Bài 1:

\(a,22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)

=\(\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)

=\(\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)

=\(\frac{70}{4}+\frac{2}{4}-\frac{5}{4}\)

=\(\frac{67}{4}\)

\(b,1,4.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)

=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{12}{15}+\frac{10}{15}\right):\frac{11}{5}\)

=\(\frac{3}{7}-\frac{22}{15}.\frac{5}{11}\)

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

=\(-\frac{5}{21}\)

\(c,125\%.\left(-\frac{1}{2}\right)^2:\left(1\frac{5}{6}-1,6\right)+2016^0\)

=\(\frac{5}{4}.\frac{1}{4}:\left(\frac{11}{6}-\frac{8}{5}\right)+1\)

=\(\frac{5}{16}:\frac{7}{30}+1\)

=\(\frac{131}{56}\)

\(d,1,4.\frac{15}{49}-\left(20\%+\frac{2}{3}\right):2\frac{1}{5}\)

=\(\frac{7}{5}.\frac{15}{49}-\left(\frac{1}{5}+\frac{2}{3}\right):\frac{11}{5}\)

=\(\frac{3}{7}-\frac{13}{15}:\frac{11}{5}\)

=\(\frac{3}{7}-\frac{13}{33}\)

=\(\frac{8}{231}\)

Bài đ làm giống hệt như bài c

Bài 2 :

\(a,\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)

=>\(\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}=\frac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}=1\\x=\frac{1}{4}:\frac{3}{4}=\frac{1}{3}\end{matrix}\right.\)

Vậy x ∈{1;\(\frac{1}{3}\)}

\(b,\frac{5}{3}.x-\frac{2}{5}.x=\frac{19}{10}\)

=>\(\frac{19}{15}.x=\frac{19}{10}\)

=>\(x=\frac{19}{10}:\frac{19}{15}=\frac{3}{2}\)

Vậy x ∈ {\(\frac{3}{2}\)}

c,\(\left|2.x-\frac{1}{3}\right|=\frac{2}{9}\)

=>\(\left[{}\begin{matrix}2.x-\frac{1}{3}=\frac{2}{9}\\2.x-\frac{1}{3}=-\frac{2}{9}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2.x=\frac{2}{9}+\frac{1}{3}=\frac{5}{9}\\2.x=-\frac{2}{9}+\frac{1}{3}=\frac{1}{9}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{5}{9}:2=\frac{5}{18}\\x=\frac{1}{9}:2=\frac{1}{18}\end{matrix}\right.\)

Vậy x∈{\(\frac{5}{18};\frac{1}{18}\)}

\(d,x-30\%.x=-1\frac{1}{5}\)

=\(70\%x=-\frac{6}{5}\)

=\(\frac{7}{10}.x=-\frac{6}{5}\)

=>\(x=-\frac{6}{5}:\frac{7}{10}=-\frac{12}{7}\)

Vậy x∈{\(-\frac{12}{7}\)}

Bài 2

a/

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x-\frac{1}{2}=\frac{1}{4}\\\frac{3}{4}.x-\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{1}{4}+\frac{1}{2}\\\frac{3}{4}.x=-\frac{1}{4}+\frac{1}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{3}{4}.x=\frac{3}{4}\\\frac{3}{4}.x=\frac{1}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}:\frac{3}{4}\\x=\frac{1}{4}:\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=\frac{1}{3}\)

b/ Đặt x làm thừa số chung rồi tính như bình thường

c/ Tương tự câu a

d/ Tương tự câu b

a)22/1/2 .7/9 + 50% -1,25

=45/2 . 7/9 + 1/2 - 5/4

=35/2 +1/2 -5/4

= 18 - 5/4

=67/4

b)1,4 . 15/49- (4/5+ 2/3):2/1/5

=7/5 . 15/49 - (12/15 + 10/15) : 11/5

=3/7 - 22/15 : 11/5

=3/7 -2/3

=9/21 -14/21

=-5/21

c)125% . (-1/2)^2 : (1/5/6 - 1,6) + 2016^0

=5/4 . (-1/4) : (11/6- 8/5) +1

=-5/16 : (55/30 - 48/30 ) +1

=-5/16 :7/30 +1

=-75/56 + 1

= 131/56

d)1,4 .15/49 -(20% + 2/3) : 2/1/5

=7/5. 15/49 -(1/5 +2/3) :11/5

=3/7 - ( 3/15 +10/15) :11/5

=3/7 - 13/15 :11/5

=3/7 - 13/33

=8/231

đ)125%. (1/2)^2 : (1/5/6 - 1,5) +2016^0

=5/4 . 1/4 : (11/6 - 3/2)+1

=5/16 : 1/3 +1

=15/16 +1

=31/16

a) \(25^3:5^2=5^6:5^2=5^4=625\)

b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)

a) \(25^3:5^2=5^6:5^2=5^{6-2}=5^4\)

