\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
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\(\frac{2^3.5.7.5^2.7^3}{2^2.5^2.7^4}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=2.5=10\)
đối với lớp 6 thì trình bày thế này dễ hiểu hơn
\(\frac{\left(2^3.5.7\right)\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}=\frac{2^3.5.5^2.7.7^3}{2.5.7^2.2.5.7^2}=\frac{2^3.5^3.7^4}{2^2.5^2.7^4}=\frac{2.5}{1}=10\)
câu trả lời của mình là 10!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
=1,7.2,3+1,7.(-3,7)-1,7.3-1,7
=1,7.[2,3+(-3,7)-3-1]
=1,7.(-5,4)
= -9,18
a, \(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}=\frac{2\cdot2^2\cdot5^2.5.7.7^2}{2\cdot5\cdot7^2}=\frac{1\cdot2^2\cdot5^2\cdot7^2}{1\cdot1\cdot1}=4900\)
b, \(1,7\cdot2,3+1,7\cdot\left(-3,7\right)-1,7.3-1,7:0,1=1,7\cdot\left(-3,7+2,3-3\right)-1,7:0,1=1,7.\left(-4,4\right)-1,7:0,1=\left(-2,7\right)-17=-19,7\)
\(\frac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\frac{2^3\left(5\cdot5^2\right)\left(7\cdot7^3\right)}{2^2\cdot5^2\cdot7^4}\)
\(=\frac{2^2\cdot2\cdot5\cdot5^2\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
Triệt tiêu ta còn \(2\cdot5=10\)
\(\frac{\left(2^3.5.7\right).\left(5^2.7^2\right)}{\left(2.5.7^2\right)^2}\)
\(=\frac{2^3.\left(5.7\right).\left(5^2.7^3\right)}{2^2.5^2.7^4}\)
\(=\frac{2^2.2.5.5^2.7.7^3}{2^2.5^2.7^4}\)
\(=\frac{2^2.2.5.5^2.7^4}{2^2.5^2.7^4}\)
\(=2.5\)
\(=10\)
\(\dfrac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{2.5.\left(7^2\right)}=\dfrac{2^3.\left(5.5^2\right).\left(7.7^3\right)}{2.5.7^2}\)
\(=\dfrac{2^3.5^3.7^4}{2.5.7^2}=\left(2^3:2\right).\left(5^3:5\right).\left(7^4:7^2\right)=2^2.5^2.7^2\)
=\(\left(2.5.7\right)^2=4900\)
(23 . 5 . 7) . (52 . 73) / 2 . 5 (72)
= (8 . 5 . 7) . (25 . 343) / 2 . 5 . 49
= 280 . 8575 / 490
= 2401000 / 490
= 4900
Bài 1:
a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)
\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)
\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)
\(=-10\cdot1.7=-17\)
b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)
\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)
\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)
\(=-\dfrac{539}{60}\)
c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)
\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)
\(=10\)
\(\dfrac{2^3.5.7(5^2.7^3)}{(2.5.7^2)^2}\)
= \(\dfrac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
= 2.5
= 10