Question

Tìm x biết :
a) \(x^2-3=1\)
b) \(2x^3+12=-4\)
c) \(\left(2x-3\right)^2=16\)
d) \(x:\left(\dfrac{-1}{2}\right)^1=\dfrac{-1}{2}\)
h) \(\left(\dfrac{1}{4}-x\right)^2=\dfrac{25}{16}\)
i) \(\left(x+\dfrac{1}{3}\right)^3=\dfrac{-1}{8}\)
k) \(\left(\dfrac{1}{2}-x\right)^2=1\)

Answer 1

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

Answer 2

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....

Answer 3

a) \(x^2\) - 3 = 1

\(\Rightarrow\) \(x^2\) = 3+1 = 4

\(\Rightarrow\) \(x^2\) = \(2^2\)

\(\Rightarrow\) \(x\) = \(\pm2\)

b) 2x\(^3\) + 12 = -4

\(\Rightarrow\) \(2x^3\) + 2.6 = -4

\(\Rightarrow\) 2 (\(x^3\) + 6)= -4

\(\Rightarrow\) \(x^3+6\) = -4:2 = -2

\(\Rightarrow\) \(x^3\) = -2-6 = -8

\(\Rightarrow\) \(x^3=-2^3\) \(\Rightarrow\) x = -2 (cách 2 ^^)

lười nha!

c) (2x-3)^2 =16

2x-3 = +-4

* 2x-3=4 => 2x=4+3 => 2x=7 => x=7:2 = 7/2

* 2x-3=(-4) => 2x=(-4)+3 => 2x=(-1) => x=(-1):2 = -1/2

d) x:(-1/2)^1 = -1/2

x.(-2)= -1/2

x= -1/2:(-2) = 1/4

x= 1/4

h) (1/4-x)^2 = 25/16

=> 1/4-x = 5/4

-x = 5/4 - 1/4 =1

x = -1

i) (x+1/3)^3 = -1/8

x+1/3 = -1/2

x= (-1/2)-1/3 =-5/6

x= -5/6

k) (1/2-x)^2 = 1

1/2-x = 1

-x= 1-1/2=1/2

x=-1/2

Chúc bạn học tốt !