Bài 1.tính hằng đẳng thức a) (3x – 2y)^3 b) (4x – 1/2)^3 c) (1/3x + 3)^3 Bài 2. tìm x a) (x – 1)^2 – (x + 1) (x + 3)= 7 b) 4 (x – 3) (x + 3) – (2x – 1

Đáp án:

1/ a/ $27x^3-54x^2y+36xy^2-8y^3$

b/ $64x^3-24x^2+3x-\dfrac{1}{8}$

c/ $\dfrac{1}{27}x^3+x^2+9x+27$

2/ a/ $x=-\dfrac{3}{2}$

b/ $x=\dfrac{37}{4}$

c/ $x=\dfrac{11}{4}$

d/ $x=\dfrac{5}{2}$

Giải thích các bước giải:

1/ a/ $(3x-2y)^3$

$=(3x)^3-3.(3x)^2.2y+3.3x.(2y)^2-(2y)^3$

$=27x^3-54x^2y+36xy^2-8y^3$

b/ `(4x-\frac{1}{2})^3`

$=(4x)^3-3.(4x)^2.\dfrac{1}{2}+3.4x.\dfrac{1}{4}-\dfrac{1}{8}$

$=64x^3-24x^2+3x-\dfrac{1}{8}$

c/ `(\frac{1}{3}x+3)^3`

$=\dfrac{1}{27}x^3+3.\dfrac{1}{9}x^2.3+3.\dfrac{1}{3}x.9+27$

$=\dfrac{1}{27}x^3+x^2+9x+27$

2/ a/ $(x-1)^2-(x+1)(x+3)=7$

$⇔ (x^2-2x+1)-(x^2+x+3x+3)=7$

$⇔ x^2-2x+1-x^2-4x-3-7=0$

$⇔ -6x-9=0$

$⇔ x=-\dfrac{9}{6}=-\dfrac{3}{2}$

b/ $4(x-3)(x+3)-(2x-1)^2=0$

$⇔ 4(x^2-9)-(4x^2-4x+1)=0$

$⇔ 4x^2-36-4x^2+4x-1=0$

$⇔ 4x-37=0$

$⇔ x=\dfrac{37}{4}$

c/ $(3x-2)^2-9(x-2)(x+2)=7$

$⇔ (9x^2-12x+4)-9(x^2-4)=7$

$⇔ 9x^2-12x+4-9x^2+36-7=0$

$⇔ -12x+33=0$

$⇔ x=\dfrac{33}{12}=\dfrac{11}{4}$

d/ $(x-3)(x+2)-(x-1)(x-4)=0$

$⇔ x^2-3x+2x-6-(x^2-x-4x+4)=0$

$⇔ x^2-x-6-x^2+5x-4=0$

$⇔ 4x-10=0$

$⇔ x=\dfrac{10}{4}=\dfrac{5}{2}$

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