Bài 1 :

\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)

\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)

\(\Rightarrow M< N\)

Bài 3 :

a) \(t^2+5t-8\) khi \(t=2\)

\(=5^2+2.5-8\)

\(=25+10-8\)

\(=27\)

b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)

\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)

\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)

c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)

\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)

\(\left(1\right)=1^3=1\)

Bài 2 :

a) \(...=2^3\left(19-14\right)+1=8.5+1=41\)

b) \(...=100-\left[60:\left(5^2-15\right)\right]=100-\left[60:10\right]=100-6=94\)

c) \(...=160:\left[17+\left(9.5-\left(14+2^3\right)\right)\right]=160:\left[17+\left(45-22\right)\right]=160:\left[17+23\right]=160:40=4\)

d) \(...=798+100\left[16-2\left(25-22\right)\right]=798+100\left[16-2.3\right]=798+100.10=798+1000=1798\)

Lời giải:
a.

$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$

b.

$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$

c.

$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$

$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$

d.

$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$

$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$

e.

$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$

$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$

mình cần gấp

bài 1 :

a)\(2,5.16,27.4+7,3=\left(2,5.4\right).16,27+7,3=10.16,27+7,3\)

\(=162,7+7,3=170\)

b) \(\frac{2}{3}+\frac{3}{4}-\frac{5}{6}=\left(\frac{2}{3}-\frac{5}{6}\right)+\frac{3}{4}=\frac{-1}{6}+\frac{3}{4}=\frac{7}{12}\)

c) \(17,6-5,3+16,8-7,6+15,3-6,8\)

\(=\left(17,6-7,6\right)+\left(-5,3+15,3\right)+\left(16,8-6,8\right)\)

\(=10+10+10=30\)

d)\(\frac{38}{11}+\left(13,16+\frac{6}{11}\right)=\frac{38}{11}+13,16+\frac{6}{11}\)

\(=\left(\frac{38}{11}+\frac{6}{11}\right)+13,16=4+13,16=17,16\)

\(2,45.46+8.0,75+54.2,45+0,5.8\\ =\left[2,45.46+54.2,45\right]\)

\(+\left[8.0,75+0,5.8\right]=\left[2,45.\left(46+54\right)\right]+\left[8.\left(0,75+0,5\right)\right]\)

\(=\left(2,45.100\right)+\left(8.1,25\right)=245+10=255\)

 bài 2 :

a) .....

b) \(1-\left(12,5+x-4,25\right):21,75=0\)

\(\Rightarrow1-\left(12,5+x-4,25\right)=21,75\)

\(\Rightarrow12,5+x-4,25=-20,75\\ \Rightarrow12,5+x=-16,5\)

\(\Rightarrow x=-29\)

cậu có thể tham khảo bài làm trên đây ạ, mn ai thấy đúng thì cho mk xin 1 t.i.c.k đúng ạ ,thank nhiều

mk ko thấy đề

c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)

\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)

\(=1-2=-1\)

Giải:

a)-1/12+4/3=-1/12+16/12=15/12=5/4

b)(-4/14-3/15)-(1/5-20/35-(-1)).7

=-17/35-22/35.7

=-17/35-22/5

=-171/35

c)3/5+-5/20+30/75+-7/4

=3/5+-1/4+2/5+-7/4

=(3/5+2/5)+(-1/4+-7/4)

=1+-2

=-1

d)5/6.-12/14+7/13

=-5/7+7/13

=-16/91

e)2/-9-5/-36-1/4

=-1/12-1/4

=-1/3

f)2/23+-5/12+7/18+21/23+-7/12

=(2/23+21/23)+(-5/12+-7/12)+7/18

=1+-1+7/18

=7/18

Giải:

a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\) 

\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\) 

\(=\dfrac{136}{9}-\dfrac{11}{2}\) 

\(=\dfrac{173}{18}\) 

b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\) 

\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\) 

\(=\dfrac{13}{9}.\dfrac{3}{4}\) 

\(=\dfrac{13}{12}\) 

c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\) 

\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\) 

\(=1+1\) 

\(=2\)

d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\) 

\(=\dfrac{53}{10}\) 

e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-2}{5}\) 

\(=1\)

Ta có : 

\(A=\frac{1}{2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(\Rightarrow A=\frac{1}{4}-\frac{1}{760}< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{380}\right)=\frac{1}{2}\left(\frac{189}{380}\right)=\frac{189}{760}< \frac{1}{4}\)

Ta có: \(A=\frac{1}{2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{18\times19\times20}\)

              \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+...+\frac{1}{18\times19}-\frac{1}{19\times20}\right)\)

               \(=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{19\times20}\right)\)

                 \(=\frac{1}{2}\times\frac{1}{1\times2}-\frac{1}{2}\times\frac{1}{19\times20}\)

                   \(=\frac{1}{4}-\frac{1}{2}\times\frac{1}{19\times20}< \frac{1}{4}\)

Vậy A < 1/4