Giải bài 4 trang 44 SGK Toán 8 – Cánh diều

Đề bài

Giải các phương trình:

a)      \(\frac{{5x – 2}}{3} = \frac{{5 – 3x}}{2}\);

b)     \(\frac{{10x + 3}}{{12}} = 1 + \frac{{6 + 8x}}{9}\);

c)      \(\frac{{7x – 1}}{6} + 2x = \frac{{16 – x}}{5}\).

Lời giải chi tiết

a)       

\(\begin{array}{l}\,\,\,\,\,\,\,\frac{{5x – 2}}{3} = \frac{{5 – 3x}}{2}\\\frac{{2\left( {5x – 2} \right)}}{6} = \frac{{3\left( {5 – 3x} \right)}}{6}\\\,2\left( {5x – 2} \right) = 3\left( {5 – 3x} \right)\\\,\,\,\,\,\,10x – 4 = 15 – 9x\\\,\,\,10x + 9x = 15 + 4\\\,\,\,\,\,\,\,\,\,\,\,\,\,19x = 19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 19:19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1.\end{array}\)

Vậy phương trình có nghiệm \(x = 1\).

b)      

\(\begin{array}{l}\,\,\,\,\,\,\frac{{10x + 3}}{{12}} = 1 + \frac{{6 + 8x}}{9}\\\frac{{3\left( {10x + 3} \right)}}{{36}} = \frac{{36}}{{36}} + \frac{{4\left( {6 + 8x} \right)}}{{36}}\\\,3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\\,\,\,\,\,\,\,30x + 9 = 36 + 24 + 32x\\\,\,\,\,\,\,\,30x + 9 = 60 + 32x\\\,30x – 32x = 60 – 9\\\,\,\,\,\,\,\,\,\,\,\,\,\, – 2x = 51\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x =  – \frac{{51}}{2}.\end{array}\)

Vậy phương trình có nghiệm \(x =  – \frac{{51}}{2}\).

c)       

\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{7x – 1}}{6} + 2x = \frac{{16 – x}}{5}\\\frac{{5\left( {7x – 1} \right)}}{{30}} + \frac{{30.2x}}{{30}} = \frac{{6\left( {16 – x} \right)}}{{30}}\\\,\,5\left( {7x – 1} \right) + 30.2x = 6\left( {16 – x} \right)\\\,\,\,\,\,\,\,\,\,\,35x – 5 + 60x = 96 – 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x – 5 = 96 – 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x + 6x = 96 + 5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,101x = 101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 101:101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1\end{array}\)

Vậy phương trình có nghiệm \(x = 1\).