Bài 124 trang 60 SBT Toán 6 – Cánh Diều Tập 2: Tìm x, biết: $- 3x + 7 = 12 - 125$ \(\frac{1}{3}: \left( {2x - 1} \right) = \frac{{...

Tìm x, biết:

a) $- 3x + 7 = 12 - 125$

b) $\frac{1}{3}:\left( {2x - 1} \right) = \frac{{ - 4}}{{21}}$

c) $\left[ {124 - \left( {20 - 4x} \right)} \right]:20 = 12$

d) $\left( {\frac{1}{{2.3}}\, + \,\frac{1}{{3.4}} + ... + \frac{1}{{8.9}} + \,\frac{1}{{9.10}}} \right)\,.\,x = \frac{1}{5}\,$

Áp dụng tính chất giao hoán, kết hợp, phân phối của phép nhân đối với phép cộng. Quy tắc dấu ngoặc

a)

$\begin{array}{l} - 3x + 7 = 12 - 125\\ - 3x + 7 = - 113\\ - 3x = - 113 - 7\\ - 3x = - 120\\x = - 120:\left( { - 3} \right)\\x = 40\end{array}$

b)

$\begin{array}{l}\frac{1}{3}:\left( {2x - 1} \right) = \frac{{ - 4}}{{21}}\\2x - 1 = \frac{1}{3}:\frac{{ - 4}}{{21}}\\2x - 1 = \frac{1}{3}.\frac{{21}}{{\left( { - 4} \right)}}\\2x - 1 = \frac{7}{{ - 4}}\\2x = 1 + \frac{{ - 7}}{4}\\2x = \frac{{ - 3}}{4}\\x = \frac{{ - 3}}{4}:2\\x = \frac{{ - 3}}{8}\end{array}$

c)

$\begin{array}{l}\left[ {124 - \left( {20 - 4x} \right)} \right]:20 = 12\\124 - \left( {20 - 4x} \right) = 12.20\\124 - \left( {20 - 4x} \right) = 240\\20 - 4x = 124 - 240\\20 - 4x = - 116\\4x = 20 - ( - 116)\\4x = 136\\x = 34\end{array}$

d)

$\begin{array}{l}\left( {\frac{1}{{2.3}}\, + \,\frac{1}{{3.4}} + ... + \frac{1}{{8.9}} + \,\frac{1}{{9.10}}} \right)\,.\,x = \frac{1}{5}\,\\\left( {\frac{1}{2}\, - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{{10}}} \right)\,.\,x = \frac{1}{5}\,\\\left( {\frac{1}{2}\, - \frac{1}{{10}}} \right)\,.\,x = \frac{1}{5}\,\\\frac{4}{{10}}.x = \frac{1}{5}\\\frac{2}{5}.x = \frac{1}{5}\\x = \frac{1}{5}:\frac{2}{5}\\x = \frac{1}{2}\end{array}$