Mn ơi giúp mih làm câu này nha
Tìm x
(x-5)*4=(x-5)*6
Cảm ơn mn nf ạ
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\(a,\dfrac{7}{12}-\left(x+\dfrac{7}{10}\right):\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{10}:\dfrac{6}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x-\dfrac{7}{12}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{5}{4}+\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}-x=\dfrac{11}{6}\)
\(\Leftrightarrow x=\dfrac{7}{12}-\dfrac{11}{6}\)
\(\Leftrightarrow\dfrac{-5}{4}\)
`48/[x+4]+48/[x-4]=5` `ĐK: x \ne +-4`
`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`
`=>48x-192+48x+192=5x^2-80`
`<=>5x^2-96x-80=0`
`<=>5x^2-100+4x-80=0`
`<=>5x(x-20)+4(x-20)=0`
`<=>(x-20)(5x+4)=0`
`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$ (t/m)
Vậy `S={-4/5;20}`
ĐK : \(x\ne\pm4\)
\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)
\(\Leftrightarrow48x+192+48x-192==5x^2-80\)
\(\Leftrightarrow96x=5x^2-80\)
\(\Leftrightarrow5x^2-96x-80=0\)
\(\Leftrightarrow5x^2+4x-100-80=0\)
\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)
\(x^{10}=25x^8\)
\(\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^8=0\\x^2-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=0;x_3=5\)
\(X^{10}=25.X^8\)
\(25=X^{10}:X^8\)
\(25=X^{10-8}\)
\(25=X^2\)
\(\Rightarrow x=5;x=-5\)
Vậy x=5 ; x=;5.
a) (-2) . ( x+7 ) + (-5) = 7
<=>(-2).(x+7)=7+5
<=>x+7=12:(-2)
<=>x+7=-6
<=>x=(-6)-7
<=>x=-13
Vậy x=-13
b)(x+4) : (-7) = 14
<=>x+4=14 x (-7)
<=>x+4=-98
<=>x=-98-4
<=>x=-102
Vậy x= -102
c) 72 : ( x+5) - 4 = -12
<=>72:(x+5)=(-12)+4
<=>x+5=72:(-8)
<=>x+5=-9
<=>x=-9-5
<=>x=-14
Vậy x= -14
d) (x+3) : (-6 ) + 12 = 8
<=>(x+3) :(-6)=8-12
<=>x+3=(-4)x(-6)
<=>x+3=24
<=>x=24-3
<=>x=21
Vậy x= 21
ĐKXĐ: \(x\notin\left\{0;-9\right\}\)
Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)
Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)
\(\Leftrightarrow9x^2+81x+180=0\)
\(\Leftrightarrow x^2+9x+20=0\)
\(\Leftrightarrow x^2+4x+5x+20=0\)
\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-4;-5}
a, 20 + 8.( x + 3 ) = 5^2 .4
20 + 8. ( x + 3 ) = 25 . 4
20 + 8. ( x + 3 ) = 100
8. ( x + 3 ) = 100 - 20
8 . ( x + 3 ) = 80
x + 3 = 80 : 8
x + 3 = 10
x = 10 - 3
x = 7
Vậy x = 7
b, /x+4/ - 12 = -6
/x+4/ = -6 + 12
/x+4/ = 6
x+4 ∈ { 6 ; -6 }
x ∈ { 2 ; -2 }
Vậy x ∈ { 2 ; -2 }
#Học tốt#
Ta có :
\(\left(x-5\right)^4=\left(x-5\right)^6\)
\(0=\left(x-5\right)^6-\left(x-5\right)^4\)
\(0=\left[\left(x-5\right)^2-1\right]\left(x-5\right)^4\)
Ta có các trường hợp :
TH1 : \(\left(x-5\right)^2-1=0\)
\(\Rightarrow\left(x-5\right)^2=1\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^2=1^2\\\left(x-5\right)^2=\left(-1\right)^2\end{matrix}\right.\)
TH2 : \(x-5=1\)
\(\Rightarrow x=6\left(TM\right)\)
TH3 : \(x-5=-1\)
\(\Rightarrow x=4\left(TM\right)\)
TH4 : \(\left(x-5\right)^4=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\left(TM\right)\)
Vậy \(x\in\left\{4,5,6\right\}\) là giá trị cần tìm
dấu * là nhân hay mũ ấy?