Bài 29 trang 13 SBT toán 7 tập 1

Đề bài

Tính giá trị của các biểu thức sau với \(\left| a \right| = 1,5;b =  - 0,75\)

\(M = a + 2ab - b\)

\(N = a: 2 - 2: b\)

\(\displaystyle P = \left( { - 2} \right):{a^2} - b.{2 \over 3}\)

Lời giải chi tiết

Vì \(\left| a \right| = 1,5\) nên \(a =1,5\) hoặc \(a = -1,5\).

* Với \(a = 1,5 ; b = -0,75\), ta có:

\(M = 1,5+ 2.1,5.(-0,75)  - (-0,75) \)

      \(= 1,5 + (-2,25) + 0,75 \)

      \(=(1,5+ 0,75) + (-2,25)\)

      \(=2,25+ (-2,25) =0\)

\(\begin{array}{l}
N = 1,5:2 - 2:\left( { - 0,75} \right)\\
\,\,\,\,\,\, = \dfrac{3}{2}.\dfrac{1}{2} - 2:\dfrac{{ - 3}}{4}\\
\,\,\,\,\,\, = \dfrac{3}{4} - \dfrac{2}{1}.\dfrac{{ - 4}}{3}\\
\,\,\,\,\,\, = \dfrac{3}{4} - \dfrac{{ - 8}}{3} = \dfrac{3}{4} + \dfrac{8}{3}\\
\,\,\,\,\,\, = \dfrac{9}{{12}} + \dfrac{{32}}{{12}} = \dfrac{{41}}{{12}} = 3\dfrac{5}{{12}}
\end{array}\)

\(\begin{array}{l}
P = \left( { - 2} \right):{\left( {1,5} \right)^2} - \left( { - 0,75} \right).\dfrac{2}{3}\\
\,\,\,\, = \left( { - 2} \right):{\left( {\dfrac{3}{2}} \right)^2} + \dfrac{3}{4}.\dfrac{2}{3}\\
\,\,\,\, = \left( { - 2} \right):\dfrac{9}{4} + \dfrac{3}{4}.\dfrac{2}{3}\\
\,\,\,\, = \left( { - 2} \right).\dfrac{4}{9} + \dfrac{1}{2} = \dfrac{{ - 8}}{9} + \dfrac{1}{2}\\
\,\,\,\, = \dfrac{{ - 16}}{{18}} + \dfrac{9}{{18}} = \dfrac{{ - 7}}{{18}}
\end{array}\)

* Với \(a = -1,5; b = -0,75\) ta có:

\(M = - 1,5 + 2.(-1,5) ( - 0,75) - (-0,75)\)

      \(= - 1,5 + ( 2,25) + 0,75 = 1,5\)

\(N = - 1,5 : 2 - 2 : ( -0,75)\)

\(\begin{array}{l}
\;\;\;= - \dfrac{3}{2}.\dfrac{1}{2} + 2:\dfrac{3}{4}\\
\;\;\;= - \dfrac{3}{4} + 2.\dfrac{4}{3}\\
\;\;\;= - \dfrac{3}{4} + \dfrac{8}{3}\\
\;\;\;= \dfrac{{ - 9}}{{12}} + \dfrac{{32}}{{12}} = \dfrac{{23}}{{12}}
\end{array}\) 

\(\begin{array}{*{20}{l}}
{P = \left( { - 2} \right):{{\left( { - 1,5} \right)}^2} - \left( { - 0,75} \right).\dfrac{2}{3}}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \left( { - 2} \right):{{\left( {\dfrac{{ - 3}}{2}} \right)}^2} + \dfrac{3}{4}.\dfrac{2}{3}}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \left( { - 2} \right):\dfrac{9}{4} + \dfrac{3}{4}.\dfrac{2}{3}}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \left( { - 2} \right).\dfrac{4}{9} + \dfrac{1}{2} = \dfrac{{ - 8}}{9} + \dfrac{1}{2}}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} = \dfrac{{ - 16}}{{18}} + \dfrac{9}{{18}} = \dfrac{{ - 7}}{{18}}}
\end{array}\)