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a)2.3.5.13/26.35
=2.3.5.13/2.13.5.7
=3/7
b)18.6-18/-36.(-35)
=18.6-18/-(36.25)=18(6-1)/-(36.35)
=18.5/-(36.35)=2.3.3.5/-(2.2.3.3.5.7)
=-1/14
c)6.4+6.7/6.5+12=6.4+6.7/6.5+6.2
=6(4+7)/6(5+2)
=11/10
d)25.9-25.17/-8.80-8.10=25(9-17)/8(-80-10)
=25.(-8)/8(-90)
=5/18
e) 34.5-36/34.13-34
=34(5-32)/34(13-1)
=5-32/13-1=-1/3
a) \(\frac{25.9-25.17}{-8.80-8.10}=\frac{25.\left(9-17\right)}{-8.\left(80+10\right)}=\frac{25.\left(-8\right)}{-8.90}=\frac{5}{18}\)
b) \(\frac{48.12-48.15}{-3.270-3.30}=\frac{48.\left(12-15\right)}{-3.\left(270+30\right)}=\frac{48.\left(-3\right)}{-3.300}=\frac{4}{25}\)
c) \(\frac{2^5.7+2^5}{2^5.5^2-2^5.3}=\frac{2^5.\left(7+1\right)}{2^5.\left(5^2-3\right)}=\frac{2^5.8}{2^5.\left(25-3\right)}=\frac{2^5.8}{2^5.22}=\frac{4}{11}\)
d) \(\frac{3^4.5-3^6}{3^4.13+3^4}=\frac{3^4.\left(5-3^2\right)}{3^4.\left(13+1\right)}=\frac{3^4.\left(5-9\right)}{3^4.14}=\frac{3^4.\left(-4\right)}{3^4.14}=\frac{-2}{7}\)
\(b,\)Để \(\frac{x+3}{x-2}\) là số nguyên thì \(x+3⋮x-2\)
Đặt \(A=\frac{x+3}{x-2}\)
Ta có :\(A=\frac{x-2+5}{x-2}\)
\(A=1+\frac{5}{x-2}\)
Do đó : \(5⋮x-2\) hoặc \(x-2\inƯ\left(5\right)\)
\(\Rightarrow x-2\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow x\in\left\{1;-3;3;7\right\}\)
Vậy .......
a)\(\frac{5}{6}-x=-\frac{7}{12}+\frac{2}{3}\)
\(\frac{5}{6}-x=\frac{1}{12}\)
\(x=\frac{5}{6}-\frac{1}{12}\)
\(\Rightarrow x=\frac{3}{4}\)
b)\(\left(2,4x-36\right):1\frac{5}{7}=-14\)
\(\left(2,4x-36\right)=-24\)
\(2,4x=12\)
\(\Rightarrow x=5\)
c)\(\left(3\frac{1}{2}+2x\right).3\frac{2}{3}=5\frac{1}{3}\)
\(3\frac{1}{2}+2x=\frac{16}{11}\)
\(2x=-\frac{45}{22}\)
\(x=-\frac{45}{44}\)
d)\(\frac{5}{6}-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{3}{8}\)
\(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{11}{24}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{11}{24}\\\frac{1}{2}x-\frac{1}{3}=-\frac{11}{24}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{19}{12}\\x=-\frac{1}{4}\end{cases}}\)
e)\(\left|\frac{1}{4}-2x\right|-\frac{3}{4}=0\)
\(\left|\frac{1}{4}-2x\right|=\frac{3}{4}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{4}-2x=\frac{3}{4}\\\frac{1}{4}-2x=-\frac{3}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{2}\end{cases}}\)
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
a ) \(\frac{-360}{450}\)
TA có : \(\frac{-360}{450}=\frac{-4}{5}\)
b ) \(\frac{-260}{1500}\)
Ta có : \(\frac{-260}{1500}\)= \(\frac{-13}{75}\)