Giúp mình với! CHUYÊN ĐỀ ÔN THI HSG TOÁN 7 (24-25),CĐ 1: THỰC HIỆN PHÉP TÍNH,Câu 1. (Để học sinh giỏi toán 7 năm học 2022-2023),Thực hiện phép tính:,$a.~A=\frac7{35}.\frac{10}{19}+\frac7{19}.\frac9{35}-\frac2{35}.$,$b.~B=\frac{\sqrt{49}}6.\frac{-12}{\sqrt{(-7)^4}}-\frac12.(\frac{-5}3)-\sqrt{(\frac{-1}2)^2}.\frac{\sqrt{(-6)^2}}7$,$c.~C=(\frac{-2}7-\frac5{12}):\frac7{12}-(\frac57+\frac7{12}):\frac7{12}$,$d.~D=\frac49:(\frac1{15}-\frac23)+\frac49:(\frac1{11}-\frac5{22})$,$e.~E=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}.$,$f.~F=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}$,$g.~G=\frac{155-\frac{10}7-\frac5{11}+\frac5{23}}{403-\frac{26}7-\frac{13}{11}+\frac{13}{23}}+\frac{\frac35+\frac3{13}-0,9}{\frac7{91}+0,2-\frac3{10}}.$,$h.~H=(\frac{0,4-\frac29+\frac2{11}}{1,4-\frac79+\frac7{11}}-\frac{\frac13-0,25+\frac15}{1\frac16-0,875+0,7}):\frac{2020}{2021}$

Question

alidade-elmir · Accepted Answer

Câu 1
$\displaystyle \begin{array}{{>{\displaystyle}l}}
f,\ F=\frac{4^{6} .9^{5} +6^{9} .120}{8^{4} .3^{12} -6^{11}}\
=\frac{\left( 2^{2} ight)^{6} .\left( 3^{2} ight)^{5} +( 2.3)^{9} .120}{\left( 2^{3} ight)^{4} .3^{12} -( 2.3)^{11}}\
=\frac{2^{12} .3^{10} +2^{9} .3^{9} .120}{2^{12} .3^{12} -2^{11} .3^{11}}\
=\frac{2^{9} .3^{9}\left( 2^{3} .3+120 ight)}{2^{11} .3^{11}( 2.3-1)}\
=\frac{144}{2^{2} .3^{2} .5} =\frac{144}{180} =\frac{4}{5}
\end{array}$

Timi · Answer

Câu 1.
a. $~A=\frac7{35}.\frac{10}{19}+\frac7{19}.\frac9{35}-\frac2{35}$
$=\frac7{35}	imes (\frac{10}{19}+\frac9{19}-\frac2{35})$
$=\frac7{35}	imes (\frac{19}{19}-\frac2{35})$
$=\frac7{35}	imes (1-\frac2{35})$
$=\frac7{35}	imes \frac{33}{35}$
$=\frac{33}{175}$
b. $~B=\frac{\sqrt{49}}6.\frac{-12}{\sqrt{(-7)^4}}-\frac12.(\frac{-5}3)-\sqrt{(\frac{-1}2)^2}.\frac{\sqrt{(-6)^2}}7$
$=\frac76	imes \frac{-12}{49}-\frac12	imes \frac{-5}3-\frac12	imes \frac67$
$=\frac76	imes \frac{-12}{49}+\frac12	imes \frac53-\frac12	imes \frac67$
$=\frac12	imes (\frac{-14}{49}+\frac53-\frac67)$
$=\frac12	imes (\frac{-2}{7}+\frac53-\frac67)$
$=\frac12	imes (-1+\frac53)$
$=\frac12	imes \frac23$
$=\frac13$
c. $~C=(\frac{-2}7-\frac5{12}):\frac7{12}-(\frac57+\frac7{12}):\frac7{12}$
$=(\frac{-2}7-\frac5{12}-\frac57-\frac7{12})	imes \frac{12}7$
$=(\frac{-7}7-\frac{12}{12})	imes \frac{12}7$
$=(-1-1)	imes \frac{12}7$
$=-2	imes \frac{12}7$
$=\frac{-24}7$
d. $~D=\frac49:(\frac1{15}-\frac23)+\frac49:(\frac1{11}-\frac5{22})$
$=\frac49:(\frac1{15}-\frac{10}{15})+\frac49:(\frac2{22}-\frac5{22})$
$=\frac49:(\frac{-9}{15})+\frac49:(\frac{-3}{22})$
$=\frac49:(\frac{-3}{5})+\frac49:(\frac{-3}{22})$
$=\frac49	imes (\frac{-5}{3}+\frac{22}{3})$
$=\frac49	imes \frac{17}{3}$
$=\frac{68}{27}$
e. $~E=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}$
$=\frac{5	imes 2^{30}	imes 3^{18}-4	imes 3^{20}	imes 2^{27}}{5	imes 2^{10}	imes 2^{19}	imes 3^{19}-7	imes 2^{29}	imes 3^{18}}$
$=\frac{2^{27}	imes 3^{18}	imes (5	imes 2^3-4	imes 3^2)}{2^{29}	imes 3^{18}	imes (5	imes 3-7)}$
$=\frac{2^{27}	imes 3^{18}	imes (40-36)}{2^{29}	imes 3^{18}	imes 8}$
$=\frac{2^{27}	imes 3^{18}	imes 4}{2^{29}	imes 3^{18}	imes 8}$
$=\frac{1}{4}$
f. $~F=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}$
$=\frac{2^{12}	imes 3^{10}+2^{11}	imes 3^{11}	imes 5}{2^{12}	imes 3^{12}-2^{11}	imes 3^{11}}$
$=\frac{2^{11}	imes 3^{10}	imes (2+3	imes 5)}{2^{11}	imes 3^{11}	imes (3-1)}$
$=\frac{2^{11}	imes 3^{10}	imes 17}{2^{11}	imes 3^{11}	imes 2}$
$=\frac{17}{6}$
g. $~G=\frac{155-\frac{10}7-\frac5{11}+\frac5{23}}{403-\frac{26}7-\frac{13}{11}+\frac{13}{23}}+\frac{\frac35+\frac3{13}-0,9}{\frac7{91}+0,2-\frac3{10}}$
$=\frac{155-(\frac{10}7+\frac5{11}-\frac5{23})}{403-(\frac{26}7+\frac{13}{11}-\frac{13}{23})}+\frac{\frac35+(\frac3{13}-\frac9{10})}{\frac1{13}+(\frac15-\frac3{10})}$
$=\frac{155-13	imes (\frac{10}{91}+\frac5{143}-\frac5{299})}{403-13	imes (\frac{26}{91}+\frac{13}{143}-\frac{13}{299})}+\frac{\frac35-\frac{69}{130}}{\frac1{13}+\frac{-1}{10}}$
$=\frac{155-13	imes \frac{130+35-35}{1001}}{403-13	imes \frac{260+65-65}{1001}}+\frac{\frac{78-69}{130}}{\frac{10-13}{130}}$
$=\frac{155-13	imes \frac{130}{1001}}{403-13	imes \frac{260}{1001}}+\frac{\frac{9}{130}}{\frac{-3}{130}}$
$=\frac{155-\frac{130}{77}}{403-\frac{260}{77}}-3$

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