Toán Lớp 7: (√ $\frac{4}{49}$ – $\frac{\frac{1}{3} -0,25+\frac{1}{5} }{\frac{7}{6} -0,875+0,7}$ ) : ( $1^{3}$ +$2^{3}$ +$3^{3}$ +….+$2021^{3}$ )
Tính
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Toán Lớp 7: (√ $\frac{4}{49}$ – $\frac{\frac{1}{3} -0,25+\frac{1}{5} }{\frac{7}{6} -0,875+0,7}$ ) : ( $1^{3}$ +$2^{3}$ +$3^{3}$ +….+$2021^{3}$ )
Tính
Comments ( 2 )
Giải đáp+Lời giải và giải thích chi tiết:
Xét:
\sqrt{4/49}=2/7
Đặt A=1/3-0,25+1/5
=1/3-1/4+1/5
Đặt B=7/6-0,875+0,7
=7/6-7/8+7/10
=7/2 . (1/3-1/4+1/5)
-> A/B=1/(7/2)=2/7
-> \sqrt{4/49}-(1/3-0,25+1/5)/(7/5-0,875+0,7)=2/7-2/7=0
-> [\sqrt{4/49}-(1/3-0,25+1/5)/(7/5-0,875+0,7)] : (1^3+2^3+3^3+…+2021^3)=0
Answer
(\sqrt{4/49} – {1/3 – 0,25 + 1/5}/{7/6 – 0,875 + 0,7}) : (1^3 + 2^3 + 3^3 + … + 2021^3)
= (\sqrt{(2/7)^2} – {1/3 – 1/4 + 1/5}/{7/6 – 7/8 + 7/10}) : (1^3 + 2^3 + 3^3 + … + 2021^3)
= 2/7 – [{1/3 – 1/4 + 1/5}/{7/2 . (1/3 – 1/4 + 1/5)}] : (1^3 + 2^3 + 3^3 + … + 2021^3)
= 2/7 – 2/7 : (1^3 + 2^3 + 3^3 + … + 2021^3)
= 0 : (1^3 + 2^3 + 3^3 + … + 2021^3)
= 0