\(|x-\frac{3}{2}|+\frac{5}{4}=\frac{x}{2}\) 

\(|\frac{4x}{4}-\frac{6}{4}|+\frac{5}{4}=\frac{2x}{4}\) 

\(|4x-6|+5=2x\) 

\(|4x-6|=2x-5\) ( 1 ) 

( ĐK \(2x-5\ge0\) ) 

\(x\ge\frac{5}{2}\) 

( 1 ) \(\Leftrightarrow\orbr{\begin{cases}4x-6=2x-5\\4x-6=5-2x\end{cases}}\) 

\(\orbr{\begin{cases}2x=1\\6x=11\end{cases}}\) 

\(\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{11}{6}\end{cases}}\) ( loại cả hai ) 

Vậy không có hoặc nói cách khác là x vô nghiệm                  

   √49 + √-5^2 - 5√1,44 + 3. √4/9 

 \(=7-5-5.1,2+3.\frac{2}{3}\)

 \(=2-6+2\)

  \(=-2\)

\(\frac{a}{b}=\frac{2a}{5}\)( chia hai vế cho a )  

\(\frac{1}{b}=\frac{2}{5}\) 

\(2\cdot b=1\cdot5\) 

\(b=\frac{5}{2}\) 

\(\frac{a}{b}=\frac{2a}{5}\)

\(\frac{a}{\frac{5}{2}}=\frac{\frac{2}{5}}{a}\) 

\(a\cdot a=\frac{5}{2}\cdot\frac{2}{5}\) 

\(a^2=1\) 

\(a=\pm\sqrt{1}=\pm1\) 

\(\orbr{\begin{cases}a=1\\a=-1\end{cases}}\) 

TH1 : 

\(\frac{a}{b}=\frac{1}{\frac{5}{2}}=\frac{2}{5}\) 

TH2 : 

 \(\frac{a}{b}=\frac{-1}{\frac{5}{2}}=\frac{-2}{5}\)

Bài làm ;

Ta có :

\(B=2020-|x-1|\le2020\) ( do \(-|x-1|\le0\) )

Dấu "=" xảy ra khi \(\Leftrightarrow|x-1|=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy GTLN của B là 2020 khi x = -3 .

Học tốt nhé

Ta có 

\(|x-1|\ge0\forall x\) 

\(2020-|x-1|\le2020\) 

Dấu = xảy ra 

\(\Leftrightarrow x-1=0\) 

\(x=0+1\) 

\(x=1\) 

Vậy GTLN của B là \(2020\Leftrightarrow x=1\)

Bài 1:
a) 0,24 = 6/25
b) 0,245 = 49/200
c) 2,5324 = 5/2
d) 0,5 = 1/2

a) \(0,\left(24\right)=\frac{24}{99}=\frac{8}{33}\)

b)\(0,2\left(45\right)=\frac{245-2}{990}=\frac{243}{990}=\frac{27}{110}\)

c)\(2,5\left(324\right)=2+0,5\left(324\right)=2+\frac{5324-5}{9990}=2+\frac{197}{370}=\frac{937}{370}\)

d) \(0,5\left(3\right)=\frac{53-5}{90}=\frac{48}{90}=\frac{8}{15}\)

Bài 2 : \(M=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)

\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{16-1}{90}}{\frac{5}{2}+\frac{5}{3}-\frac{83-8}{90}}\)

\(M=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}=\frac{1}{5}\)

\(\left|x\right|+\left|-x\right|=3-x\)

\(x+x=3-x\)

\(2.x=3-x\)

\(2.x-3+x=0\)

\(2.x+x-3=0\)

\(3.x=0+3\)

\(3.x=3\)

\(x=3:3\)

\(x=1\)

Vậy \(x=1\)

ko biết đâu

\(x^2-\frac{7}{6}x+\frac{1}{3}=0\)

=> \(\left[x^2-2\cdot x\cdot\frac{7}{12}+\left(\frac{7}{12}\right)^2\right]-\frac{1}{144}=0\)

=> \(\left(x-\frac{7}{12}\right)^2-\frac{1}{144}=0\)

=> \(\left(x-\frac{7}{12}\right)^2-\left(\frac{1}{12}\right)^2=0\)

=> \(\left(x-\frac{7}{12}+\frac{1}{12}\right)\left(x-\frac{7}{12}-\frac{1}{12}\right)=0\)

=> \(\left(x-\frac{1}{2}\right)\left(x-\frac{2}{3}\right)=0\)

=> x = 1/2 hoặc x = 2/3

Bài làm :

\(\frac{125^6.3^{61}.8^{10}}{4^{15}.25^9.9^{30}}\)

\(=\frac{\left(5^3\right)^6.3^{61}.\left(2^3\right)^{10}}{\left(2^2\right)^{15}.\left(5^2\right)^9.\left(3^2\right)^{30}}\)

\(=\frac{5^{18}.3^{61}.2^{30}}{2^{30}.5^{18}.3^{60}}\)

\(=3\)

Học tốt nhé

        Bài làm :

Ta có :

\(\frac{125^6.3^{61}.8^{10}}{4^{15}.25^9.9^{30}}\)

\(=\frac{\left(5^3\right)^6.3^{61}.\left(2^3\right)^{10}}{\left(2^2\right)^{15}.\left(5^2\right)^9.\left(3^2\right)^{30}}\)

\(=\frac{5^{18}.3^{61}.2^{30}}{2^{30}.5^{18}.3^{60}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

               Bài làm :

Ta có hình vẽ :

Ta có :

\(\hept{\begin{cases}\widehat{AOD}+\widehat{BOC}=100^o\\\widehat{AOD}=\widehat{BOC}\left(\text{2 góc đối đỉnh}\right)\end{cases}\Rightarrow\widehat{AOD}=\widehat{BOC}=\frac{100}{2}=50^O}\)

\(\Rightarrow\widehat{BOD}=\widehat{COA}=180-50=130^O\)

Vì \(\widehat{AOD}=\widehat{BOC}\)(2 góc đối đỉnh) mà \(\widehat{AOD}+\widehat{BOC}=100^0\Rightarrow\widehat{AOD}=\widehat{BOC}=\frac{100^0}{2}=50^0\)

Tương tự: \(\widehat{AOC}=\widehat{BOD}\)(2 góc đối đỉnh) mà \(\widehat{AOC}+\widehat{AOD}=180^0\)(2 góc kề bù)

\(\Rightarrow\widehat{BOD}=\widehat{AOC}=180^0-50^0=130^0\)