\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(3-x\right)-x\left(3+x\right)=8x-6\)

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

\(\Leftrightarrow-x^2-x^2+3x-2x-3x-8x=-6+6\)

\(\Leftrightarrow-2x^2-10x=0\)

\(\Leftrightarrow-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;5\right\}\)

\(\frac{-5}{6}\)\(+\)\(\frac{4}{9}\)\(\times\)\(\left(\frac{5}{4}-\frac{2}{3}\right)\)\(\times\)\(\left(-3\right)^2\)\(+\)\(\frac{5}{9}\)\(\times\)\(30\%\)

\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{4}{9}\)\(\times\)\(\frac{7}{12}\)\(\times\)\(9\)\(+\)\(\frac{5}{9}\)\(\times\)\(\frac{3}{10}\)

\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{7}{3}\)\(+\)\(\frac{5}{9}\)\(\times\)\(\frac{3}{10}\)

\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{7}{3}\)\(+\)\(\frac{1}{6}\)

\(=\)\(\frac{-5}{6}\)\(+\)\(\frac{1}{6}\)\(+\)\(\frac{7}{3}\)

\(=\)\(\frac{-2}{3}\)\(+\)\(\frac{7}{3}\)

\(=\)\(\frac{5}{3}\)

a) \(\dfrac{13}{15}< \dfrac{14}{15}\)  ( vì 13 < 14 )

b) Ta có:

\(\dfrac{2}{3}=\dfrac{2\times3}{3\times3}=\dfrac{6}{9}\)

Vì 6 = 6 nên \(\dfrac{6}{9}=\dfrac{6}{9}\)

Vậy \(\dfrac{2}{3}=\dfrac{6}{9}\)

c) Ta có:

\(\dfrac{4}{7}=\dfrac{4\times6}{7\times6}=\dfrac{24}{42}\)

\(\dfrac{5}{6}=\dfrac{5\times7}{6\times7}=\dfrac{35}{42}\)

Vì 24 < 35 nên \(\dfrac{24}{42}< \dfrac{35}{42}\)

Vậy \(\dfrac{4}{7}< \dfrac{5}{6}\)

d) Vì 9 < 13 nên \(\dfrac{4}{9}>\dfrac{4}{13}\) 

Vậy \(\dfrac{4}{9}>\dfrac{4}{13}\)

1)<
2)=
3)<
4)>

a: 14/5-7/5=7/5

b: 7/8-1/3+5/4

=21/24-8/24+30/24

=43/24

c; =7/6+5/6+2/15+13/15

=2+1

=3

d: =4*5/3*11=20/33

e: =2/9*1/6*1/4=2/9*1/24=1/108

2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)

b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)

c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)

14/5-7/5=7/5 

7/8-1/3+5/4=13/24+5/4=43/24 

7/6+2/15+5/6+13/15=13/10+5/6+13/15=32/15+13/15=3

4/3*5/11=20/35

2/9:6*1/4=1/27*1/4=1/108

Xin lỗi m.n nhé gửi nhầm tí

a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

 =  \(\dfrac{5}{6}\)  + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)

= 1 + \(\dfrac{1}{23}\)

 = \(\dfrac{24}{23}\) 

b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)

= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5

= 1 - 1 + 0,5

= 0,5 

c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)

=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)

=0

d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)

= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)

= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)

= 1

e, 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)

 =    (0,5 + 0,4) + ( \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + (\(\dfrac{5}{7}\)  - \(\dfrac{4}{35}\))

 = 0,9 + \(\dfrac{1}{2}\) + \(\dfrac{3}{5}\)

= 0,9 + 0,5 + 0,6

= 2

#include <bits/stdc++.h>

using namespace std;

long long a[1000],n,i,tam,j;

int main()

{

cin>>n;

for (i=1; i<=n; i++)

cin>>a[i];

for (i=1; i<=n-1; i++)

for (j=i+1; j<=n; j++)

if (a[i]>a[j]) swap(a[i],a[j]);

for (i=1; i<=n; i++)

cout<<a[i]<<" ";

return 0;

}

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