Giúp mk vs
Tìm x
a, ( x - 5 ) ^4 = ( x - 5 ) ^ 6
b , ( 2x - 15 ) ^ 5 = ( 2x -15 ) ^ 3
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a)\(x-15\%x=\frac{1}{3}\)
\(x.\left(1-15\%\right)=\frac{1}{3}\)
\(x.\frac{-280}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{-280}{3}\)
\(x=\frac{-1}{280}\)
Vậy \(x=\frac{-1}{280}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{6}{5}=\frac{1}{2}-\frac{4}{3}\)
\(-\frac{17}{10}x+\frac{6}{5}=\frac{-5}{6}\)
\(-\frac{17}{10}x=-\frac{5}{6}-\frac{6}{5}\)
\(-\frac{17}{10}x=\frac{-61}{30}\)
\(x=\frac{-61}{30}:\frac{-17}{10}\)
\(x=\frac{61}{51}\)
Vậy \(x=\frac{61}{51}\)
15 - 3 . (-6) - 2x = 4 . (-2) + 3x - 32
<=> 15 + 18 - 2x = -8 + 3x - 9
<=> -2x - 3x = -15 - 18 - 8 - 9
<=> - 5x = -50
<=> x = 10
Vậy ...
b ) - 8 + 30(x+2)-6(x-5)-24 = 10
<=> - 8 + 30x + 60 - 6x + 30 - 24 = 10
<=> 30x - 6x = 8 - 60 - 30 + 24 + 10
<=> 24x = -48
<=> x = - 2
Vậy ...
b. x^10 = x
x^10 = x^1
x^10 - x^1 = 0
x^1.x^9 - x^1.1 = 0
x^1 . (x^9 - 1 ) = 0
x^1 = 0 hoặc x^9 - 1 = 0
x^1 = 0 = 0^1 x^9 = 0 + 1
=> x = 0 x^9 = 1
x^9 =1^9
=> x = 1
Vậy x = 0 ; 1
=> là suy ra
. là dấu nhân
\(1,\)
\(2x\left(x-3\right)-\left(3-x\right)=0\)
\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
\(2,\)
\(3x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
\(3,\)
\(x^4-x^2=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(4,\)
\(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(5,\)
\(x\left(x+6\right)-10\left(x-6\right)=0\)
\(\Leftrightarrow x^2+6x-10x+60=0\)
\(\Leftrightarrow x^2-4x+60=0\)
\(\Leftrightarrow x^2-4x+4+56=0\)
\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)
=> Phương trình vô nghiệm
a, <=> 6x-3-5x-5-4.5=0
<=> x=28
b, 5|3x+1|-4|3x+1|=19
<=> |3x+1|=19
<=>\(\orbr{\begin{cases}3x+1=19\\3x+1=-19\end{cases}}\)
<=>\(\orbr{\begin{cases}x=6\\x=\frac{-20}{3}\end{cases}}\)
Chúc hok tốt!!
a/(x-5)^4=9x-5)^6\(\Rightarrow\) TH1:x-5=1\(\rightarrow\) x=6
\(\Rightarrow\) TH2:x-5=-1\(\rightarrow\) x=4
\(\Rightarrow\) TH3:x-5=0\(\rightarrow\) x=5
vậy....................
b/(2x-15)^5=(2x-15)^6\(\Rightarrow\) 2x-15=1\(\rightarrow\) 2x=16\(\rightarrow\)x=8
\(\Rightarrow\) 2x-15=0\(\rightarrow\)2x=15\(\rightarrow\)x=7.5
vậy................
a) Có : ( x-5 ) \(^4\) = ( x - 5 ) \(^6\)
\(\Rightarrow\left(x-5\right)^6:\left(x-5\right)^4=1\Rightarrow\left(x-5\right)^2=1\)
\(\Rightarrow x-5=1\) hoặc \(x-5=-1\)
\(\Rightarrow x=1+5\) hoặc \(x=-1+5\)
\(\Rightarrow x=6\) hoặc \(x=4\)
b) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5:\left(2x-15\right)^3=1\Rightarrow\left(2x-15\right)^2=1\)
\(\Rightarrow2x-15=1\) hoặc \(2x-15=-1\)
\(\Rightarrow2x=16\) hoặc \(2x=14\)
\(\Rightarrow x=8\) hoặc \(x=7\)