TÍNH BẰNG CÁCH THUẬN TIỆN A,20^5.5^7/100^5 B, 6^3+3.6^2+3^3/-13 C,4^6.9^5+6^9>120/8^4.3^12-6^11 18/08/2021 Bởi Adalyn TÍNH BẰNG CÁCH THUẬN TIỆN A,20^5.5^7/100^5 B, 6^3+3.6^2+3^3/-13 C,4^6.9^5+6^9>120/8^4.3^12-6^11
Đáp án: $A = 25$; $B=-27$ Giải thích các bước giải: $\begin{array}{l}A)\dfrac{{{{20}^5}{{.5}^7}}}{{{{100}^5}}} = \dfrac{{{{20}^5}{{.5}^7}}}{{{{\left( {20.5} \right)}^5}}} = \dfrac{{{{20}^5}{{.5}^7}}}{{{{20}^5}{{.5}^5}}} = {5^2}A = 25\\B)\dfrac{{{6^3} + {{3.6}^2} + {3^3}}}{{ – 13}}\\ = \dfrac{{{6^2}\left( {6 + 3} \right) + {3^3}}}{{ – 13}}\\ = \dfrac{{{2^2}{{.3}^2}.9 + {3^3}}}{{ – 13}}\\ = \dfrac{{{2^2}{{.3}^4} + {3^3}}}{{ – 13}}\\ = \dfrac{{{3^3}\left( {{2^2}.3 + 1} \right)}}{{ – 13}}\\ = \dfrac{{{3^3}.13}}{{ – 13}}\\ = – {3^3}\\ = – 27\\C){4^6}{.9^5} + {6^9} = {2^{12}}{.3^{10}} + {2^9}{.3^9} = {2^9}{.3^9}\left( {{2^3}.3 + 1} \right) = {2^9}{.3^9}.25\\\dfrac{{120}}{{{8^4}{{.3}^{12}} – {6^{11}}}} = \dfrac{{{2^3}.3.5}}{{{2^{12}}{{.3}^{12}} – {2^{11}}{{.3}^{11}}}} = \dfrac{{{2^3}.3.5}}{{{2^{11}}{{.3}^{11}}\left( {2.3 – 1} \right)}} = \dfrac{{{2^3}.3.5}}{{{2^{11}}{{.3}^{11}}.5}} = \dfrac{1}{{{2^8}{{.3}^{10}}}}\\{2^9}{.3^9}.25 > 1 > \dfrac{1}{{{2^8}{{.3}^{10}}}} \Rightarrow {4^6}{.9^5} + {6^9} > \dfrac{{120}}{{{8^4}{{.3}^{12}} – {6^{11}}}}\end{array}$ Bình luận
Đáp án:
Giải thích các bước giải:
Đáp án:
$A = 25$; $B=-27$
Giải thích các bước giải:
$\begin{array}{l}
A)\dfrac{{{{20}^5}{{.5}^7}}}{{{{100}^5}}} = \dfrac{{{{20}^5}{{.5}^7}}}{{{{\left( {20.5} \right)}^5}}} = \dfrac{{{{20}^5}{{.5}^7}}}{{{{20}^5}{{.5}^5}}} = {5^2}A = 25\\
B)\dfrac{{{6^3} + {{3.6}^2} + {3^3}}}{{ – 13}}\\
= \dfrac{{{6^2}\left( {6 + 3} \right) + {3^3}}}{{ – 13}}\\
= \dfrac{{{2^2}{{.3}^2}.9 + {3^3}}}{{ – 13}}\\
= \dfrac{{{2^2}{{.3}^4} + {3^3}}}{{ – 13}}\\
= \dfrac{{{3^3}\left( {{2^2}.3 + 1} \right)}}{{ – 13}}\\
= \dfrac{{{3^3}.13}}{{ – 13}}\\
= – {3^3}\\
= – 27\\
C){4^6}{.9^5} + {6^9} = {2^{12}}{.3^{10}} + {2^9}{.3^9} = {2^9}{.3^9}\left( {{2^3}.3 + 1} \right) = {2^9}{.3^9}.25\\
\dfrac{{120}}{{{8^4}{{.3}^{12}} – {6^{11}}}} = \dfrac{{{2^3}.3.5}}{{{2^{12}}{{.3}^{12}} – {2^{11}}{{.3}^{11}}}} = \dfrac{{{2^3}.3.5}}{{{2^{11}}{{.3}^{11}}\left( {2.3 – 1} \right)}} = \dfrac{{{2^3}.3.5}}{{{2^{11}}{{.3}^{11}}.5}} = \dfrac{1}{{{2^8}{{.3}^{10}}}}\\
{2^9}{.3^9}.25 > 1 > \dfrac{1}{{{2^8}{{.3}^{10}}}} \Rightarrow {4^6}{.9^5} + {6^9} > \dfrac{{120}}{{{8^4}{{.3}^{12}} – {6^{11}}}}
\end{array}$