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Trả lời
568-{5.[143-3.(14-1)]+10}:10
=568-{5.[143-3.13]+10}:10
=568-{5.[143-39]+10}:10
=568-{5.104+10}:10
=568-530:10
=568-53
=515
568-{5x[143-3x13+10}:10
=568-{5x[143-39+10}:10
=568-(5x114):10
=568-570:10
=568-57
=511
\(3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\)
\(\Rightarrow\) Hệ số của \(x^5\) là 6
`1,14 . 6,4 + 1,14 . 3,6 + 0,2 . 1,14 . 5`
`= 1,14 . (3,6 + 6,4 + 1)`
`= 1,14 . 11`
`= 12,54`
\(\dfrac{20^3.\left(-49\right)^2}{14^3.5^4}=\dfrac{2^3.2^3.5^3.7^4}{2^3.7^3.5^4}=\dfrac{2^3.7}{5}=\dfrac{8.7}{5}=\dfrac{56}{5}\)
b)1024(5)-132(5)=892(5)
c)214(5).32(5)=6848(5)
d)211(5):13(5)=17(5)
2,1001(2):11(2)=91(2)
hehe làm bừa 
sai xin lỗi
vì tui ko hiểu
Ta có: \(A=1\cdot99+2\cdot98+3\cdot97+\cdots+98\cdot2+99\cdot1\)
\(=2\left(1\cdot99+2\cdot98+\cdots+49\cdot51\right)+50\cdot50\)
\(=2\left\lbrack1\left(100-1\right)+2\left(100-2\right)+\cdots+49\left(100-49\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\left(1+2+\cdots+49\right)-\left(1^2+2^2+\cdots+49^2\right)\right\rbrack+2500\)
\(=2\cdot\left\lbrack100\cdot\frac{49\cdot50}{2}-\frac{49\cdot\left(49+1\right)\left(2\cdot49+1\right)}{6}\right\rbrack+2500\)
\(=2\left\lbrack50\cdot49\cdot50-\frac{49\cdot50\cdot99}{6}\right\rbrack+2500\)
\(=2\cdot\left\lbrack49\cdot50\cdot50-49\cdot25\cdot33\right\rbrack+2500\)
\(=2\cdot49\cdot25\cdot\left(2\cdot50-33\right)+2500\)
\(=49\cdot50\cdot67+2500=166650\)
Ta có: \(B=1\cdot2\cdot3+2\cdot3\cdot4+\ldots+17\cdot18\cdot19\)
\(=2\left(2-1\right)\left(2+1\right)+3\left(3-1\right)\left(3+1\right)+\cdots+18\left(18-1\right)\left(18+1\right)\)
\(=2\cdot\left(2^2-1\right)+3\left(3^2-1\right)+\cdots+18\left(18^2-1\right)\)
\(=\left(2^3+3^3+\cdots+18^3\right)-\left(2+3+\cdots+18\right)\)
\(=\left(1^3+2^3+\cdots+18^3\right)-\left(1+2+3+\cdots+18\right)\)
\(=\left(1+2+\cdots+18\right)^2-\left(1+2+\cdots+18\right)\)
\(=\left(18\cdot\frac{19}{2}\right)^2-18\cdot\frac{19}{2}=\left(9\cdot19\right)^2-9\cdot19=29070\)
Ta có: \(C=1\cdot4+2\cdot5+\cdots+100\cdot103\)
\(=1\left(1+3\right)+2\left(2+3\right)+\cdots+100\cdot\left(100+3\right)\)
\(=\left(1^2+2^2+\cdots+100^2\right)+3\left(1+2+\cdots+100\right)\)
\(=\frac{100\left(100+1\right)\left(2\cdot100+1\right)}{6}+\frac{3\cdot100\cdot101}{2}\)
\(=\frac{100\cdot101\cdot201}{6}+\frac{3\cdot100\cdot101}{2}=50\cdot101\cdot67+3\cdot50\cdot101\)
\(=50\cdot101\cdot70=3500\cdot101=353500\)
Ta có: \(D=1\cdot3+2\cdot4+3\cdot5+\cdots+97\cdot99+98\cdot100\)
\(=1\left(1+2\right)+2\left(2+2\right)+3\left(3+2\right)+\cdots+97\cdot\left(97+2\right)+98\cdot\left(98+2\right)\)
\(=\left(1^2+2^2+\cdots+98^2\right)+2\cdot\left(1+2+3+\cdots+98\right)\)
\(=\frac{98\cdot\left(98+1\right)\left(2\cdot98+1\right)}{6}+2\cdot\frac{98\cdot99}{2}\)
\(=\frac{98\cdot99\cdot197}{6}+98\cdot99=49\cdot33\cdot197+98\cdot99=49\cdot33\left(197+2\cdot3\right)\)
\(=49\cdot33\cdot203=328251\)
\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)
x(1 - 3x)(4 - 3x) - (x - 4)(3x + 5)
= (x - 3x2)(4 - 3x) - 3x2 - 5x + 12x + 20
= 4x - 3x2 - 12x2 + 9x3 - 3x2 - 5x + 12x + 20
= 9x3 - 18x2 + 11x + 20