câu này mik làm rồi ở dưới kìa

cho đúng nha

a) \(\left(2x-1\right)^2-2x+1=0\)

\(\Leftrightarrow4x^2-4x+1-2x+1=0\)

\(\Leftrightarrow4x^2-6x+2=0\)

\(\Delta=\left(-6\right)^2-4.4.2=4\)

\(x_1=\dfrac{1}{2};x_2=1\)

b) \(\left(4x-3\right)^2-12x+9=0\)

\(\Leftrightarrow16x^2-24x+9-12x+9=0\)

\(\Leftrightarrow16x^2-36x+18=0\)

\(\Delta=\left(-36\right)^2-4.16.18=144\)

\(x_1=\dfrac{3}{4};x_2=\dfrac{3}{2}\)

c) \(3x^2-48=0\)

\(\Leftrightarrow3x^2=48\)

\(\Leftrightarrow x^2=16\)

\(x_1=4;x_2=-4\)

Tick cho mk nha

a: Ta có: \(x^3+1=0\)

\(\Leftrightarrow x^3=-1\)

hay x=-1

b: Ta có: \(6x^2-12x-48=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)

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

(a) Phương trình tương đương: \(\left|2x-5\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}2x-5=7\\2x-5=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\).

Vậy: \(S=\left\{-1;6\right\}\)

(b) Điều kiện: \(x\ge0\).

Phương trình tương đương: \(\sqrt{3x}-2\sqrt{3x}=3\sqrt{3}-4\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}\left(\sqrt{x}-2\sqrt{x}\right)=-\sqrt{3}\)

\(\Leftrightarrow\sqrt{x}-2\sqrt{x}=-1\)

\(\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\left(TM\right)\).

Vậy: \(S=\left\{1\right\}\)

\(\frac{12x-5y}{7}=\frac{20z-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20x}{7+9+11}=0\)\(=0\)

=> 12x - 15y =0

=> 12x = 15y

=> \(\frac{x}{15}=\frac{y}{12}\)

=> \(\frac{x}{60}=\frac{y}{48}\)

20z - 12x = 0 

=> 20z = 12x 

=> \(\frac{x}{20}=\frac{z}{12}\)

=> \(\frac{x}{60}=\frac{z}{36}\)

=> \(\frac{x}{60}=\frac{y}{48}=\frac{z}{36}=\frac{x+y+z}{60+48+36}=\frac{48}{144}=\frac{1}{3}\)

=> x = 1 . 60 : 3 = 20

y = 1 . 48 : 3 = 16

z = 1 . 36 : 3 = 12

(12x - 48)^2 = 114

(12x - 48)^2 = 12^2

(12x - 48)= 12

12x = 12+48

12x = 60

x = 60/12

x = 5

(12x-48)²=144

(12x-48)²= 12² ( cũng có thể nói 12x-48 = \(\sqrt{144}\))

=>12x-48=12

12x = 12 + 48 = 60

x = 60:12

x = 5