Giúp mình giải bài này với
A,(2x+1)mũ2/5 -(x-1)mũ2/3=7x mũ2-14x-5/15
B,7x-1/6+2x=16-x/5
C,(x-2)mũ2/3-(2x-3)(2x+3)/8+(x-4)mũ2/6=0
Giúp mình giải bài này với
A,(2x+1)mũ2/5 -(x-1)mũ2/3=7x mũ2-14x-5/15
B,7x-1/6+2x=16-x/5
C,(x-2)mũ2/3-(2x-3)(2x+3)/8+(x-4)mũ2/6=0
A, $\frac{(2x + 1)^{2}}{5} – \frac{(x – 1)^{2}}{3} = \frac{7x^{2} – 14x – 5}{15}$
$⇔ 3(4x^{2} + 4x + 1) – 5(x^{2} – 2x + 1) = 7x^{2} – 14x – 5$
$⇔ 12x^{2} + 12x + 3 – 5x^{2} + 10x – 5 = 7x^{2} – 14x – 5$
$⇔ 36x = -3$
$⇔ x = \frac {-1}{12}$.
B, $\frac {7x – 1}{6} + 2x = \frac {16 – x}{5}$
$⇔ 5(7x – 1) + 30.2x = 6(16 – x)$
$⇔ 35x – 5 + 60x = 96 – 6x$
$⇔ 101x = 101$
$⇔ x = 1$.
C, $\frac{(x – 2)^{2}}{3} – \frac{(2x – 3)(2x + 3)}{8} + \frac{(x – 4)^{2}}{6} = 0$
$⇔ 8(x^{2} – 4x + 4) – 3(4x^{2} – 9) = 4(x^{2} – 8x + 16) = 0$
$⇔ 8x^{2} – 32x + 32 – 12x^{2} + 27+ 4x^{2} – 32x + 64 = 0$
$⇔ -64x = 123$
$⇔ x = \frac{-123}{64}$.
XIN HAY NHẤT
CHÚC EM HỌC TỐT
A,(2x+1)^2/5 – (x-1)^2/3 = 7x^2-14x-5/15
3(4x^2+4x+1) – 5(x^2-2x+1) = 7x^2-14x-5
12x^2+12x+3-5x^2+10x-5 =7x^2-14x-5
12x^2+12x-5x^2+10x-7x^2+14x=-5-3+5
36x =-3
x =-12
B,7x-1/6+2x=16-x/5
5(7x-1) + 60x = 6(16-x)
35x-5 + 60x =96 – 6x
35x+60x +6x =96+5
101x =101
x =1
C,(x-2)^2/3 – (2x-3)(2x+3)/8 + (x-4)^2/6 = 0
8(x^2-4x+4) – 3(2x-3)(2x+3) + 4(x^2-8x+16) = 0
8x^2-32x+32-12x^2-18x+18x+27+4x^2-32x+64=0
(8x^2-32x-12x^2-18x+18x+4x^2-32x)+(32+27+64)=0
-64x + 123 =0
=>-64x = -123
x =123/64