Tìm x :
\(_{\left(2-x\right)^8+\left(2-x\right)^9=0}\)
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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):}`
KK
5 tháng 9 2021
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
KK
5 tháng 9 2021
d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
30 tháng 10 2021
1) (x−1):0,16=−9:(1−x)
\(\Rightarrow\)(x-1):0,16= 9:(-1):(x-1)
\(\Rightarrow\)(x-1):0,16=9:(x-1)
\(\Rightarrow\)(x-1).(x-1)= 9. 0,16
\(\Rightarrow\)(x-1)\(^2\)= 1,44=1,2\(^2\)=(-1,2)\(^2\)
\(\Rightarrow\)x-1=1,2\(\Rightarrow\)x=2,2
hoặc x-1= -1,2\(\Rightarrow\)x= -0,2
Vậy x =2,2 ; x=0,2
...............................
PT
20 tháng 9 2018
a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=27\)
\(\Rightarrow x^3+3^3-x\left(x^2-4\right)=27\)
\(\Rightarrow x^3+27-x^3+4x=27\)
\(\Rightarrow27+4x=27\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\)
PT
20 tháng 9 2018
b) \(2x^2+7x+3=0\)
\(\Rightarrow2x^2+x+6x+3=0\)
\(\Rightarrow x\left(2x+1\right)+3\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
26 tháng 12 2021
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
TL
30 tháng 6 2016
a) \(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(\left(2x-2\right)+\left(3x+6\right)\right)\left(\left(2x-2\right)-\left(3x+6\right)\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)
b) \(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow4x+13=11\)
\(\Leftrightarrow x=-\frac{1}{2}\)
DK
30 tháng 6 2016
a) \(4\left(x-1\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-1\right)\right]^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x-2+3x+6\right)\left(2x-2-3x-6\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow4x+13=11\)
\(\Leftrightarrow4x=-2\)
\(\Leftrightarrow x=-\frac{2}{4}=-\frac{1}{2}\)
(Nhớ k cho mình với nhé!)
W
6 tháng 3 2020
a) (2x-2)(3x-7)=0
<=> 2x-2=0 hoặc 3x-7=0
<=> x=1 hoặc x=7/3
Vậy \(x\in\left\{1;\frac{7}{3}\right\}\)
b) (x-9)(x+8)<0
<=>(x-9)<0 hoặc (x+8)<0
<=>x<9 hoặc x<-8
<=>x<-8
6 tháng 3 2020
a(2x-2)(3x-7)=0
<=>\(\orbr{\begin{cases}2x-2=0\\3x-7=0\end{cases}}\)
<=>\(\orbr{\begin{cases}2x=2\\3x=7\end{cases}}\)
<=>\(\orbr{\begin{cases}x=1\\x=\frac{3}{7}\end{cases}}\)
Vậy x thuộc {1;3/7}
\(\left(x-2\right)^8+\left(2-x\right)^9=0.\)
\(\Rightarrow\left(2-x\right)^8\left(1+2-x\right)=0\)
\(\Rightarrow\left(2-x\right)^8\left(3-x\right)=0\)
\(\Rightarrow x\in\left\{2;3\right\}\)
(2 - x)8 + (2 - x)9 = 0
=> (2 - x)8. (1 + 2 - x) = 0
=> (2 - x)8 . (3 - x) = 0
=> \(\orbr{\begin{cases}\left(2-x\right)^8=0\\3-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2-x\right)^8=0^8\\x=3-0\end{cases}}\)
=> \(\orbr{\begin{cases}2-x=0\\x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=2-0\\x=3\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=3\end{cases}}\)