Question

Bài 3. (3 điểm) Tìm x, biết :

b) 12x(3 – 4x) + 7(4x – 3) = 0

c) (x + 2)(x² – 2x + 4) + x(5 – x)(x + 5) = -17 d) 9x² – 4 – 2(3x – 2)² = 0

Answer 1

b,
\(12x\left(3-4x\right)+7\left(4x-3\right)=0\)
\(\Leftrightarrow12x\left(3-4x\right)-7\left(3-4x\right)=0\)
\(\Leftrightarrow\left(3-4x\right)\left(12x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-4x=0\\12x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=3\\12x=7\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{7}{12}\end{matrix}\right.\)Vậy...

Answer 2

b) 12x(3 – 4x) + 7(4x – 3) = 0

⇔ 12x( 3 - 4x ) - 7( 3 - 4x) = 0

⇔ ( 12x - 7 ) ( 3 - 4x ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}12x-7=0\\3-4x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{12}\\x=\frac{3}{4}\end{matrix}\right.\)

d,

9x2 – 4 – 2(3x – 2)2 = 0

⇔ 9x\(^2\) - 4 - 2( 9x\(^2\) -12x + 4 ) = 0

⇔ 9x\(^2\) - 4 - 18x\(^2\) + 24x -8 = 0

⇔ -9x\(^2\) + 24x - 12 = 0

⇔ 3x\(^2\) - 8x + 4 = 0

⇔ 3x\(^2\) - 6x - 2x +4 = 0

⇔ 3x ( x - 2 ) - 2 ( x - 2 ) = 0

⇔ ( 3x - 2 ) ( x - 2 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=2\end{matrix}\right.\)

Answer 3

b, 12x(3-4x)+ 7(4x-3)=0

\(\Leftrightarrow\)12x(3-4x) -7(3-4x)=0

\(\Leftrightarrow\)(3-4x)(12x-7)=0

\(\Leftrightarrow\)\(\left[{}\begin{matrix}3-4x=0\\12x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{7}{12}\end{matrix}\right.\)