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Bài 2:
(1 + x)3 + (1 - x)3 - 6x(x + 1) = 6
<=> x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 - 6x = 6
<=> -6x + 2 = 6
<=> -6x = 6 - 2
<=> -6x = 4
<=> x = -4/6 = -2/3
Bài 3:
a) (7x - 2x)(2x - 1)(x + 3) = 0
<=> 10x3 + 25x2 - 15x = 0
<=> 5x(2x - 1)(x + 3) = 0
<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0
<=> x = 0 hoặc x = 1/2 hoặc x = -3
b) (4x - 1)(x - 3) - (x - 3)(5x + 2) = 0
<=> 4x2 - 13x + 3 - 5x2 + 13x + 6 = 0
<=> -x2 + 9 = 0
<=> -x2 = -9
<=> x2 = 9
<=> x = +-3
c) (x + 4)(5x + 9) - x2 + 16 = 0
<=> 5x2 + 9x + 20x + 36 - x2 + 16 = 0
<=> 4x2 + 29x + 52 = 0
<=> 4x2 + 13x + 16x + 52 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> 4x + 13 = 0 hoặc x + 4 = 0
<=> x = -13/4 hoặc x = -4
T a c ó : A = B ⇔ ( x + 2 ) ( x – 2 ) + 3 x 2 = ( 2 x + 1 ) 2 + 2 x ⇔ x 2 – 4 + 3 x 2 = 4 x 2 + 4 x + 1 + 2 x ⇔ x 2 + 3 x 2 – 4 x 2 – 4 x – 2 x = 1 + 4 ⇔ - 6 x = 5 ⇔ x = - 5 / 6 V ậ y v ớ i x = - 5 / 6 t h ì A = B .
\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)
(ĐK a\(\ne-3\))
\(\Leftrightarrow40\left(a+3\right)-3\left(3a-1\right)-2\left(7a+2\right)=24\left(a+3\right)\)
\(\Leftrightarrow40a+120-9a+3-14a-4=24a+72\)
\(\Leftrightarrow7a=47\)
\(\Leftrightarrow a=\frac{47}{7}\)
\(\frac{10}{3}-\frac{3a-1}{4a+12}-\frac{7a+2}{6a+18}=2\)
\(\frac{10}{3}-\frac{3a-1}{4\left(a+3\right)}-\frac{7a+2}{6\left(a+3\right)}=2\)
\(40\left(a+3\right)-3\left(a-1\right)-2\left(7a+2\right)=24\left(a+3\right)\)
\(17a+119=24a+27\)
\(17a-24a=72-119\)
\(-7a=-47\)
\(a=\frac{47}{7}\)
T a c ó : A = B ⇔ ( x – 3 ) ( x + 4 ) – 2 ( 3 x – 2 ) = ( x – 4 ) 2 ⇔ x 2 + 4 x – 3 x – 12 – 6 x + 4 = x 2 – 8 x + 16 ⇔ x 2 – x 2 + 4 x – 3 x – 6 x + 8 x = 16 + 12 – 4 ⇔ 3 x = 24 ⇔ x = 8 V ậ y v ớ i x = 8 t h ì A = B