Tìm x thuộc Q, biết rằng
a: (x-1/20)^2 = 0
b: (x-2)^2 = 1
c: (2x-1)^3 = -8
d: (x+1/2)^2 = 1/16
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19 tháng 9 2016
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) (x - 2)2 = 1
=> \(\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
Vậy \(x\in\left\{3;1\right\}\)
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=16\)
=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{array}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}\)
P
14 tháng 11 2018
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) (x−12)2=0(x−12)2=0
=> x−12=0x−12=0
=> x=12x=12
Vậy x=12x=12
b) (x - 2)2 = 1
=> [x−2=1x−2=−1[x−2=1x−2=−1=> [x=3x=1[x=3x=1
Vậy x∈{3;1}x∈{3;1}
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> x=−12x=−12
Vậy x=−12x=−12
d) (x+12)2=16(x+12)2=16
=> [x+12=14x+12=−14[x+12=14x+12=−14=> [x=−14x=−34[x=−14x=−34
Vậy x∈{−14;−34}
27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=1^2\)
\(x-2=1\)
\(x=1+2\)
\(x=3\)
c) \(\left(2x-1\right)^3=\left(-8\right)\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=\left(-2\right)\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}\)
27 tháng 6 2017
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^2=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)
NT
1
11 tháng 7 2016
a) \(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
c)\(\left(2x-1\right)^3=-8=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
14 tháng 7 2016
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=-\frac{1}{4}\)
PA
14 tháng 7 2016
a.
\(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
b.
\(\left(x-2\right)^2=1\)
\(x-2=\pm1\)
TH1:
\(x-2=1\)
\(x=1+2\)
\(x=3\)
TH2:
\(x-2=-1\)
\(x=-1+2\)
\(x=1\)
Vậy x = 3 hoặc x = 1
c.
\(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=-2\)
\(2x=-2+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d.
\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\pm\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\pm\frac{1}{4}\)
TH1:
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{1}{4}\)
TH2:
\(x+\frac{1}{2}=-\frac{1}{4}\)
\(x=-\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}-\frac{2}{4}\)
\(x=-\frac{3}{4}\)
Vậy \(x=-\frac{1}{4}\) hoặc \(x=-\frac{3}{4}\)
LD
3
20 tháng 9 2018
a) ( x - 1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 1/2
b) ( x - 2 )2 = 1
<=> x - 2 = { 1; -1 }
<=> x = { 3; 1 }
c) ( 2x - 1 )3 = -8 = ( -2 )3
<=> 2x - 1 = -2
<=> 2x = -1
<=> x = -1/2
d) ( x + 1/2 )2 = 1/16
<=> x + 1/2 = { 1/4; -1/4 }
<=> x = { -1/4; -3/4 }
20 tháng 9 2018
a.
(x-1/2)^2=0\(\Rightarrow\)x-1/2=0\(\Rightarrow\)x=1/2
b
(x-2)^2=1\(\Rightarrow\)x-2\(\in\){1;-1}
\(\Rightarrow\)x\(\in\){3;1}
c
(2x-1)^3=-8\(\Rightarrow\)2x-1=-2
\(\Rightarrow\)2x=-1
\(\Rightarrow\)x=-1/2
d
(x+1/2)^2=1/16\(\Rightarrow\)x+1/2\(\in\){1/4;-1/4}
\(\Rightarrow\)x\(\in\){-1/4;-3/4}
20 tháng 9 2021
Bài 2:
a: Ta có: \(x\left(2x-1\right)-2x+1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\end{matrix}\right.\)
6 tháng 10 2020
Bài 1 :
a) 72x-1 = 343
=> 72x-1 = 73
=> 2x - 1 = 3 => 2x = 4 => x = 2
b) (7x - 11)3 = 25.32 + 200
=> (7x - 11)3 = 32.9 + 200
=> (7x - 11)3 = 488
xem kĩ lại đề này :vvv
c) 174 - (2x - 1)2 = 53
=> (2x - 1)2 = 174 - 53
=> (2x - 1)2 = 174 - 125 = 49
=> (2x - 1)2 = (\(\pm\)7)2
=> \(\orbr{\begin{cases}2x-1=7\\2x-1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-3\end{cases}}\)
Mà x \(\in\)N nên x = 4( thỏa mãn điều kiện)
Bài 2 :
a) x5 = 32 => x5 = 25 => x = 2
b) (x + 2)3 = 27
=> (x + 2)3 = 33
=> x + 2 = 3 => x = 3 - 2 = 1
c) (x - 1)4 = 16
=> (x - 1)4 = 24
=> x - 1 = 2 => x = 3 ( vì đề bài cho x thuộc N nên thỏa mãn)
d) (x - 1)8 = (x - 1)6
=> (x - 1)8 - (x - 1)6 = 0
=> (x - 1)6 [(x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(x-1\right)^6=0\\\left(x-1\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^2=\left(\pm1\right)^2\end{cases}}\)
+) x - 1 = 1 => x = 2 ( tm)
+) x - 1 = -1 => x = 0 ( tm)
Vậy x = 1,x = 2,x = 0
a. (x-1/20)2=0
=> x-1/20=0
=> x=1/20
b. (x-2)2=1
=> (x-2)2=12=(-1)2
+) x-2=1
=> x=3
+) x-2=-1
=> x=1
Vậy x \(\in\){1;3}
c. (2x-1)3=-8
=> (2x-1)3=(-2)3
=> 2x-1=-2
=> 2x=-1
=> x=-1/2
d. (x+1/2)2=1/16
=> (x+1/2)2=(1/4)2=(-1/4)2
+) x+1/2=1/4
=> x=-1/4
+) x+1/2=-1/4
=> x=-3/4
Vậy x \(\in\){-3/4; -1/4}