Khi bớt đi mỗi kho 168 bao thì kho thứ nhất vẫn nhiều hơn kho thứ hai 12576 bao 

Hiệu số phần bằng nhau:

\(9-5=4\) (phần)

Số bao có trong kho thứ nhất khi đã bớt 168 bao:

\(12576:4\times9=28296\) (bao)

Số bao ban đầu có trong khi thứ nhất:

\(28296+168=28464\left(bao\right)\)

Số bao ban đầu có trong kho thứ hai:

\(28464-12576=15888\left(bao\right)\)

\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)

\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)

\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)

\(\Rightarrow4\times y=\dfrac{16}{81}\)

\(\Rightarrow y=\dfrac{16}{81}:4\)

\(\Rightarrow y=\dfrac{4}{81}\)

Ta có:

\(5^{75}=\left(5^5\right)^{15}=3125^{15}\)

\(7^{60}=\left(7^4\right)^{15}=2401^{15}\)

Mà: \(3125^{15}>2401^{15}\)

\(\Rightarrow5^{75}>7^{60}\)

_______________

Ta có:

\(3^{39}< 3^{42}\); \(3^{42}=\left(3^6\right)^7=729^7\)

\(11^{21}=\left(11^3\right)^7=1331^7\)

Mà: \(729^7< 1331^7\)

\(\Rightarrow3^{42}< 11^{21}\)

\(\Rightarrow3^{39}< 11^{21}\)

\(x^3:\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow x^3:\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)

\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)

\(\Rightarrow x^3=\left(\dfrac{1}{2}\right)^3\)

\(\Rightarrow x=\dfrac{1}{2}\)

1m2dm=.1,2.m

1m2dm=..1,2.m

\(36\cdot33-105\cdot11+22\cdot15\)

\(=11\cdot\left(36\cdot3-105+2\cdot15\right)\)

\(=11\cdot\left(108-105+30\right)\)

\(=11\cdot33\)

\(=33\cdot\left(10+1\right)\)

\(=330+33\)

\(=363\)

a) \(\left(-17\right)\cdot32+17\cdot\left(-68\right)-17\)

\(=17\cdot\left(-32-68-1\right)\)

\(=17\cdot\left(-101\right)\)

\(=17\cdot\left(-100-1\right)\)

\(=-1700-17\)

\(=-1717\)

b) \(25\cdot\left(34-89\right)+25\cdot89\)

\(=25\cdot\left(34-89+89\right)\)

\(=25\cdot34\)

\(=850\)

c) \(-145\cdot\left(13-57\right)+57\cdot\left(10-145\right)\)

\(=-145\cdot13+145\cdot57+57\cdot10-145\cdot57\)

\(=-145\cdot\left(13+57\right)+57\cdot\left(145+10\right)\)

\(=-145\cdot70+57\cdot155\)

\(=-10150+8835\)

\(=-1315\)

Cho: \(A=\dfrac{2}{2^2}+\dfrac{2}{3^2}+\dfrac{2}{4^2}+....+\dfrac{2}{100^2}\)

\(A=2\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)\)

Và cho \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

Mà: 

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)

....

\(\dfrac{1}{100^2}< \dfrac{1}{99\cdot100}\)

Nên: \(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Rightarrow B< 1-\dfrac{1}{100}\)

\(\Rightarrow B< \dfrac{99}{100}\)

Mà: \(\dfrac{99}{100}< 1\) (tử nhỏ hơn mẫu)

\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< 1\)

\(\Rightarrow A=2\cdot\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+..+\dfrac{1}{100^2}\right)< 2\) (đpcm)

Ta có:

\(n^2+n=n\left(n+1\right)\)

Mà: \(n\left(n+1\right)\) ⋮ \(n+1\forall n\)

\(\Rightarrow n^2+n\) ⋮ \(n+1\forall n\) 

\(347+\left(-40\right)+3150+\left(-307\right)\)

\(=\left(347-40-3-7\right)+3150\)

\(=0+3150\)

\(=3150\)