Қысқаша көбейту формулаларын пайдаланып сандық өрнектерді ықшамдау
Қысқаша көбейту формулалары көбіне сандық өрнектерді ықшамдау кезінде қолданылады.
1.6. Есептеңіз:
1) $\sqrt {\dfrac{{{{149}^2} – {{76}^2}}}{{{{457}^2} – {{384}^2}}}} $
Шешуі
$$\sqrt {\frac{{{{149}^2} – {{76}^2}}}{{{{457}^2} – {{384}^2}}}} = \sqrt {\frac{{(149 – 76)(149 + 76)}}{{(457 – 384)(457 + 384)}}} = \sqrt {\frac{{73 \cdot 225}}{{73 \cdot 841}}} = \sqrt {\frac{{225}}{{841}}} = \frac{{15}}{{29}}$$
2) $\dfrac{{{{71}^2} – {{23}^2} + 94 \cdot 42}}{{{{62}^2} – {{32}^2}}}$
Шешуі
$$\frac{{{{71}^2} – {{23}^2} + 94 \cdot 42}}{{{{62}^2} – {{32}^2}}} = \frac{{(71 – 23)(71 + 23) + 94 \cdot 42}}{{(62 – 32)(62 + 32)}} = $$ $$ = \frac{{48 \cdot 94 + 94 \cdot 42}}{{30 \cdot 94}} = \frac{{94(48 + 42)}}{{94 \cdot 30}} = 3$$
3) $\sqrt {{9^2} \cdot {{5,3}^2} – {{25,2}^2}} $
Шешуі
$$\sqrt {{9^2} \cdot {{5,3}^2} – {{25,2}^2}} = \sqrt {{9^2} \cdot {{5,3}^2} – {9^2} \cdot {{2,8}^2}} = $$ $$ = \sqrt {{9^2} \cdot \left( {{{5,3}^2} – {{2,8}^2}} \right)} = \sqrt {{9^2} \cdot (5,3 – 2,8)(5,3 + 2,8)} = $$ $$ = 9\sqrt {2,5 \cdot 8,1} = 9\sqrt {\frac{{25 \cdot 81}}{{100}}} = \frac{{9 \cdot 5 \cdot 9}}{{10}} = 40,5$$
4) $0,3\sqrt {\dfrac{{{{115}^2} – {{15}^2}}}{{130}}} + \dfrac{1}{{17}}\sqrt {\dfrac{{{{273}^2} – {{16}^2}}}{{257}}} $
Шешуі
$$0,3\sqrt {\frac{{{{115}^2} – {{15}^2}}}{{130}}} + \frac{1}{{17}}\sqrt {\frac{{{{273}^2} – {{16}^2}}}{{257}}} = $$ $$ = 0,3\sqrt {\frac{{100 \cdot 130}}{{130}}} + \frac{1}{{17}}\sqrt {\frac{{257 \cdot 289}}{{257}}} = $$ $$0,3 \cdot 10 + \frac{1}{{17}} \cdot 17 = 4$$
5) $\sqrt {\dfrac{{{{(\sqrt {11} )}^3} – {{(\sqrt 7 )}^3}}}{{\sqrt {11} – \sqrt 7 }} + \sqrt {77} } \cdot (\sqrt {11} – \sqrt 7 )$
Шешуі
$$\sqrt {\frac{{{{(\sqrt {11} )}^3} – {{(\sqrt 7 )}^3}}}{{\sqrt {11} – \sqrt 7 }} + \sqrt {77} } \cdot (\sqrt {11} – \sqrt 7 ) = $$ $$\sqrt {\frac{{(\sqrt {11} – \sqrt 7 )(11 + \sqrt {77} + 7)}}{{\sqrt {11} – \sqrt 7 }} + \sqrt {77} } \times (\sqrt {11} – \sqrt 7 ) = $$ $$ = \sqrt {11 + 2\sqrt {77} + 7} \cdot (\sqrt {11} – \sqrt 7 ) = \sqrt {{{(\sqrt {11} + \sqrt 7 )}^2}} \cdot (\sqrt {11} – \sqrt 7 ) = $$ $$ = (\sqrt {11} + \sqrt 7 ) \cdot (\sqrt {11} – \sqrt 7 ) = 4$$
6) $\dfrac{{27 \cdot \left( {{{1,7}^3} – {{1,5}^3}} \right)}}{{{{5,1}^2} + 5,1 \cdot 4,5 + {{4,5}^2}}}$
Шешуі
$$\dfrac{{27 \cdot \left( {{{1,7}^3} – {{1,5}^3}} \right)}}{{{{5,1}^2} + 5,1 \cdot 4,5 + {{4,5}^2}}} = \frac{{27 \cdot \left( {{{1,7}^3} – {{1,5}^3}} \right)}}{{{3^2}\left( {{{1,7}^2} + 1,7 \cdot 1,5 + {{1,5}^2}} \right)}} = $$ $$ = \frac{{3(1,7 – 1,5)\left( {{{1,7}^2} + 1,7 \cdot 1,5 + {{1,5}^2}} \right)}}{{{{1,7}^2} + 1,7 \cdot 1,5 + {{1,5}^2}}} = 3 \cdot 0,2 = 0,6$$
7) $\left( {\dfrac{{\sqrt {{{561}^2} – {{459}^2}} }}{{4\frac{2}{7} \cdot 0,15 + 4\frac{2}{7}:\frac{{20}}{3}}} + 4\sqrt {10} } \right):\dfrac{1}{3}\sqrt {40} $
