Bài 132 trang 55 SGK Toán 6 tập 2

Đề bài

Tìm x, biết:

a) \( \displaystyle 2{2 \over 3}.x + 8{2 \over 3} = 3{1 \over 3}\)

b) \( \displaystyle 3{2 \over 7}.x - {1 \over 8} = 2{3 \over 4}\) 

Lời giải chi tiết

a)

\(\begin{array}{l}
2\dfrac{2}{3}.x + 8\dfrac{2}{3} = 3\dfrac{1}{3}\\
\dfrac{8}{3}x + \dfrac{{26}}{3} = \dfrac{{10}}{3}
\end{array}\) 

\(\begin{array}{l}
\dfrac{8}{3}x = \dfrac{{10}}{3} - \dfrac{{26}}{3}\\
\dfrac{8}{3}x = \dfrac{{ - 16}}{3}\\
x = \dfrac{{ - 16}}{3}:\dfrac{8}{3}\\
x = \dfrac{{ - 16}}{3}.\dfrac{3}{8}\\
x = - 2
\end{array}\)

b) 

\(\begin{array}{l}
3\dfrac{2}{7}.x - \dfrac{1}{8} = 2\dfrac{3}{4}\\
\dfrac{{23}}{7}x - \dfrac{1}{8} = \dfrac{{11}}{4}\\
\dfrac{{23}}{7}x = \dfrac{{11}}{4} + \dfrac{1}{8}\\
\dfrac{{23}}{7}x = \dfrac{{22}}{8} + \dfrac{1}{8}\\
\dfrac{{23}}{7}x = \dfrac{{23}}{8}\\
x = \dfrac{{23}}{8}:\dfrac{{23}}{7}\\
x = \dfrac{{23}}{8}.\dfrac{7}{{23}}\\
x = \dfrac{7}{8}
\end{array}\) 

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