\(A=\frac{16}{10}:\frac{4}{3}=\frac{16}{10}\cdot\frac{3}{4}=\frac{6}{5}\)

\(B=\frac{14}{10}\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)

\(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)\(=\frac{3}{7}-\frac{2}{3}\)\(=\frac{-5}{21}\)

a. A= \(\frac{16}{10}:\frac{4}{3}=\frac{16}{10}.\frac{3}{4}=\frac{6}{5}\)

b. B= \(\frac{14}{10}.\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}=\frac{3}{7}-\frac{2}{3}=\frac{-5}{21}\)

a. \(25^3:5^2\)
\(=\left(5^2\right)^3:5^2\)
\(=5^6:5^2=5^4\)
b. \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21-\left(2+6\right)}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)

\(a,25^3:5^2\)

=\(\left(5^2\right)^3:5^2\)

=\(5^6:5^2\)

=\(5^4\)

\(b,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)

=\(\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)

\(=\left(\frac{3}{7}\right)^9\)

\(c,3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)

=\(3-1+\frac{1}{4}:2\)

\(=2+\frac{1}{4}\cdot\frac{1}{2}\)

\(=2+\frac{1}{8}\)

\(=\frac{17}{8}\)

\(d,\left(-\frac{7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}\)

\(=\left(-\frac{7}{4}\cdot\frac{8}{5}\right)\cdot\frac{11}{16}\)

\(=-\frac{14}{5}\cdot\frac{11}{16}\)

\(=-\frac{77}{40}\)

\(e,\frac{2}{3}+\frac{1}{3}\cdot\frac{-6}{10}\)

\(=\frac{2}{3}-\frac{1}{5}\)

\(=\frac{7}{15}\)

c. \(3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3+1+\frac{1}{4}.\frac{1}{2}\)
\(=4+\frac{1}{8}\)
\(=\frac{33}{8}\)
 

\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)

\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)

\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)

\(\left(\frac{x}{-5}+1\frac{1}{2}\right):\frac{28}{75}-1,4.\frac{15}{49}=\left|-\frac{2}{3}\right|.\left(-\frac{3}{2}\right)^3\)

\(\left(\frac{x}{-5}+\frac{3}{2}\right).\frac{75}{28}-\frac{14}{10}.\frac{15}{49}=\frac{2}{3}.\frac{-27}{8}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}-\frac{3}{7}=\frac{-9}{4}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-9}{4}+\frac{3}{7}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-63}{28}+\frac{12}{28}\)

\(\left(\frac{-x}{5}+\frac{3}{2}\right).\frac{75}{28}=\frac{-51}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}:\frac{75}{28}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-51}{28}.\frac{28}{75}\)

\(\frac{-x}{5}+\frac{3}{2}=\frac{-17}{25}\)

\(\frac{-x}{5}=\frac{-17}{25}-\frac{3}{2}\)

\(\frac{-x}{5}=\frac{-34}{50}-\frac{75}{50}\)

\(\frac{-x}{5}=\frac{-109}{50}\)

\(\frac{-10x}{50}=\frac{-109}{50}\)

Hình như đề sai thì phải

ko biết nữa

mik chỉ biết là cô cho vậy thôi

bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi

Bài nỳ tuy rất dài nhưng cũng dễ

Chí cần cậu cuyển VT sang VP rồi tìm x bình thường

Chúc cậu học tốt

\(\left(0,66-0,012:0,2\right):\left(1-1\frac{4}{4}\times0,4\right)=2\frac{9}{13}.x:\left(3,125-5,6:2\frac{2}{3}\right)\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{3}{5}:\frac{13}{35}\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{21}{13}\)

\(2\frac{9}{13}.x=\frac{861}{520}\)

\(x=\frac{123}{200}\)

1/2.x-3/5=-4/5

1/2.x=-4/5+3/5

1/2.x=-1/5

x=-1/5:1/2

x=-2/5

kl:.....

câu đầu mik tính ra sốn to lắm

câu cuối mik tính ko chia hết nên chỉ làm đc câu giữa

Mk sửa đề nha :

2020²⁰²⁰ x ( 7¹⁰ : 7⁸ - 3 x 2⁴ - 2²⁰²⁰ : 2²⁰²⁰ )

= 2020²⁰²⁰ x ( 7² - 48 - 2⁰ )

= 2020²⁰²⁰ x ( 49 - 48 - 1 )

= 2020²⁰²⁰ x 0

= 0

Study well ! >_<

Mk sửa đề nha :

2020²⁰²⁰ x ( 7¹⁰ : 7⁸ - 3 x 2⁴ - 2²⁰²⁰ : 2²⁰²⁰ )

= 2020²⁰²⁰ x ( 7² - 48 - 2⁰ )

= 2020²⁰²⁰ x ( 49 - 48 - 1 )

= 2020²⁰²⁰ x 0

= 0

Study well ! >_<