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電子課本網(wǎng) 第79頁(yè)

第79頁(yè)

信息發(fā)布者:
解:$3(2y + 1) = -2(1 + y) + 3(y + 3)$
$6y + 3 = -2 - 2y + 3y + 9$
$6y + 2y - 3y = -2 + 9 - 3$
$5y = 4$
$y = \frac{4}{5}$
解:$x - 2[x - 3(x - 1)] = 8$
$x - 2[x - 3x + 3] = 8$
$x - 2[-2x + 3] = 8$
$x + 4x - 6 = 8$
$5x = 14$
$x = \frac{14}{5}$
解:根據(jù)題意,得
$3(2 - x) = 2×2(x + 1)$
$6 - 3x = 4(x + 1)$
$6 - 3x = 4x + 4$
$-3x - 4x = 4 - 6$
$-7x = -2$
$x = \frac{2}{7}$
答:當(dāng)$x = \frac{2}{7}$時(shí),代數(shù)式$3(2 - x)$的值是$2(x + 1)$的值的2倍。
解:方程兩邊同乘2,得$\frac{1}{2}[ \frac{1}{2}(y + 1) - \frac{1}{2} ] - \frac{1}{2} = 1$
方程兩邊同加$\frac{1}{2},$得$\frac{1}{2}[ \frac{1}{2}(y + 1) - \frac{1}{2} ] = \frac{3}{2}$
方程兩邊同乘2,得$\frac{1}{2}(y + 1) - \frac{1}{2} = 3$
方程兩邊同加$\frac{1}{2},$得$\frac{1}{2}(y + 1) = \frac{7}{2}$
方程兩邊同乘2,得$y + 1 = 7$
方程兩邊同減1,得$y = 6$
解:解方程$5m + 2x = \frac{1}{2} + x,$
移項(xiàng)得$2x - x = \frac{1}{2} - 5m,$
解得$x = \frac{1}{2} - 5m。$
解方程$x(m + 1) = m(x + 1),$
去括號(hào)得$mx + x = mx + m,$
移項(xiàng)得$mx + x - mx = m,$
解得$x = m。$
由題意得$\frac{1}{2} - 5m - m = 2,$
合并同類(lèi)項(xiàng)得$\frac{1}{2} - 6m = 2,$
移項(xiàng)得$-6m = 2 - \frac{1}{2},$
即$-6m = \frac{3}{2},$
解得$m = -\frac{1}{4}。$
答:當(dāng)$m = -\frac{1}{4}$時(shí),滿(mǎn)足條件。