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ta có : (ghi lại đề)
=6+12+18+24+30/3+6+9+12+15
=2*(3/3+6/6+9/9+12/12+15/15)
=2*(1+1+1+1+1)
=2*5=10
chúc main học tốt nhé
\(\left(3-\dfrac{1}{2}\right)^3-\dfrac{5^4}{25}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)
\(=\left(\dfrac{5}{2}\right)^3-\dfrac{5^4}{5^2}+\left[\left(\dfrac{1}{2}\right)^2\right]^3\)
\(=\left(\dfrac{5}{2}\right)^3+\left[\left(\dfrac{1}{2}\right)^2\right]^3-5\)
\(=\left(\dfrac{5}{2}+\dfrac{1}{2}\right)\left[\left(\dfrac{5}{2}\right)^2-\dfrac{5}{2}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-5\)
\(=3\left[\dfrac{25}{4}+\dfrac{1}{4}-\dfrac{5}{4}\right]-5\)
\(=3.\dfrac{21}{4}-5\)
\(=\dfrac{63}{4}-5=\dfrac{43}{4}\)
Đính chính \(\dfrac{5^4}{5^2}=25\)
\(...=\dfrac{63}{4}-25=-\dfrac{37}{4}\)
Gọi số xe loại 40 tấn, 25 tấn, 5 tấn lần lượt là x, y, z (xe) (x,y,z thuộc N*)
Theo bài ra ta có: x + y + z = 114
2/3a=2/5b=3/7c(1)
Chia cả 3 vế của (1) cho BCNN(2,2,3) = 6 được:
2a/3.6=2b/5.6=3c/7.6
=>a/9=b/15=c/14=a+b+c/9+15+14=114/38=3
=> a = 3.9 = 27
b = 3.15 = 45
c = 3.14 = 42
Vậy số xe 40 t, 25 t, 5 t lần lượt là 27,45,42
a: =-15/24-18/24+60/24
=27/24=9/8
b: =6/12-9/12-10/12-7/12=-20/12=-5/3
c: =17/2+3/7-5/3=305/42
c: =-3-2/3-10/9-25/3-5/6
=-10-19/9-5/6
=-180/18-38/18-15/18=-233/18
1)
a. \(\left(3x^2-50\right)^2=5^4\)
\(\Leftrightarrow3x^4-50=625\)
\(\Leftrightarrow3x^4=675\)
\(\Leftrightarrow x^4=225\)
\(\Leftrightarrow x=\sqrt{15}\)
2)
a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)
\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)
b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)
\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)
\(A=\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\dfrac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\dfrac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)
\(=\dfrac{2^{12}.3^4.2}{2^{12}.3^5.4}-\dfrac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)
\(=\dfrac{1}{6}-\dfrac{-10}{3}\)
\(=\dfrac{7}{2}\)
Bài 1 :
a) \(\frac{12}{21}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{4}{7}-\frac{3}{7}+\left(-\frac{2}{3}\right)=\frac{1}{7}-\frac{2}{3}=-\frac{11}{21}\)
b) \(\left(-\frac{25}{13}\right)+\left(-\frac{9}{17}\right)+\frac{12}{13}+\left(-\frac{25}{17}\right)\)
\(=\left[\left(-\frac{25}{13}\right)+\frac{12}{13}\right]+\left[\left(-\frac{9}{17}\right)+\left(-\frac{25}{17}\right)\right]\)
\(=-1+\left(-2\right)=-1-2=-3\)
c) \(\frac{5}{9}\cdot\frac{7}{13}+\frac{5}{9}\cdot\frac{9}{13}-\frac{5}{9}\cdot\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\cdot1=\frac{5}{9}\)
Bài 2 :
a) \(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
=> \(\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}=-\frac{29}{70}\)
=> \(x=\left(-\frac{29}{70}\right):\frac{2}{3}=\left(-\frac{29}{70}\right)\cdot\frac{3}{2}=-\frac{87}{140}\)
b) \(x:\frac{5}{2}-\frac{1}{2}=-\frac{2}{3}\)
=> \(x:\frac{5}{2}=-\frac{2}{3}+\frac{1}{2}=-\frac{1}{6}\)
=> \(x=\left(-\frac{1}{16}\right)\cdot\frac{5}{2}=-\frac{5}{32}\)
c) Bạn chỉ cần xét hai trường hợp âm và dương thôi :>
25 3 : 5 2 = 25 3 : 25 = 25 2 = 625