Giải bài 3 trang 44 SGK Toán 8 – Cánh diều

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)      

\(\begin{array}{l}\,3y – 1 =  – y + 19\\3y + y = 19 + 1\\\,\,\,\,\,\,\,4y = 20\\\,\,\,\,\,\,\,\,\,\,y = 20:5\\\,\,\,\,\,\,\,\,\,\,y = 4.\end{array}\)

Vậy phương trình có nghiệm \(y = 4\).

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 \(t = 10\).