b)  ( 2/5 )^ x-1 : ( 2/5 )^2 = ( 2/5 )^3

     (2/5)^ x-1                  = ( 2/5) ^3 . (2/5)^2

     (2/5)^ x-1                  = ( 2/5 ) ^ 6

     => x-1                      = 6

     => x= 7

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

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\cdot\cdot\cdot\left(\frac{1}{2009}-1\right)\)

\(=\frac{-1}{2}\cdot\frac{-2}{3}\cdot\cdot\cdot\cdot\frac{-2008}{2009}\)

\(=\frac{\left(-1\right)\cdot\left(-2\right)\cdot\cdot\cdot\left(-2008\right)}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1\cdot2\cdot\cdot\cdot2008}{2\cdot3\cdot\cdot\cdot2009}\)

\(=\frac{1}{2009}\)

1,

\(| x - \frac{2}{7} | = \frac{-1}{5}.\frac{-5}{7}\)

\(|x- \frac{2}{7}|=\frac{1}{7}\)

<=> \(x- \frac{2}{7} = \frac{1}{7} => x= \frac{3}{7} \)

Và \(x - \frac{2}{7} =\frac{-1}{7} => x= \frac{1}{7}\)

Học tốt

\(5^{61}+25^{31}+125^{21}\)

\(=5^{61}+\left(5^2\right)^{31}+\left(5^3\right)^{21}\)

\(=5^{61}\cdot5^{2\cdot31}\cdot5^{3\cdot21}\)

\(=5^{61}+5^{62}+5^{63}\)

\(=5^{61}\cdot\left(1+5+5^2\right)\)

\(=5^{61}\cdot\left(6+5^2\right)\)

\(=5^{61}\cdot\left(6+25\right)\)

\(=5^{61}\cdot31\)

Vì \(5^{61}\inℤ\)

\(\Rightarrow5^{61}\cdot31⋮31\)

\(\Rightarrow5^{61}+25^{31}+125^{21}⋮31\)

Vậy bài toán đã được chứng minh . 

Bài 2:

a) \(\frac{4}{9}+x=\frac{-5}{3}\)

\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)

\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)

Vậy: \(x=\frac{-19}{9}\)

b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)

\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)

\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)

c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)

\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-5\right\}\)

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=3\sqrt{3}\)

vậy \(A=3\sqrt{3}\)

\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)  \(ĐKXĐ:x>0;x\ne1\)

\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)

\(B=1-x\)

vậy \(B=1-x\)

\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)

\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)

\(C=4+5+6\)

\(C=15\)

vậy \(C=15\)

Cho mk giải câu a:

\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)

\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)

\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)

\(A=\left(18-10-5\right)\sqrt{3}\)

\(A=3\sqrt{3}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

Hình như đề sai thì phải

ko biết nữa

mik chỉ biết là cô cho vậy thôi

a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)

\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)

\(\Leftrightarrow3x-1=3\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

b) Đề sai ! Sửa :

\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)

\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)

\(\Leftrightarrow2x+5=4\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\frac{1}{2}\)

c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)

\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)

\(\Leftrightarrow x+3=\frac{4}{3}\)

\(\Leftrightarrow x=-\frac{5}{3}\)

d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)

\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)

\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{4}{15}\)