\(\left(3.4^x-3\right).\left(x^3-125\right)=0\\ \rightarrow\left[{}\begin{matrix}3.4^x-3=0\\x^3-125=0\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}3.4^x=3\\x^3=125\end{matrix}\right.\)

\(\rightarrow\left[{}\begin{matrix}4^x=1=4^0\\x^3=5^3\end{matrix}\right.\\ \rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

\(\left(3\cdot4^x-3\right)\left(x^3-125\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3\cdot4^x-3=0\\x^3-125=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3\left(4^x-1\right)=0\\x^3-5^3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4^x=1\\x=5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

x=5 và 0

(4x-3)(2x-5) +(3-4x)(x-1)=0

(4x-3)(2x-5)-(4x-3)(x-1)=0

(4x-3)(2x-5-x+1)=0

(4x-3)(x-4)=0

4x-3=0 hoặc x-4=0

x=\(\frac{3}{4}\)hoặc x=4

\(\left(2x-1\right)^2-\left(4x^2-3\right)=0\)

\(\Rightarrow4x^2-4x+1-4x^2+3=0\)

\(\Rightarrow-4x+4=0\)

\(\Rightarrow-4x=-4\)

\(\Rightarrow x=1\)

\(\left(12-4x\right)\left(x+2014\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}12-4x=0\\x+2014=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2014\end{cases}}\)

Ta có (12-4x).(x+2014) = 0

Ta thấy một số nhân với 0 đều bằng 0 . Mà 2014 > 0 mà (12 - 4x ).(x+2014) = 0

=> x +2014 ko thể bằng 0

Ta thấy 12 = 4.3

=> 4x = 4.3

Ta có phép tính

(12-4.3).(3+2014)= 0 (hợp lí)

=> x= 3

Đúng k cho mk nha

A={5}

vi 3x. voi phep tinh trong ngoac =0 nen 4x-20 phai =0, x phai =5 de 4.5 =20 -20=0

cảm ơn bạn

Bài làm:

Ta có: \(4x^2-4x-3=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)

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

\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)

Ta có : \(4x^2-4x-3=0\)

\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)

\(\Leftrightarrow\left(2x-1\right)^2=4\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)

a/ 3x^2 - 8x + 4 = 0
<=> (3x-2)(x-2)=0
<=> x=2/3 or x=2
b/ 4x^2 - 4x - 3 = 0
<=> (2x+1)(2x-3)=0
<=> x=-1/2 or x=3/2