( 7,13 + 8,13 ) : (9 2/3 - x) =39

763/50:(29/3-x)=39

           (29/3-x)=763/50:39

           29/3-x=763/1950

                  x=29/3-763/1950

                  x=6029/650

( 7,13 + 8,13 ) : ( \(9\frac{2}{3}\)- x ) = 39

15,26 : ( 29/3 -x ) = 39

            ( 29/3 - x ) = 15,26 : 39

            ( 29/3 - x ) = 0,391

                        x   =  29/3 - 0,391

                        x   = 9.27566666666667

a)

\(\begin{array}{l}x:{\left( {\frac{{ - 1}}{2}} \right)^3} =  - \frac{1}{2}\\x =  - \frac{1}{2}.{\left( {\frac{{ - 1}}{2}} \right)^3}\\x = {\left( {\frac{{ - 1}}{2}} \right)^4}\\x = \frac{1}{{16}}\end{array}\)              

Vậy \(x = \frac{1}{{16}}\).

 b)

\(\begin{array}{l}x.{\left( {\frac{3}{5}} \right)^7} = {\left( {\frac{3}{5}} \right)^9}\\x = {\left( {\frac{3}{5}} \right)^9}:{\left( {\frac{3}{5}} \right)^7}\\x = {\left( {\frac{3}{5}} \right)^2}\\x = \frac{9}{{25}}\end{array}\)

Vậy \(x = \frac{9}{{25}}\).

c)

\(\begin{array}{l}{\left( {\frac{{ - 2}}{3}} \right)^{11}}:x = {\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^{11}}:{\left( {\frac{{ - 2}}{3}} \right)^9}\\x = {\left( {\frac{{ - 2}}{3}} \right)^2}\\x = \frac{4}{9}.\end{array}\)         

Vậy \(x = \frac{4}{9}\).

d)

\(\begin{array}{l}x.{\left( {0,25} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x.{\left( {\frac{1}{4}} \right)^6} = {\left( {\frac{1}{4}} \right)^8}\\x = {\left( {\frac{1}{4}} \right)^8}:{\left( {\frac{1}{4}} \right)^6}\\x = {\left( {\frac{1}{4}} \right)^2}\\x = \frac{1}{{16}}\end{array}\)

Vậy \(x = \frac{1}{{16}}\).

a, |x-7|=|9-234|

=> |x-7|=|-225|

=> |x-7|=225

=>\(\orbr{\begin{cases}x-7=225\\x-7=-225\end{cases}}\)=>\(\orbr{\begin{cases}x=232\\x=-218\end{cases}}\)

b, \(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

=>\(\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

=>\(\left(\frac{x}{8}\right)^2=\left(\frac{1}{2}\right)^2\)

=>\(\orbr{\begin{cases}\frac{x}{8}=\frac{1}{2}\\\frac{x}{8}=\frac{-1}{2}\end{cases}}\)=>\(\orbr{\begin{cases}x.2=8\\x.2=-8\end{cases}}\)=>\(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

c, (8-3²)²+(x-39)²=10

=>(-1)²+(x-39)²=10

=>(x-39)²=10-1

=>(x-39)²=9

=>(x-39)²=3²

=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)

a)\(\left|x-7\right|=\left|9-234\right|\)

\(\Rightarrow\left|x-7\right|=\left|-225\right|\)

\(\Rightarrow\left|x-7\right|=225\)

\(\Rightarrow x-7=\pm225\)

Vậy \(x\in\left\{232;-218\right\}\)

b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)

\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)

\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)

Vậy \(x\in\left\{\pm4\right\}\)

c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)

\(\Rightarrow1+\left(x-39\right)^2=10\)

\(\Rightarrow\left(x-39\right)^2=9\)

\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow x-39=\pm3\)

Vậy \(x\in\left\{42;36\right\}\)

Mình tách thành hai phần nhìn cho dễ hiểu nhé !

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)

\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)

+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{9-x+x-9-x+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)

\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)

x = từ 1 đến 10000....0

bạn ơi trả lời được câu này kông

( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35