Tìm x:
\(x^2=\frac{5}{7}x\)
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\(bn\)\(xem\)\(lai\)\(giup\)\(mk\)\(cho\)\(\frac{x+522}{7}\)\(neu\)\(thay\)\(bang\)\(\frac{x+552}{7}\)\(thi\)\(dug\)\(hon\)
thế thì bạn giải thử xem cô t ra đề thế mà ừ thì cứ cho là x + 552 cx đc
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
8 - 6x - 10x - 7 = 4
-16x + 1 = 4
-16x = 4 - 1 = 3
x = \(-\frac{3}{16}\)
2(4 - 3x) - 5(2x - 7) = -(-4)
=> 8 - 6x - 10x + 35 = 4
=> -16x + 43 = 4
=> -16x = 4 - 43
=> -16x = -39
=> x = -39 : (-16)
=> x = 39/16
a -5<x<-5 + 15
-5<x<10
=>x thuộc ( -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9)
b |x|< 2
Mà |x|> hoạc = 0 => x thuộc tập hợp 0 1=> x thuộc tập hợp 0 1 -1
a, \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{7}{6}\)
\(\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=\frac{47}{30}\)
\(\frac{3}{2}-\frac{4}{15}x=\frac{47}{50}\)
\(\frac{4}{15}x=\frac{14}{25}\)
\(x=\frac{21}{10}\)
\(\left(2x-3\right)^2=\left(x+7\right)^2\)
<=> \(2x-3=x+7\)
<=> \(x=10\)
Vậy \(x=10\)
\(\left(2x-3\right)^2=\left(x+7\right)^2\)
\(\Leftrightarrow\left|2x-3\right|=\left|x+7\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+7\\2x-3=-7-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=10\\x=\frac{-4}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{-4}{3};10\right\}\)
1.Vì \(\frac{x}{-2}=\frac{-8}{x}\Rightarrow-2.\left(-8\right)=x.x\)
\(16=x.x\)hay \(4^2=x^2\Rightarrow x=4\)
2. Rút gọn : \(\frac{20}{28}=\frac{5}{7}=\frac{-5}{-7}\)
\(\Rightarrow x=-7\)
3. \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)
\(\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)
Mà \(\frac{8}{7}+\frac{11}{5}=\frac{502}{35}\)
\(\Rightarrow x=\frac{234}{35}\)
1) \(\frac{x}{-2}=\frac{-8}{x}\)
\(\Rightarrow x\times x=\left(-2\right)\times\left(-8\right)\)
\(\Rightarrow x^2=16\)
\(\Rightarrow\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
Vậy x = 4 hoặc x = -4
2) \(\frac{-5}{x}=\frac{20}{28}\)
\(\Rightarrow\frac{-5}{x}=\frac{5}{7}\)
\(\Rightarrow5\times x=\left(-5\right)\times7\)
\(\Rightarrow5\times x=-35\)
\(\Rightarrow x=\left(-35\right):5\)
\(\Rightarrow x=-7\)
Vậy x = -7
3) \(\frac{x}{2}-\frac{11}{5}=\frac{7}{8}\times\frac{64}{49}\)
\(\Rightarrow\frac{x}{2}-\frac{11}{5}=\frac{8}{7}\)
\(\Rightarrow\frac{x}{2}=\frac{8}{7}+\frac{11}{5}\)
\(\Rightarrow\frac{x}{2}=\frac{117}{35}\)
\(\Rightarrow35x=117\times2\)
\(\Rightarrow35x=234\)
\(\Rightarrow x=234:35\)
\(\Rightarrow x=\frac{234}{35}\)
Vậy \(x=\frac{234}{35}\)
4) \(\frac{x}{5}+\frac{9}{2}=\frac{6}{7}\times\frac{36}{48}\)
\(\Rightarrow\frac{x}{5}+\frac{9}{2}=\frac{9}{14}\)
\(\Rightarrow\frac{x}{5}=\frac{9}{14}-\frac{9}{2}\)
\(\Rightarrow\frac{x}{5}=\frac{-27}{7}\)
\(\Rightarrow7x=\left(-27\right)\times5\)
\(\Rightarrow7x=-135\)
\(\Rightarrow x=\left(-135\right):7\)
\(\Rightarrow x=\frac{-135}{7}\)
Vậy \(x=\frac{-135}{7}\)
5) \(\frac{3}{x-5}=\frac{-4}{x+2}\)
\(\Rightarrow\frac{3}{x-5}+\frac{4}{x+2}=0\)
\(\Rightarrow3\left(x+2\right)+4\left(x-5\right)=0\)
\(\Rightarrow3x+6+4x-20=0\)
\(\Rightarrow\left(3x+4x\right)+\left(6-20\right)=0\)
\(\Rightarrow7x-14=0\)
\(\Rightarrow7x=14\)
\(\Rightarrow x=14:7\)
\(\Rightarrow x=2\)
Vậy x = 2
_Chúc bạn học tốt_
a ) Ta có : - 12 . ( x - 5 ) + 7( 3 - x ) = 5
Suy ra - 12x - ( - 12 ) . 5 + 7 . 3 - 7x = 5
Suy ra - 12x + 60 + 21 - 7x = 5
Suy ra - 12x - 7x = 5 - 60 - 21
Suy ra - 19x = - 76
Suy ra x = -76 : ( - 19 )
Vậy x = 4
b ) Ta có : 30 . ( x + 2 ) - 6 . ( x - 5 ) - 24x = 100
Suy ra 30x + 30 . 2 - 6x - ( - 6 ) . 5 - 24x = 100
Suy ra 30x + 60 - 6x + 30 - 24x = 100
Suy ra 30x - 6x - 24x = 100 - 60 - 30
Suy ra 0x = 10
Vậy x = 0
\(x^2=\frac{5}{7}x\Leftrightarrow x^2-\frac{5}{7}x=0\)
\(\Leftrightarrow x\left(x-\frac{5}{7}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}\)
\(x^2=\frac{5}{7}x\)
\(\Rightarrow x^2-\frac{5}{7}x=0\)
\(\Rightarrow x\left(x-\frac{5}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-\frac{5}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{7}\end{cases}}\)