Bài 3 trang 100 SBT toán 7 tập 2

Đề bài

Tìm \(x\) biết \(\dfrac{{{{\left( { - 0,7} \right)}^2}.{{\left( { - 5} \right)}^3}}}{{{{\left( { - 2\dfrac{1}{3}} \right)}^3}.{{\left( {1\dfrac{1}{2}} \right)}^4}.{{\left( { - 1} \right)}^5}}}.\,x \)\(= \dfrac{{{{\left( { - 40} \right)}^2}}}{{{{3.7.2}^2}}}\) 

Lời giải chi tiết

Ta có:

 \(\begin{array}{l}\dfrac{{{{\left( { - 0,7} \right)}^2}.{{\left( { - 5} \right)}^3}}}{{{{\left( { - 2\dfrac{1}{3}} \right)}^3}.{{\left( {1\dfrac{1}{2}} \right)}^4}.{{\left( { - 1} \right)}^5}}}\\ = \dfrac{{0,49.\left( { - 125} \right)}}{{{{\left( { - \dfrac{7}{3}} \right)}^3}.{{\left( {\dfrac{3}{2}} \right)}^4}.\left( { - 1} \right)}}\\ = \dfrac{{0,49.\left( { - 125} \right)}}{{\dfrac{{ - {7^3}}}{{{3^3}}}.{{\dfrac{3}{{{2^4}}}}^4}.\left( { - 1} \right)}}\\ = \dfrac{{ - 0,49.125.16}}{{{7^3}.3}}\\ = \dfrac{{ - 0,49.5.25.4.4}}{{49.7.3}} \\ = \dfrac{{ - 0,49.100.5.4}}{{49.7.3}}\\ = \dfrac{{ - 49.20}}{{49.21}}=  - \dfrac{{20}}{{21}}\end{array}\)

Và \(\dfrac{{{{\left( { - 40} \right)}^2}}}{{{{3.7.2}^2}}} = \dfrac{{{{\left( {{{5.2}^3}} \right)}^2}}}{{{{3.7.2}^2}}} = \dfrac{{{5^2}{{.2}^6}}}{{{{3.7.2}^2}}} = \dfrac{{400}}{{21}}\)

Từ đó, ta có:

\(\dfrac{{{{\left( { - 0,7} \right)}^2}.{{\left( { - 5} \right)}^3}}}{{{{\left( { - 2\dfrac{1}{3}} \right)}^3}.{{\left( {1\dfrac{1}{2}} \right)}^4}.{{\left( { - 1} \right)}^5}}}.\,x \)\(= \dfrac{{{{\left( { - 40} \right)}^2}}}{{{{3.7.2}^2}}}\)

\(\begin{array}{l} - \dfrac{{20}}{{21}}x = \dfrac{{400}}{{21}}\\x = \dfrac{{400}}{{21}}:\left( { - \dfrac{{20}}{{21}}} \right)\\x =  - 20.\end{array}\)

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