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\(A=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{10^2-1}{10^2}.\)
A là tổng của 9 số hạng; mỗi số hạng đều nhỏ hơn 1 nên A<9*1<50.
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\(a,9.5+8.10-27\)
\(=45+80-27\)
\(=125-27\)
\(=98\)
\(b,36:9+100-27.3\)
\(=4+100-81\)
\(=104-81\)
\(=23\)
\(c,29-\left[16+3.\left(51-49\right)\right]\)
\(=29-\left[16+3.2\right]\)
\(=29-\left[16+6\right]\)
\(=29-22\)
\(=7\)
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5) 24*(15+49)+12*(50+42)
=24*64+12*92
=24*64+12*2*46
=24*64+24*46
=24*(64+46)
=24*110
=2640
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a)
\({5^7}{.5^5} = 5^{7+5}={5^{12}}\)
\({9^5} :{8^0} = {9^5}:1 = {9^5}\)
\(2^{10}:64.16 = 2^{10}:2^6.2^4 = 2^{10-6+4} = 2^8\)
b)
\(\begin{array}{l}54297 = 5.10000 + 4.1000 + 2.100 + 9.10 + 7\\ = {5.10^4} + {4.10^3} + {2.10^2} + 9.10 + 7\end{array}\)
\(\begin{array}{l}2023 = 2.1000 +0.100+2.10 + 3\\ = {2.10^3}+ 2.10 +3\end{array}\)
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7: =-1+8=7
6: \(=-\left(8\cdot125\right)\cdot\left(2\cdot5\right)\cdot\left(25\cdot4\right)=-1000000\)
5: =-50+19+143+79-25-48
=-75+98+95
=23+95
=118
4: =37x9+17x9-35x9-35x11
=18+153-385
=-214
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3: =187-23-20+180
=367-43
=324
4: \(=37\cdot9+17\cdot9-35\cdot9-35\cdot11=19\cdot9-385\)
\(=171-385=-214\)
5: =-50+19+143+79-25-48
=-75+94+95
=20+94
=114
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\(A=2+\frac{3}{4}+\frac{8}{9}+......+\frac{2499}{2500}\)
\(A=2+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+.....+\left(1-\frac{1}{2500}\right)\)
\(A=2+1-\frac{1}{4}+1-\frac{1}{9}+.........+1-\frac{1}{2500}\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{4}+\frac{1}{9}+....+\frac{1}{2500}\right)\)
\(A=2+\left(1+1+....+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{50^2}\right)\)
Vì mỗi số 1 đều đi với 1 phân số nên có số số 1 là: (50-1)/1+1=50(số)
\(A=52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.......+\frac{1}{50^2}\right)\)
\(\frac{1}{2^2}<\frac{1}{1\cdot2}\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}\)
.........
\(\frac{1}{50^2}<\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+.....+\frac{1}{49\cdot50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{1}{1}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}<\frac{49}{50}\)
\(\Rightarrow52-\left(\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{50^2}\right)>52-\frac{49}{50}\)
\(\Rightarrow A>51\frac{1}{50}\)
Vì\(51\frac{1}{50}>50\Rightarrow A>50\)
\(\text{c ) }34.30+34.60+340\)
\(=34.30+34.60+34.10\)
\(=34\left(30+60+10\right)\)
\(=34.100=3400\)