Question

Giải phương trình
1) 27x²(x+3)-12(x²+3x)=0
2) 16x²-8x+1=4(x+3)(4x-1)
3) (2x-1)²=49
4) (5x-3)²-(4x-7)²=0

Answer 1

1, \(27x^2\left(x+3\right)-12x\left(x+3\right)=0\Leftrightarrow\left(x+3\right)\left(27x^2-12x\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\Leftrightarrow x=0;x=-3;x=\dfrac{4}{9}\)

2, \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)

\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\Leftrightarrow x=\dfrac{1}{4}\)

3, \(\left(2x-1\right)^2=49\Leftrightarrow\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

4, \(\left(5x-3\right)^2-\left(4x-7\right)^2=0\Leftrightarrow\left(5x-3\right)^2=\left(4x-7\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=4x-7\\5x-3=7-4x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{10}{9}\end{matrix}\right.\)

Answer 2

1) \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)

\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)

\(\Leftrightarrow3x\left(x+3\right)\left(9x-4\right)=0\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\9x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{4}{9}\end{matrix}\right.\)

Vậy: ... 

2)  \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\) 

\(\Leftrightarrow\left(4x\right)^2-2\cdot4x\cdot1+1^2=4\left(x+3\right)\left(4x-1\right)\)

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

\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-12\right)=0\)

\(\Leftrightarrow-13\left(4x-1\right)=0\)

\(\Leftrightarrow4x-1=0\\ \Leftrightarrow4x=1\\ \Leftrightarrow x=\dfrac{1}{4}\)

Vậy: ...