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\(=\left(\frac{5}{7}\cdot0,6-5:3\frac{1}{2}\right)\cdot\left(40\%-1,4\right)\left(-2\right)^3\)
\(=\left(\frac{5}{7}\cdot\frac{3}{5}-5:\frac{7}{2}\right)\left(\frac{2}{5}-\frac{1}{4}\right)\cdot\left(-8\right)\)
\(=\left(\frac{25}{21}-\frac{10}{7}\right)\cdot\frac{3}{20}\cdot\left(-8\right)\)
\(=-\frac{5}{21}\cdot\frac{3}{20}\cdot\left(-8\right)\)
\(=-\frac{2}{7}\)
Nhớ k nha các bạn gái làm vợ đi
\(B=\left(\frac{3}{7}-5:\frac{7}{2}\right).\left(\frac{2}{5}-1,4\right).\left(-8\right)\)
\(B=\left(\frac{3}{7}-\frac{35}{2}\right).\left(-1\right).\left(-8\right)\)
\(B=-\frac{239}{14}.\left(-1\right).\left(-8\right)\)
\(B=-\frac{956}{7}\)
Ps: Không làm quá chi tiết vì mất thời gian. Mà làm vậy là quá chi tiết so với bình thường tui làm rồi!
\(\frac{2^8\times6}{3^3\times5^4}\div\frac{8^3\times9}{5^3\times3^3}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-4^2\right)\)
\(=\frac{2^9\times3}{3^3\times5^4}\times\frac{5^3\times3^3}{2^9\times3^2}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\left(2^4-2^4\right)\)
\(=\frac{2^9\times3^4\times5^3}{3^5\times5^4\times2^9}-\left(2^{14}+3^{19}\right)\left(3^{81}+5^{64}\right)\times0\)
\(=\frac{1}{3\times5}-0\)
\(=\frac{1}{15}\)
\(\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^{2013}=\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)=-\dfrac{1}{6}\)
\(\left(\dfrac{1}{5}\right)^{15}.\left(\dfrac{1}{4}\right)^{20}=\dfrac{1}{5^{12}}.\dfrac{1}{4^{20}}=5^{-12}.4^{-20}=125^{-4}.1024^{-4}=\left(125.1024\right)^{-4}=128000^{-4}\)
\(\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.2^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3+1\right)}=\dfrac{2.6}{3.7}=\dfrac{4}{7}\)
Bài làm:
Ta có: \(\frac{9^2.\left(-4\right)^5.72}{5.\left(-6\right)^6}\cdot\frac{6^3.81.\left(-4\right)}{\left(-8\right)^6.3^9}\)
\(=\frac{3^2.-2^{10}.2^3.3^2.2^3.3^3.3^4.-2^2}{5.2^3.3^3.2^{18}.3^9}\)
\(=\frac{2^{18}.3^{11}}{2^{21}.3^{12}.5}\)
\(=\frac{1}{2^3.2.5}=\frac{1}{120}\)
\(\frac{25^5.2^{10}}{20^4.5^4}\)
=\(\frac{25^5.2^{10}}{20^4.25^2}\)
=\(\frac{25^3.2^{10}}{20^4}\)
=\(\frac{25^3.1024}{160000}\)
=\(\frac{25^3.4}{25^2}\)
=\(25.4\)
=100
# Mik làm ý A trước nhé, mik sợ dài :
- Với n = 1 \(\Rightarrow1=\frac{1.2.3}{6}\)( đúng )
- Giả sử đẳng thức cũng đúng với\(n=k\)hay :
\(1^2+2^2+3^2+...+k^2=\)\(\frac{k\left(k+1\right)\left(2k+1\right)}{6}\)
Ta cần chứng minh nó cũng đúng với\(n=k+1\)hay :
\(1^2+2^2+3^2+...+k^2+\left(k+1\right)^2=\)\(\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)}{6}\)
Thật vậy, ta có:
\(1^2+2^2+3^2+...+k^2+\left(k+1\right)^2=\)\(\frac{k\left(k+1\right)\left(2k+1\right)}{6}+\left(k+1\right)^2\)
\(\Rightarrow\left(k+1\right)\left(\frac{k\left(2k+1\right)}{6}+k+1\right)=\)\(\left(k+1\right)\left(\frac{2k^2+k+6k+6}{6}\right)\)
\(\Rightarrow\)\(\left(k+1\right)\left(\frac{2k^2+7k+6}{6}\right)=\)\(\frac{\left(k+1\right)\left(k+2\right)\left(2k+3\right)}{6}\)( đpcm )
# giờ mik làm ý B nha !
- Với n = 1 \(\Rightarrow\)1 = 1 ( đúng )
Giả sử bài toán đúng với\(n=k\left(n\inℕ^∗\right)\)thì ta có :
1 + 23 + 33 + .... + k3 = \(\left[\frac{n\left(n+1\right)}{2}\right]^2\left(1\right)\)
Ta cần chứng minh đề bài đúng với\(n=k+1\)tức là :
13 + 23 + 33 + ...... + n3 = \(\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\left(2\right)\)
Đặt \(B=1^3+2^3+...+\left(k+1\right)^3\)
\(=\left(\frac{k\left(k+1\right)}{2}\right)^2+\left(k+1\right)^3\)theo ( 1 )
\(=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)theo ( 2 )
\(\Rightarrow\left(1\right),\left(2\right)\)đều đúng
Mà \(\left[\frac{n\left(n+1\right)}{2}\right]^2=\)\(\frac{n^2\left(n+1\right)^2}{4}\)
\(\Rightarrow\)\(1^3+2^3+...+n^3=\)\(\frac{n^2\left(n+1\right)^2}{4}\)( đpcm )
