Giải bài 3 trang 44 SGK Toán 8 – Cánh diều — Không quảng cáo

Đề bài

Lời giải chi tiết

a)

$\begin{array}{l}6x + 4 = 0\\\,\,\,\,\,\,\,\,6x =  - 4\\\,\,\,\,\,\,\,\,\,\,x = \left( { - 4} \right):6\\\,\,\,\,\,\,\,\,\,\,x =  - \frac{2}{3}.\end{array}$

Vậy phương trình có nghiệm $x =  - \frac{2}{3}.$

b)

$\begin{array}{l} - 14x - 28 = 0\\\,\,\,\,\,\,\,\,\, - 14x = 28\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 28:\left( { - 14} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x =  - 2\end{array}$

Vậy phương trình có nghiệm $x =  - 2.$

c)

$\begin{array}{l}\frac{1}{3}x - 5 = 0\\\,\,\,\,\,\,\,\frac{1}{3}x = 5\\\,\,\,\,\,\,\,\,\,\,\,x = 5:\frac{1}{3}\\\,\,\,\,\,\,\,\,\,\,\,x = 15.\end{array}$

Vậy phương trình có nghiệm $x = 15$.

d)

$\begin{array}{l}\,3y - 1 =  - y + 19\\3y + y = 19 + 1\\\,\,\,\,\,\,\,4y = 20\\\,\,\,\,\,\,\,\,\,\,y = 20:4\\\,\,\,\,\,\,\,\,\,\,y =5.\end{array}$

Vậy phương trình có nghiệm $y = 5$.

e)

$\begin{array}{l} - 2\left( {z + 3} \right) - 5 = z + 4\\\,\,\, - 2z - 6 - 5 = z + 4\\\,\,\,\,\,\,\,\, - 2z - 11 = z + 4\\\,\,\,\,\,\,\,\,\,\, - 2z - z = 4 + 11\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3z = 15\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z = 15:\left( { - 3} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,z =  - 5.\end{array}$

Vậy phương trình có nghiệm $z =  - 5$.

g)

$\begin{array}{l}3\left( {t - 10} \right) = 7\left( {t - 10} \right)\\\,\,\,\,3t - 30 = 7t - 70\\\,\,\,\,\,3t - 7t =  - 70 + 30\\\,\,\,\,\,\,\,\,\,\, - 4t =  - 40\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = \left( { - 40} \right):\left( { - 4} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = 10.\end{array}$

Vậy phương trình có nghiệm $t = 10$.