Câu hỏi của Lê Hoàng Thảo Nhi

Question

tìm x thuộc Q bt:

a)$\left(x-\dfrac{1}{2}\right)^2=0$

b) $\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

c) $\left(2.x-3\right)^3=-8$

d) $\left(3.x-2\right)^5=-243$

e) \(\left(7.x+2\right)^{...

Lê Minh Anh · Accepted Answer

a)(x − 12)2 = 0

=>x − 12 = 0

=> x = 12

b) (x+12)2 = 0,25

=> x + 12 = 0,5 hoặc x + 12= -0,5

=> x = -11,5 hoặc x = -12,5

c) (2x−3)3 = -8

=> 2x - 3 = -2

=> x = 0,5

d) (3x−2)5 = −243

=> 3x - 2 = -3

=> x = -1/3

e) (7x+2)-1 = 3-2

=> $\dfrac{1}{7x+2}=\dfrac{1}{9}$

=> 7x + 2 = 9

=> x = 1

f) (x−1)3 = −125

=> (x−1) = −5

=> x = -4

g) (2x−1)4 = 81

=> 2x - 1 = 3

=> x = 2

h) (2x−1)6 = (2x−1)8

=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1

=> x = 1/2 hoặc x = 1 hoặc x = 0Câu trả lời: a)(x − 12)2 = 0

=>x − 12 = 0

=> x = 12

b) (x+12)2 = 0,25

=> x + 12 = 0,5 hoặc x + 12= -0,5

=> x = -11,5 hoặc x = -12,5

c) (2x−3)3 = -8

=> 2x - 3 = -2

=> x = 0,5

d) (3x−2)5 = −243

=> 3x - 2 = -3

=> x = -1/3

e) (7x+2)-1 = 3-2

=> $\dfrac{1}{7x+2}=\dfrac{1}{9}$

=> 7x + 2 = 9

=> x = 1

f) (x−1)3 = −125

=> (x−1) = −5

=> x = -4

g) (2x−1)4 = 81

=> 2x - 1 = 3

=> x = 2

h) (2x−1)6 = (2x−1)8

=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1

=> x = 1/2 hoặc x = 1 hoặc x = 0

Nguyễn Thanh Hằng · Answer

a/ $\left(x-\dfrac{1}{2}\right)^2=0$

$\Leftrightarrow x-\dfrac{1}{2}=0$

$\Leftrightarrow x=\dfrac{1}{2}$

Vậy ...

b/ $\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

$\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

Vậy ..

c/ $\left(2x-3\right)^3=-8$

$\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3$

$\Leftrightarrow2x-3=-2$

$\Leftrightarrow x=\dfrac{1}{2}$

Vậy ...

d/ $\left(3x-2\right)^5=-243$

$\left(3x-2\right)^5=\left(-3\right)^5$

$\Leftrightarrow3x-2=-3$

$\Leftrightarrow x=-\dfrac{1}{3}$

Vậy ...

e/ $\left(x-1\right)^3=-125$

$\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3$

$\Leftrightarrow x-1=-5$

$\Leftrightarrow x=-4$

Vậy..

f/ $\left(2x-1\right)^4=81$

$\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=3\2x-1=-3\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x=-1\end{matrix}\right.$

Vậy...

g/ $\left(2x-1\right)^6=\left(2x-1\right)^8$

$\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0$

$\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\left(2x-1\right)^2-1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\left[{}\begin{matrix}2x-1=1\2x-1=-1\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\left[{}\begin{matrix}x=1\x=0\end{matrix}\right.\end{matrix}\right.$

Vậy..Câu trả lời: a/ $\left(x-\dfrac{1}{2}\right)^2=0$

$\Leftrightarrow x-\dfrac{1}{2}=0$

$\Leftrightarrow x=\dfrac{1}{2}$

Vậy ...

b/ $\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}$

$\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=0\x=-1\end{matrix}\right.$

Vậy ..

c/ $\left(2x-3\right)^3=-8$

$\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3$

$\Leftrightarrow2x-3=-2$

$\Leftrightarrow x=\dfrac{1}{2}$

Vậy ...

d/ $\left(3x-2\right)^5=-243$

$\left(3x-2\right)^5=\left(-3\right)^5$

$\Leftrightarrow3x-2=-3$

$\Leftrightarrow x=-\dfrac{1}{3}$

Vậy ...

e/ $\left(x-1\right)^3=-125$

$\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3$

$\Leftrightarrow x-1=-5$

$\Leftrightarrow x=-4$

Vậy..

f/ $\left(2x-1\right)^4=81$

$\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=3\2x-1=-3\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x=-1\end{matrix}\right.$

Vậy...

g/ $\left(2x-1\right)^6=\left(2x-1\right)^8$

$\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0$

$\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\left(2x-1\right)^2-1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\left[{}\begin{matrix}2x-1=1\2x-1=-1\end{matrix}\right.\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\left[{}\begin{matrix}x=1\x=0\end{matrix}\right.\end{matrix}\right.$

Vậy..

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