8}\)
e) pt <=> \(\frac{7}{8}:x=\frac{7}{2}< =>\frac{7}{8}.\frac{1}{x}=\frac{7}{2}< =>\frac{7}{8x}=\frac{7}{2}< =>56x=14< =>x=\frac{14}{56}=\frac{1}{4}\)
vậy x = \(\frac{1}{4}\)
a) pt <=> \(x+\frac{11}{4}=\frac{17}{3}< =>x=\frac{17}{3}-\frac{11}{4}=\frac{35}{12}\)
vậy x = \(\frac{35}{12}\)
b) pt <=> \(\frac{x.7}{2}=\frac{19}{4}< =>x=\frac{19.2}{4.7}=\frac{38}{28}=\frac{19}{14}\)
vậy x = \(\frac{19}{14}\)
Giải:
Ta có:
\(x+8⋮x+1\)
\(\Rightarrow\left(x+1\right)+7⋮x+1\)
\(\Rightarrow7⋮x+1\)
\(\Rightarrow x+1\in\left\{1;-1;7;-7\right\}\)
+) \(x+1=1\Rightarrow x=0\)
+) \(x+1=-1\Rightarrow x=-2\)
+) \(x+1=7\Rightarrow x=6\)
+) \(x+1=-7\Rightarrow x=-8\)
Vậy \(x\in\left\{0;-2;6;-8\right\}\)
\(\frac{x+8}{x+1}=\frac{x+7+1}{x+1}=\frac{x+1}{x+1}+\frac{7}{x+1}\)
\(\Rightarrow x+1\in\text{Ư}\left(7\right)=\left\{1;7;-1;-7\right\}\)
\(\Rightarrow x\in\left\{0;6;-2;-8\right\}\)