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4 6 . 9 5 + 6 9 . 120 8 4 . 3 12 − 6 11 = 2 2 6 . 3 2 5 + 2 9 .3 9 .2 3 .3.5 2 3 4 .3 12 − 2 11 .3 11 = 2 12 .3 10 + 2 12 .3 10 .5 2 12 .3 12 − 2 11 .3 11 = 2 12 .3 10 1 + 5 2 11 .3 11 2.3 − 1 = 2.6 3.5 = 4 5
a: \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=>\(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=>\(2A+A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2+2^{100}-2^{99}+...+2^2-2\)
=>\(3A=2^{101}-2\)
=>\(A=\dfrac{2^{101}-2}{3}\)
b: Sửa đề: \(A=\dfrac{2\cdot8^4\cdot27^2+4\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(A=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot2^9\cdot3^9}{2^7\cdot2^7\cdot3^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^3+3^3\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2}{3}\cdot\dfrac{4+27}{16+15}=\dfrac{2}{3}\)
c: \(B=\dfrac{4^5\cdot9^4-2\cdot6^4}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^4\cdot3^4}{2^{10}\cdot3^8+2^8\cdot2^2\cdot5\cdot3^8}\)
\(=\dfrac{2^5\cdot3^4\left(2^5\cdot3^4-1\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1}{2^5\cdot3^4}\cdot\dfrac{32\cdot81-1}{6}\)
\(=\dfrac{2591}{2^6\cdot3^5}\)
cho tam giác ABC có góc A=120 đọ .Trên tia pg của góc A lấy điểm E sao cho AE=AB+AC.cm BCElà tam giác đều
\(311-x+82=46\left(x-21\right)\)
<=> \(311+82-46+21=x+x\)
<=> \(2x=368\)
<=> \(x=184\)
\(-x+821+534=499+x-84\)
<=> \(-x-x=499-84-821-534\)
<=> \(-2x=-940\)
<=> \(x=470\)
\(-\left(x-3+85\right)=x+70-71-5\)
<=> \(-x+3-85=x-6\)
<=> \(-x-x=-6-3+85\)
<=> \(-2x=76\)
<=> \(x=-38\)
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{15}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\cdot\left(3^5+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=\frac{2\cdot248}{3\cdot5}\)
\(=\frac{496}{15}\)