tìm x \(^{\left(x-5^{ }\right)^{12}}\)=\(\left(x-5\right)^9\)
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Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
26 tháng 8 2018
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9\)
\(\Rightarrow\left(x+2\right)^2=3^2\)
\(\Rightarrow x+2=3\)
\(\Rightarrow x=3-2=1\)
SM
26 tháng 8 2018
a) ( x + 2 )2 = 9
=> ( x + 2 ) 2 = 9
=> ( x + 2 )2 = 32
=> x + 2 = + 3
=> \(\orbr{\begin{cases}x+2=-3\\x+2=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
Vậy x = -1; 5
b) ( x + 2 )2 - x2 + 4 = 0
=> ( x + 2 )2 - ( x2 - 4 ) = 0
=> ( x + 2 )2 - ( x + 2 ) ( x - 2 ) = 0
=> ( x + 2 ) ( x + 2 - x + 2 ) = 0
=> ( x + 2 ) . 4 = 0
=> x + 2 = 0
=> x = - 2
Vậy x = - 2
c) 5 ( 2x - 3 )2 - 5 ( x + 1 )2 - 15( x + 4 ) ( x - 4 ) = - 10
=> 5 ( 4x2 - 12x + 9 ) - 5 ( x2 + 2x + 1 ) - 15 ( x2 - 42 ) = - 10
=> 20x2 - 60x + 45 - 5x2 - 10x - 5 - 15x2 + 240 = -10
=> - 70x + 280 = - 10
=> - 70x = - 290
=> x = \(\frac{29}{7}\)
Vậy x = \(\frac{29}{7}\)
d) x ( x + 5 ) ( x - 5 ) - ( x + 2 ) ( x2 - 2x + 4 ) = 3
=> x ( x2 - 25 ) - ( x3 - 8 ) = 3
=> x3 - 25x - x3 + 8 = 3
=> - 25x + 8 = 3
=> - 25x = -5
=> x = \(\frac{1}{5}\)
Vậy x = \(\frac{1}{5}\)
DT
Dang Tung
CTVHS
10 tháng 10 2023
a) (-5/9)^10 : x = (-5/9)^8
=> x = (-5/9)^10 : (-5/9)^8
=> x = (-5/9)^10-8 = (-5/9)^2
=> x = 25/81
b ) x : (-5/9)^8 = (-9/5)^8
=> x = (-9/5)^8 . (-5/9)^8
=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )
=> x = 1
C) x^3 = -8 =(-2)^3
=> x = -2
KV
10 tháng 10 2023
a) (-5/9)¹⁰ : x = (-5/9)⁸
x = (-5/9)¹⁰ : (-5/9)⁸
x = (-5/9)²
x = 25/81
b) x : (-5/9)⁸ = (-9/5)⁸
x = (-9/5)⁸ . (-5/9)⁸
x = [-9/5 . (-5/9)]⁸
x = 1⁸
x = 1
c) x³ = -8
x³ = (-2)³
x = -2
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
26 tháng 3 2022
Ta có:\(\left|x-1\right|\ge0;\forall x\)
\(\left|x+2\right|\ge0;\forall x\)
\(\left|x-3\right|\ge0;\forall x\)
\(\left|x+4\right|\ge0;\forall x\) ......
Cộng tất cả ta được:
\(\left|x-1\right|+\left|x+2\right|+\left|x-3\right|+\left|x+4\right|+...+\left|x-9\right|\ge0\)
\(\Rightarrow Min_T=0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}x=1\\x=-2\\x=3\\x=-4.....\end{matrix}\right.\)
Ta có:
\(\left(x-5\right)^{12}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{12}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9.\left[\left(x-5\right)^3-1\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^9=0\\\left(x-5\right)^3+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=6\end{cases}}}\)
(x - 5)\(^3\).(x - 5)\(^9\)= (x - 5)\(^9\)
(x - 5)\(^3\)= 1
x -5 = 1
x= 6