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Để \(5n+19⋮n+3\)
\(\Rightarrow5n+15+4⋮n+3\)
\(\Rightarrow5\left(n+3\right)+4⋮n+3\)
Vì \(5\left(n+3\right)⋮n+3\Rightarrow4⋮n+3\Rightarrow n+3\inƯ\left(4\right)\Rightarrow n+3\in\left\{1;2;4\right\}\Rightarrow n\in\left\{-2;-1;1\right\}\)
Mà n là só tự nhiên => n = 1
Vậy n = 1
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Ok, mk sẽ làm rõ ra cho bạn !
Bài 1: 25-(45-x)=13
45-x =25-13
45-x =12
x =45-12
x =33.
Bài 2: 10+(2x-4)=16
2x-4 =16-10
2x-4 =6
2x =6+4
2x =10
x =10:2
x =5
Bài 3: 24+3(5-x)=27
3(5-x)=27-24
3(5-x)=3
5-x =3:3
5-x =1
x =5-1
x =4
Bài 1
\(25-\left(45-x\right)=13\)
\(45-x=25-13\)
\(45-x=12\)
\(x=45-12\)
\(x=33\)
Bài 2
\(10+\left(2x-4\right)=16\)
\(2x-4=16-10\)
\(2x-4=6\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
Bài 3
\(24+3\left(5-x\right)=27\)
\(27\left(5-x\right)=27\)
\(5-x=27:27\)
\(5-x=1\)
\(x=5-1\)
\(x=4\)
Chi tiết lắm rồ đóa !!!
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a: \(\Leftrightarrow\left(\dfrac{13}{4}:x\right)\cdot\left(-\dfrac{5}{4}\right)=\dfrac{-10}{6}-\dfrac{5}{6}=\dfrac{-15}{6}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{13}{4}:x=\dfrac{5}{2}\cdot\dfrac{5}{4}=\dfrac{25}{8}\)
hay \(x=\dfrac{13}{4}:\dfrac{25}{8}=\dfrac{13}{4}\cdot\dfrac{8}{25}=\dfrac{26}{25}\)
b: \(\Leftrightarrow\dfrac{3}{4}:x=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{2}{36}=\dfrac{1}{18}\)
=>\(x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{54}{4}=\dfrac{27}{2}\)
c: \(\Leftrightarrow\left(-\dfrac{6}{5}+x\right):\left(-3.6\right)=-\dfrac{7}{4}+\dfrac{1}{4}\cdot8=\dfrac{1}{4}\)
=>x-6/5=-9/10
=>x=3/10
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\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}\) \(\frac{3}{5}.\frac{2}{8}+\frac{-6}{16}.\frac{2}{5}+\frac{-6}{15}:\left(-16\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)\) \(=\frac{3}{20}+\frac{-3}{20}+\frac{1}{40}\)
\(=\frac{-5}{7}.1=\frac{-5}{7}\) \(=0+\frac{1}{40}=\frac{1}{40}\)
\(x-\frac{2}{5}=0,24\) \(\left(\frac{7}{3}x-0,6\right):3\frac{2}{5}=1\)
\(\Rightarrow x=0,24+\frac{2}{5}=\frac{16}{25}\) \(\Rightarrow\left(\frac{7}{3}x-0,6\right):\frac{17}{5}=1\)
vậy x = 16/25 \(\Rightarrow\frac{7}{3}x-0,6=\frac{17}{5}\)
\(\Rightarrow\frac{7}{3}x=\frac{17}{5}+0,6=4\)
\(\Rightarrow x=4:\frac{7}{3}=\frac{12}{7}\)
vậy x = 12/7
\(3\frac{1}{3}x+16\frac{3}{4}=-13,25\)
\(=>\frac{10}{3}x+\frac{67}{4}=\frac{-53}{4}\)
\(\frac{10}{3}x=-30\)
x = -30 : \(\frac{10}{3}\)
= > x = -9