đây là bài giải phương trình lớp 8

Đáp án + Giải thích các bước giải:

 a) $(2x - 1)^2 = 49$

$\Leftrightarrow$ \(\left[ \begin{array}{l}2x-1=7\\2x-1=-7\end{array} \right.\) 
$\Leftrightarrow$ \(\left[ \begin{array}{l}2x=8\\2x=-6\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=4\\x=-3\end{array} \right.\) 

Vậy $S = $`{-3; 4}`

$---------------------$

b) $(2x + 5)(x - 4) = (x - 5)(4 - x)$

$\Leftrightarrow (2x + 5)(x - 4) - (x - 5)(4 - x) = 0$

$\Leftrightarrow (2x + 5)(x - 4) + (x - 5)(x - 4) = 0$

$\Leftrightarrow (2x + 5 + x - 5)(x - 4) = 0$

$\Leftrightarrow 3x(x - 4) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}3x=0\\x-4=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=0\\x=4\end{array} \right.\) 

Vậy $S = $`{0; 4}`

$---------------------$

c) $9x^2 - 1 = (3x + 1)(4x + 1)$
$\Leftrightarrow (3x + 1)(3x - 1) = (3x + 1)(4x + 1)$

$\Leftrightarrow (3x + 1)(3x - 1) - (3x + 1)(4x + 1) = 0$

$\Leftrightarrow (3x + 1)(3x - 1 - 4x - 1) = 0$

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

$\Leftrightarrow$ \(\left[ \begin{array}{l}3x+1=0\\-x-2=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}3x=-1\\-x=2\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=\dfrac{-1}{3}\\x=-2\end{array} \right.\)

Vậy $S = $`{(-1)/3; -2}`

$---------------------$

d) $(x + 7)(3x - 1) = 49 - x^2$

$\Leftrightarrow (x + 7)(3x - 1) = (7 + x)(7 - x)$

$\Leftrightarrow (x + 7)(3x - 1) - (7 + x)(7 - x) = 0$

$\Leftrightarrow (x + 7)(3x - 1 - 7 + x) = 0$

$\Leftrightarrow (x + 7)(4x - 8) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}x+7=0\\4x-8=0\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=-7\\4x=8\end{array} \right.\) 

$\Leftrightarrow$ \(\left[ \begin{array}{l}x=-7\\x=2\end{array} \right.\) 

Vậy $S = $`{-7; 2}`

$---------------------$

e) $(2x + 1)^2 = (x - 1)^2$

$\Leftrightarrow (2x + 1)^2 - (x - 1)^2 = 0$

$\Leftrightarrow (2x + 1 + x - 1)(2x + 1 - x + 1) = 0$

$\Leftrightarrow 3x(x + 2) = 0$

$\Leftrightarrow$\(\left[ \begin{array}{l}3x=0\\x+2=0\end{array} \right.\) 

$\Leftrightarrow$\(\left[ \begin{array}{l}x=0\\x=-2\end{array} \right.\) 

Vậy $S = $`{-2; 0}`

$---------------------$

g) $x^2 - 2x - 8 = 0$

$\Leftrightarrow x^2 + 2x - 4x - 8 = 0$

$\Leftrightarrow x(x + 2) - 4(x + 2) = 0$

$\Leftrightarrow (x - 4)(x + 2) = 0$

$\Leftrightarrow$ \(\left[ \begin{array}{l}x-4=0\\x+2=0\end{array} \right.\)

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

Vậy $S = $`{-2; 4}`