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Bài 4:
a; \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) = \(\dfrac{5}{20}\) - \(\dfrac{4}{20}\) = \(\dfrac{1}{20}\)
b; \(\dfrac{3}{5}\) - \(\dfrac{-1}{2}\) = \(\dfrac{6}{10}\) + \(\dfrac{5}{10}\) = \(\dfrac{11}{10}\)
c; \(\dfrac{3}{5}\) - \(\dfrac{-1}{3}\) = \(\dfrac{9}{15}\) + \(\dfrac{5}{15}\) = \(\dfrac{14}{15}\)
d; \(\dfrac{-5}{7}\) - \(\dfrac{1}{3}\)= \(\dfrac{-15}{21}\) - \(\dfrac{7}{21}\)= \(\dfrac{-22}{21}\)
Bài 5
a; 1 + \(\dfrac{3}{4}\) = \(\dfrac{4}{4}\) + \(\dfrac{3}{4}\) = \(\dfrac{7}{4}\) b; 1 - \(\dfrac{1}{2}\) = \(\dfrac{2}{2}\) - \(\dfrac{1}{2}\) = \(\dfrac{1}{2}\)
c; \(\dfrac{1}{5}\) - 2 = \(\dfrac{1}{5}\) - \(\dfrac{10}{5}\) = \(\dfrac{-9}{5}\) d; -5 - \(\dfrac{1}{6}\) = \(\dfrac{-30}{6}\) - \(\dfrac{1}{6}\) = \(\dfrac{-31}{6}\)
e; - 3 - \(\dfrac{2}{7}\)= \(\dfrac{-21}{7}\) - \(\dfrac{2}{7}\)= \(\dfrac{-23}{7}\) f; - 3 + \(\dfrac{2}{5}\) = \(\dfrac{-15}{5}\) + \(\dfrac{2}{5}\)= - \(\dfrac{13}{5}\)
g; - 3 - \(\dfrac{2}{3}\) = \(\dfrac{-9}{3}\) - \(\dfrac{2}{3}\) = \(\dfrac{-11}{3}\) h; - 4 - \(\dfrac{-5}{7}\) = \(\dfrac{-28}{7}\)+ \(\dfrac{5}{7}\) = - \(\dfrac{23}{7}\)
Bài 5:
a. Gọi $d=ƯCLN(n-2, n+1)$
$\Rightarrow n-2\vdots d; n+1\vdots d$
$\Rightarrow (n+1)-(n-2)\vdots d$
$\Rightarrow 3\vdots d\Rightarrow d\in \left\{1; 3\right\}$
Để ps tối giản thì $n-2\not\vdots 3$
$\Leftrightarrow n\neq 3k+2$ với $k$ là số tự nhiên bất kỳ.
b.
Gọi $d=ƯCLN(n+5, n-2)$
$\Rightarrow n+5\vdots d; n-2\vdots d$
$\Rightarrow (n+5)-(n-2)\vdots d$
$\Rightarrow 7\vdots d$
$\Rightarrow d\in \left\{1; 7\right\}$
Để ps tối giản thì $n-2\not\vdots 7$
$\Rightarrow n\neq 7k+2$ với $k$ là số tự nhiên bất kỳ.
Bài 2:
\(\dfrac{12}{-24}=\dfrac{12:12}{-24:12}=\dfrac{1}{-2}\)
\(\dfrac{-39}{75}=\dfrac{-39:3}{75:3}=\dfrac{-13}{25}\)
\(\dfrac{132}{-264}=\dfrac{132:132}{-264:132}=\dfrac{1}{-2}\)
Bài 3:
\(\dfrac{1}{-2}=\dfrac{-1}{2};\dfrac{-3}{-5}=\dfrac{3}{5};\dfrac{2}{-7}=\dfrac{-2}{7}\)
Bài 4:
\(15p=\dfrac{1}{4}h;20p=\dfrac{1}{3}h;45p=\dfrac{3}{4}h;50p=\dfrac{5}{6}h\)
Lời giải:
\(E=\frac{\frac{2013}{1}.\frac{2014}{2}.\frac{2015}{3}....\frac{3012}{1000}}{\frac{1001}{1}.\frac{1002}{2}.\frac{1003}{3}....\frac{3012}{2012}}\\ =\frac{2013.2014.2015....3012}{1001.1002.1003....3012}.\frac{1.2.3...2012}{1.2.3..1000}\\ =\frac{1}{1001.1002...2012}.(1001.1002....2012)=1\)
Có quá nhiều bài, thứ nhất em đăng tách ra, thứ hai chụp gần cận cho rõ, thứ ba em chỉ đăng bài cần giúp
\(a,MSC:180\\ Có:-5=\dfrac{-5.180}{180}=\dfrac{-900}{180};\dfrac{17}{-20}=\dfrac{17.\left(-9\right)}{\left(-9\right).\left(-20\right)}=\dfrac{-153}{180};\dfrac{-16}{9}=\dfrac{-16.20}{9.20}=\dfrac{-320}{180}\\ ---\\ b.MSC:75\\ Có:\dfrac{13}{-15}=\dfrac{13.\left(-5\right)}{\left(-15\right).\left(-5\right)}=\dfrac{-65}{75};\dfrac{-18}{25}=\dfrac{-18.3}{25.3}=\dfrac{-54}{75};-3=\dfrac{-3.75}{75}=\dfrac{-225}{75}\)
Bài 1:
a; 24 ⋮ \(x\); 30 ⋮ \(x\); 48 \(⋮\) \(x\) và \(x\) lớn nhất.
vì 24 \(⋮\) \(x\); 30 ⋮ \(x\); 48 ⋮ \(x\) ⇒ \(x\) \(\in\) ƯC(24; 30; 48)
Vì \(x\) là lớn nhât nên \(x\) \(\in\) ƯCLN(24; 30; 48)
24 = 22.33; 30 = 2.3.5; 48 = 24.3
ƯCLN(24; 30; 48) = 2.3 = 6
⇒ \(x\) = 6
Vậy \(x\) = 6
b; 120 ⋮ \(x\); 180 ⋮ \(x\); 30 ⋮ \(x\)
⇒ \(x\) \(\in\) ƯC(120; 180; 390)
120 = 23.3.5; 180 = 22.32.5; 390 = 2.3.5.13
ƯC(120; 180; 390) = 2.3.5 = 30
⇒ \(x\in\) Ư(30) = {1; 2; 3; 5; 6; 10;15; 30}
Vì 5 ≤ \(x\) ≤ 15 nên \(x\) \(\in\) {5; 6; 10; 15}
Bài 2:
\(x^2-xy-y+2=0\)
=>\(x^2-1+3-y\left(x+1\right)=0\)
=>\(\left(x-1\right)\left(x+1\right)-y\left(x+1\right)=-3\)
=>(x+1)(x-1-y)=-3
=>\(\left(x+1;x-1-y\right)\in\left\{\left(1;-3\right);\left(3;-1\right);\left(-1;3\right);\left(-3;1\right)\right\}\)
TH1: \(\left\{{}\begin{matrix}x+1=1\\x-1-y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0\\y=x-1-\left(-3\right)=x-1+3=x+2=0+2=2\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+1=3\\x-1-y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=x-1+1=x=2\end{matrix}\right.\)
TH3: \(\left\{{}\begin{matrix}x+1=-1\\x-1-y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-2\\y=x-1-3=x-4=-2-4=-6\end{matrix}\right.\)
TH4: \(\left\{{}\begin{matrix}x+1=-3\\x-1-y=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-4\\y=x-1-1=x-2=-4-2=-6\end{matrix}\right.\)