Шешуі
$$\left( {\frac{{\sqrt {{{561}^2} – {{459}^2}} }}{{4\frac{2}{7} \cdot 0,15 + 4\frac{2}{7}:\frac{{20}}{3}}} + 4\sqrt {10} } \right):\frac{1}{3}\sqrt {40} = \left( {\frac{{\sqrt {102 \cdot 1020} }}{{4\frac{2}{7}(0,15 + 0,15)}} + 4\sqrt {10} } \right):\frac{2}{3}\sqrt {10} = $$ $$ = \left( {\frac{{102\sqrt {10} }}{{\frac{{30}}{7} \cdot \frac{3}{{10}}}} + 4\sqrt {10} } \right) \cdot \frac{3}{{2\sqrt {10} }} = \left( {\frac{{238\sqrt {10} }}{3} + 4\sqrt {10} } \right) \cdot \frac{3}{{2\sqrt {10} }} = \frac{{250\sqrt {10} }}{3} \cdot \frac{3}{{2\sqrt {10} }} = 125$$
8) $0,125 \cdot \left( {{{2,1}^3} + 12 \cdot 2,1 \cdot 1,9 + {{1,9}^3}} \right)$
Шешуі
$$0,125 \cdot \left( {{{2,1}^3} + 12 \cdot 2,1 \cdot 1,9 + {{1,9}^3}} \right) = \left| {{a^3} + 3ab(a + b) + {b^3} = {{(a + b)}^3}} \right| = $$ $$ = 0,125 \cdot ({2,1^3} + 3 \cdot 2,1 \cdot 1,9 \cdot \underbrace {(2,1 + 1,9)}_{ = 4} + {1,9^3}) = \frac{1}{8}{(2,1 + 1,9)^3} = \frac{1}{8} \cdot {4^3} = 8$$
9) ${0,298^3} + 3 \cdot 0,298 \cdot 0,702 + {0,702^3}$
Шешуі
$${0,298^3} + 3 \cdot 0,298 \cdot 0,702 + {0,702^3} = \left| {{{(a + b)}^3} = {a^3} + 3ab(a + b) + {b^3}} \right| = $$ $$ = {0,298^3} + 3 \cdot 0,298 \cdot 0,702 \cdot \underbrace {(0,298 + 0,702)}_{ = 1} + {0,702^3} = {(0,298 + 0,702)^3} = 1$$
10) ${27^5} – 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29 – 15 \cdot {81^2}$
Шешуі
$${27^5} – 25 \cdot 26 \cdot 27 \cdot 28 \cdot 29 – 15 \cdot {81^2} = |27 = x| = $$ $$ = {x^5} – (x – 2)(x – 1)x(x + 1)(x + 2) – 15 \cdot {3^2} \cdot {x^2} = $$ $$ = {x^5} – x\left( {{x^2} – 4} \right)\left( {{x^2} – 1} \right) – 15 \cdot 9{x^2} = {x^5} – {x^5} + 5{x^3} – 4x – 15 \cdot 9{x^2} = $$ $$ = 5 \cdot {27^3} – 4 \cdot 27 – 3 \cdot 5 \cdot 9 \cdot {27^2} = 5 \cdot {3^9} – 108 – 5 \cdot {3^9} = – 108$$
11) $3 \cdot 5 \cdot \left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right) – {16^8} – 4$
Шешуі
$3 \cdot 5 \cdot \left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)$ өрнегін ${4^2} – 1$ айырмасына бөліп, қысқаша көбейту формуласы бойынша көбейткіштерге жіктейміз: $$3 \cdot 5 \cdot \left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right) \cdot \frac{{\left( {{4^2} – 1} \right)}}{{\left( {{4^2} – 1} \right)}} – {16^8} – 4 = $$ $$ = \frac{{15 \cdot \left( {{4^2} – 1} \right)\left( {{4^2} + 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)}}{{\left( {{4^2} – 1} \right)}} – {16^8} – 4 = $$ $$ = \frac{{15 \cdot \left( {{4^4} – 1} \right)\left( {{4^4} + 1} \right)\left( {{4^8} + 1} \right)}}{{15}} – {16^8} – 4 = $$ $$ = \left( {{4^8} – 1} \right)\left( {{4^8} + 1} \right) – {16^8} – 4 = {4^{16}} – 1 – {16^8} – 4 = $$ $$ = {4^{16}} – 1 – {4^{16}} – 4 = – 5.$$