Rút gọn:
a, 4.25:(23.1/16)
b, (85.104.253):164.6253
c, C= 2200-2199+2198-2197+...+22-2
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\(21^{100}-21^{98}< 21^{99}-21^{98}< 21^{99}-21^9\)
\(\Rightarrow21^{100}-21^{98}< 21^{99}-21^9\)
\(a,112882:\left\{22.\left[743-\left(2009-1999\right)\right]\right\}\\ =112882:\left\{22.\left[743-10\right]\right\}\\ =112882:\left(22.733\right)\\ =112882:16126=7\\ b,2200:\left\{320:\left[88-\left(72-16.4\right)\right]\right\}\\ =2200:\left\{320:\left[88-\left(72-64\right)\right]\right\}\\ =2200:\left\{320:\left[88-8\right]\right\}\\ =2200:\left\{320:80\right\}\\ =2200:4=550\\ c,2^4.5-\left\{140-\left[868-12.\left(3087:7^2+4^0\right)\right]\right\}\\ =16.5-\left\{140-\left[868-12.\left(3087:49+1\right)\right]\right\}\\ =80-\left\{140-\left[868-12.\left(63+1\right)\right]\right\}\\ =80-\left\{140-\left[868-12.64\right]\right\}\\ =80-\left\{140-\left[868-768\right]\right\}\\ =80-\left\{140-100\right\}\\ =80-40=40\)
a) 112882 : {22 nhân [743 - (2009 - 1999)]}
Đầu tiên, ta tính trong ngoặc nhọn: 2009 - 1999 = 10
Tiếp theo, ta tính 743 - 10 = 733
Sau đó, ta tính 22 nhân 733 = 16126
Cuối cùng, ta tính 112882 : 16126 = 7
Vậy kết quả của phép tính a) là 7.
b) 2200 : {320 : [88 - (72 - 16 nhân 4)]}
Đầu tiên, ta tính trong ngoặc nhọn: 72 - 16 nhân 4 = 8
Tiếp theo, ta tính 88 - 8 = 80
Sau đó, ta tính 320 : 80 = 4
Cuối cùng, ta tính 2200 : 4 = 550
Vậy kết quả của phép tính b) là 550.
c) 2 mũ 4 nhân 5 - {140 - [868 - 12 nhân (3087 : 7 mũ 2 + 4 mũ 0)]}
Đầu tiên, ta tính trong ngoặc nhọn: 3087 : 7 mũ 2 = 3087 : 49 = 63
Tiếp theo, ta tính 4 mũ 0 = 1
Sau đó, ta tính 12 nhân (63 + 1) = 12 nhân 64 = 768
Tiếp theo, ta tính 868 - 768 = 100
Sau đó, ta tính 140 - 100 = 40
Cuối cùng, ta tính 2 mũ 4 nhân 5 - 40 = 16 nhân 5 - 40 = 80 - 40 = 40
Vậy kết quả của phép tính c) là 40.
\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)
a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=199+195+...+3\)
\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)
b. \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)
\(=2c^2\)
Giải:
a) \(4.2^5:\left(2^3.\dfrac{1}{16}\right)\)
\(=4.2^5:\dfrac{2^3}{16}\)
\(=2^2.2^5:\dfrac{2^3}{2^4}\)
\(=2^7:\dfrac{1}{2}\)
\(=2^6=64\)
Vậy ...
b) \(\dfrac{8^5.10^4.25^3}{16^4.625^3}\)
\(=\dfrac{2^{15}.2^4.5^4.5^6}{2^8.5^{12}}\)
\(=\dfrac{2^{19}.5^{10}}{2^8.5^{12}}\)
\(=\dfrac{2^{11}}{5^2}\)
Vậy ...
c) \(C=2^{200}-2^{199}+2^{198}-2^{197}+...+2^2-2\)
\(\Leftrightarrow C=\left(2^{200}-2^{199}\right)+\left(2^{198}-2^{197}\right)+...+\left(2^2-2\right)\)
\(\Leftrightarrow C=2^{199}\left(2-1\right)+2^{197}\left(2-1\right)+...+2\left(2-1\right)\)
\(\Leftrightarrow C=2^{199}+2^{197}+...+2\)
\(\Leftrightarrow4C=2^{201}+2^{199}+...+2^3\)
\(\Leftrightarrow3C=4C-C=2^{201}-2\)
\(\Leftrightarrow C=\dfrac{2^{201}-2}{3}\)
Vậy ...
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