Đáp án:
a. \(\left[ \begin{array}{l}x=6+4\sqrt[]{2} \\x=6-4\sqrt[]{2}\end{array} \right.\)
b. \(\left[ \begin{array}{l}x=5/2\\x=1/2\end{array} \right.\)
c. x = 1/7
d. \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
e. \(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)
f. \(\left[ \begin{array}{l}x=3\\x=0\end{array} \right.\)
g. \(\left[ \begin{array}{l}x=-1\\x=9/2\end{array} \right.\)
h. \(\left[ \begin{array}{l}x=-7\\x=-6\end{array} \right.\)
i. \(\left[ \begin{array}{l}x=-3\\x=1/5\end{array} \right.\)
Giải thích các bước giải:
a. (x - 2)(x + 3) - 3(4x - 2) = x - 4
=> x^2 + x - 6 - 12x + 6 = x - 4
=> x^2 - 12x + 4 = 0
\(\left[ \begin{array}{l}x=6+4\sqrt[]{2} \\x=6-4\sqrt[]{2}\end{array} \right.\)
b. (2x-2)(2x-4)=3
=> 4$x^{2}$ - 8x - 4x + 8 = 3
=> 4x^2 - 12x + 5 = 0
=>\(\left[ \begin{array}{l}x=5/2\\x=1/2\end{array} \right.\)
c. 5 - (x - 6) = 4(3 - 2x)
=> 5 - x + 6 = 12 - 8x
=> 7x = 1
=> x = 1/7
d. x^3 + 1 = x(x+1)
=> (x+1)(x^2 - x + 1)= x(x+1)
=> (x+1)(x^2 - x + 1 - x)= 0
=> (x+1)(x^2 - 2x+ 1)= 0
=> \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
e. x^6 - x^2 = 0
=> x^2(x^4 - 1) = 0
=> x^2(x^2 - 1)(x^2 + 1)=0
=> X^2(x - 1)(x + 1)(x^2 + 1) = 0
=> \(\left[ \begin{array}{l}x=0\\x=±1\end{array} \right.\)
f. (x - 3)(x - 5) = (2x - 5)(x -3)
=> (x - 3) (x - 5 - 2x +5)= 0
=> (x - 3)(-x)= 0
=>\(\left[ \begin{array}{l}x=3\\x=0\end{array} \right.\)
g. 2x^2 - 7x - 9 = 0
=> (x + 1)(2x - 9) = 0
=> \(\left[ \begin{array}{l}x=-1\\x=9/2\end{array} \right.\)
h. (2x - 1)(x + 7) = x^2 - 49
=> (2x - 1)(x + 7) = (x -7)(x + 7)
=> (x + 7 )(2x -1 - x + 7) = 0
=> (x + 7)(x + 6) = 0
=> \(\left[ \begin{array}{l}x=-7\\x=-6\end{array} \right.\)
i. (3x + 1)^2 = 4. (x - 1)^2
=> (3x + 1 - 2x + 2)(3x + 1 + 2x -2) = 0
=> (x + 3)( 5x - 1)= 0
=> \(\left[ \begin{array}{l}x=-3\\x=1/5\end{array} \right.\)
