a: \(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left[30-4^2\right]\)

\(=20-14=6\)

b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)

\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)

\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)

c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)

\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)

\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)

d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)

\(=411-\left[\dfrac{110}{5}-4\right]\)

=410-22+4

=410-18

=392

e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)

\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)

\(=450-5\cdot\left[9\cdot8-12\right]+18\)

=468-5*60

=468-300

=168

f:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

1)(-6-25).(-6+2)=-31*(-4)=124

2) \(\frac{7\cdot3-9}{-6}=\frac{12}{-6}=-2\)

3) (-5+19).(-4)=14*(-4)=-56

4) \(\frac{872}{-6\cdot2+4}=\frac{872}{-12+4}=\frac{872}{-8}=-109\)

5) -3*7-4*(-5)+19=-21+20+19=18

6)\(18-\frac{10}{2}-71=18-5-71=-58\)

7) \(\frac{115}{-5}\cdot\left(-3\right)-8=69-8=61\)

8) \(\left(6\cdot8-\frac{20}{5}\right)+3\cdot\left(-7\right)=48-4-21=23\)

1+2+3+4+5+6+7+8+9+11+22+33+44+55+66+77+88+99=540

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8+ 9 + 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88+ 99

= ( 1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 + 9 ) + ( 11 + 22 + 33 + 44 + 55 +66 +77 + 88 + 99)

= 45                                                 +                   495

=      540

\(m.\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)\)

\(=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}\)

\(=\left(\frac{31}{23}-\frac{8}{23}\right)-\frac{7}{32}\)

\(=1-\frac{7}{32}=\frac{25}{32}\)

\(n.\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}+\left(-1\right)+1\)

\(=\frac{1}{3}\)

\(o.\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)

\(=\frac{38}{45}-\frac{8}{45}+\frac{17}{51}+\frac{3}{11}\)

\(=\left(\frac{38}{45}-\frac{8}{45}\right)+\left(\frac{17}{51}+\frac{3}{11}\right)\)

\(=\frac{2}{3}+\frac{20}{33}\)

\(=\frac{22}{33}+\frac{20}{33}=\frac{42}{33}=\frac{14}{11}\)

Cảm ơn bạn nhiề nha.

Tui không biết nha giúp tui làm

550 bạn nhé

1, <=>33+x=45
    <=>    x  = 12
Vậy x=12
2, <=>x+73 =102
    <=>   x    =29
3,x=2
4, -23+x=25<=>x=2
6, x-8=20<=>x=28

hoi vo van

đúng òi vớ va vớ vẩn

câu nầy hơn khó nên mình ko giải đc bạn đợ mình giải đã

1) 2763 + 152 = 2951

2)(-7) + (-14) = -21

1−2−3+4+5−6−7+8+...+21−22−23+24+25

= (1 - 2 - 3 + 4) + (5 - 6 - 7 + 8) + ... + (21 - 22 - 23 + 24) + 25=(1−2−3+4)+(5−6−7+8)+...+(21−22−23+24)+25

= 0 + 0 + ... + 0 + 25=0+0+...+0+25

= 25

\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)

quá đỉnh luôn