Bài 1:
a, (x + 8)² - 2 . (x + 8) . (x - 2) + (x - 2)²
= [(x + 8) - (x - 2)]²
= (x + 8 - x + 2)²
= 10²
= 100
b, x² (x - 4)(x + 4) - (x² + 1)(x² - 1)
= x² (x² - 16) - ($x^{4}$ - 1)
= $x^{4}$ - 16x² - $x^{4}$ + 1
= -16x² + 1
c, (x + 1)(x² - x + 1) - (x - 1)(x² + x + 1)
= x³ + 1 - (x³ - 1)
= x³ + 1 - x³ + 1
= 2
d, (x + y)² - (x - y)²
= (x + y - x + y)(x + y + x - y)
= 2y . 2x
= 4xy
e, (x + y + z)² - 2(x + y + z)(x + y) + (x + y)²
= [(x + y + z) - (x + y)]²
= (x + y + z - x - y)²
= z²
Bài 2:
a, Thay x = 7, ta có:
7² + 6 . 7 + 9
= 49 + 42 + 9
= 100
b, Thay x = -11, ta có:
(-11)³ + 3 . (-11)² + 3 . (-11) + 1
= (-11 + 1)³
= (-10)³
= -1000
Bài 3:
a, (2x - 1)² - (4x + 3)²=0
(2x - 1 - 4x - 3)(2x - 1 + 4x + 3) = 0
(-2x - 4)(6x + 2) = 0
-(2x + 4)(6x + 2) = 0
(2x + 4)(6x + 2) = 0
⇔ \(\left[ \begin{array}{l}2x+4=0\\6x+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}2x=-4\\6x=-2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-2\\x=-2 : 6 =\dfrac{-1}{3}\end{array} \right.\)
Vậy x ∈ {-2; $\dfrac{-1}{3}$}
b, (x + 5)² - (3x - 2)² = 0
(x + 5 - 3x + 2)(x + 5 + 3x - 2) = 0
(-2x + 7)(4x + 3) = 0
⇔ \(\left[ \begin{array}{l}-2x + 7=0\\4x+3=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}-2x=-7\\4x=-3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\dfrac{-7}{-2}=\dfrac{7}{2}\\x=\dfrac{-3}{4}\end{array} \right.\)
Vậy x ∈ {$\dfrac{7}{2}$; $\dfrac{-3}{4}$}
