Tìm x
\(\left(7,13+8,13\right):\left(9\frac{2}{3}-x\right)=39\)
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HN
2
30 tháng 8 2021
( 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
30 tháng 8 2021
( 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
HN
0
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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 \(x = \frac{1}{{16}}\).
11 tháng 10 2020
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
=>\(\orbr{\begin{cases}x-39=3\\x-39=-3\end{cases}}\)=>\(\orbr{\begin{cases}x=42\\x=36\end{cases}}\)
11 tháng 10 2020
a)\(\left|x-7\right|=\left|9-234\right|\)
\(\Rightarrow\left|x-7\right|=\left|-225\right|\)
\(\Rightarrow\left|x-7\right|=225\)
\(\Rightarrow x-7=\pm225\)
- \(x-7=225\Rightarrow x=232\)
- \(x-7=-225\Rightarrow x=-218\)
Vậy \(x\in\left\{232;-218\right\}\)
b)\(\left(\frac{x}{8}\right)^2+\frac{3}{16}=\frac{7}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{7}{16}-\frac{3}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{4}{16}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\frac{1}{4}\)
\(\Rightarrow\left(\frac{x}{8}\right)^2=\left(\pm\frac{1}{2}\right)^2\)
\(\Rightarrow\frac{x}{8}=\pm\frac{1}{2}\)
- \(\Rightarrow\frac{x}{8}=\frac{1}{2}\Rightarrow\frac{x}{8}=\frac{4}{8}\Rightarrow x=4\)
- \(\Rightarrow\frac{x}{8}=\frac{-1}{2}\Rightarrow\frac{x}{8}=\frac{-4}{8}\Rightarrow x=-4\)
Vậy \(x\in\left\{\pm4\right\}\)
c)\(\left(8-3^2\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow\left(8-9\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow\left(-1\right)^2+\left(x-39\right)^2=10\)
\(\Rightarrow1+\left(x-39\right)^2=10\)
\(\Rightarrow\left(x-39\right)^2=9\)
\(\Rightarrow\left(x-39\right)^2=\left(\pm3\right)^2\)
\(\Rightarrow x-39=\pm3\)
- \(\Rightarrow x-39=3\Rightarrow x=42\)
- \(\Rightarrow x-39=-3\Rightarrow x=36\)
Vậy \(x\in\left\{42;36\right\}\)
19 tháng 9 2020
Mình tách thành hai phần nhìn cho dễ hiểu nhé !
ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)
+) \(\frac{x-3\sqrt{x}}{x-9}-1=\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-1\)
\(=\frac{\sqrt{x}}{\sqrt{x}+3}-1=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{-3}{\sqrt{x}+3}\)
+) \(\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}+3}\)
\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\frac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\frac{x-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{9-x+x-9-x+4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
=> \(\frac{-3}{\sqrt{x}+3}\div\frac{4-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{-3}{\sqrt{x}+3}\times\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{4-x}\)
\(=\frac{3\left(\sqrt{x}-2\right)}{x-4}=\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3}{\sqrt{x}+2}\)
4 tháng 6 2019
bạn ơi trả lời được câu này kông
( x + 1 ) + ( x - 3 ) + ( x + 5 ) + ............ + ( x +9) = 35
\(\left(7,13+8,13\right):\left(9\frac{2}{3}-x\right)=39\)
\(15,26:\left(\frac{29}{3}-x\right)=39\)
\(\frac{29}{3}-x=15,26:39\)
\(\frac{29}{3}-x=\frac{763}{1950}\)
\(x=\frac{29}{3}-\frac{763}{1950}\)
\(x=\frac{6029}{650}\)