tìm x
(2x - 15)5 = (2x-15)3
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( 2x - 15 ) ^5 = ( 2x - 15 ) ^3
=>( 2x - 15 ) ^5 - ( 2x - 15 ) ^3 = 0
=>( 2x - 15 ) ^2 =0
=> 2x-15 = 0
=> x = \(\frac{15}{2}\) =7 \(\frac{1}{2}\)
Mình không viết lại đề nhé!
a) 27 - x + 15 + x = x + 2
-x + x - x = 2 - 15 - 27
-x = -40
x = 40
b) 8x - 75 = 5x +21
8x - 5x = 21 + 75
3x = 96
x = 32
c) 15 - |2x - 1| = |-8|
15 - 2x + 1 = 8
-2x = 8 -1 -15
-2x = -8
x = 4
d) 9x + 25 = -(2x - 58)
9x + 25 = -2x + 58
9x +2x = 58 - 25
7x = 33
x = 33/7
a)
\(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)
b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)
\(2x^2+6x-8=0\)
<=> \(2x^2-2x+8x-8=0\)
<=> \(2x\left(x-1\right)+8\left(x-1\right)=0\)
<=> \(\left(2x+8\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}2x+8=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-4\\x=1\end{cases}}\)
\(2x^2-x-1=0\)
<=> \(2x^2-2x+x-1=0\)
<=> \(2x\left(x-1\right)+\left(x-1\right)=0\)
<=> \(\left(2x+1\right)\left(x-1\right)=0\)
<=> \(\hept{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
\(4x^2-5x-9=0\)
<=> \(4x^2+4x-9x-9=0\)
<=> \(4x\left(x+1\right)-9\left(x+1\right)=0\)
<=> \(\left(4x-9\right)\left(x+1\right)=0\)
<=> \(\hept{\begin{cases}4x-9=0\\x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)
học tốt
\(2x^2+6x-8=0\)
\(< =>2x^2-2x+8x-8=0\)
\(\Leftrightarrow2x\left(x-1\right)+8\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow2x+8=0\)hoặc \(x-1=0\)
\(\Leftrightarrow x=-4\)hoặc \(x=1\)
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 \(x=7\)
Vì \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow2x-15=\orbr{\begin{cases}0\\1\end{cases}}\)vì chỉ có \(0^5=0^3;1^5=1^3\)
\(\Rightarrow2x=\orbr{\begin{cases}15\\16\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}\frac{15}{2}\\8\end{cases}}\)
\(\left(2x-15\right)^5=\left(2x-15\right)^3.\)
\(\Rightarrow\hept{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow\hept{\begin{cases}2x=15\\2x=16\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=8\end{cases}}}\)