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\(A=1+2+2^2+...+2^{51}\)
\(2A=2+2^2+2^3+...+2^{52}\)
\(2A-A=\left(2+2^2+2^3+...+2^{52}\right)-\left(1+2+2^2+...+2^{51}\right)\)
\(A=2^{52}-1\)
\(B=5+5^2+5^3+...+5^{100}\)
\(5B=5^2+5^3+5^4+...+5^{101}\)
\(5B-B=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(4B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
A= 1+5+5^2+5^3+...+5^51
=> 5A= 5+5^2+5^3+5^4+...+5^52
=> 5A - A= ( 5+5^2+5^3+5^4+...+5^52) -(1+5+5^2+5^3+...+5^51)
=> 4A = 5^52-1
=>A=(5^52-1)/4
a: \(A=\dfrac{2\cdot8^4\cdot27^2+44\cdot6^9}{2^7\cdot6^7+2^7\cdot40\cdot9^4}\)
\(=\dfrac{2\cdot2^{12}\cdot3^6+2^2\cdot11\cdot2^9\cdot3^9}{2^7\cdot3^7\cdot2^7+2^7\cdot2^3\cdot5\cdot3^8}\)
\(=\dfrac{2^{13}\cdot3^6+2^{11}\cdot3^9\cdot11}{2^{14}\cdot3^7+2^{10}\cdot5\cdot3^8}\)
\(=\dfrac{2^{11}\cdot3^6\left(2^2+3^3\cdot11\right)}{2^{10}\cdot3^7\left(2^4+5\cdot3\right)}\)
\(=\dfrac{2\cdot301}{3\cdot31}=\dfrac{602}{93}\)
\(a,-3x^2+7x-9+\left(x-1\right)\left(x+2\right)\\ =-3x^2+7x-9+x^2-x+2x-2\\ =\left(-3x^2+x^2\right)+\left(7x-x+2x\right)-\left(9+2\right)\\ =-2x^2+8x-11\\ b,x\left(x-5\right)-2x\left(x+1\right)\\ =x^2-5x-2x^2-2x\\ =\left(x^2-2x^2\right)-\left(5x+2x\right)\\ =-3x^2-7x\\ c,4x\left(x^2-x+1\right)-\left(x-1\right)\left(x^2-x\right)\\ =4x^3-4x^2+4x-x\left(x^2-x\right)+x^2-x\\ =4x^3-4x^2+4x-x^3+x^2+x^2-x\\ =\left(4x^3-x^3\right)+\left(-4x^2+x^2+x^2\right)+\left(4x-x\right)\\ =3x^3-2x^2+3x\\ =x\left(3x^2-2x+3\right)\)
\(d,-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\\ =-5x^2+25x+x\left(x^2-7\right)-3\left(x^2-7\right)\\ =-5x^2+25x+x^3-7x-3x^2+21\\ =\left(-5x^2-3x^2\right)+\left(25x-7x\right)+x^3+21\\ =-8x^2+x^3+18x+21\)
Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{50}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{50}}\)
\(\Rightarrow A=1-\frac{1}{2^{50}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}\)
\(A=\frac{2^{50}-1}{2^{50}}.\)
2S=2(1+2+22+...+250)
2S=2+22+...+251
2S-S=(2+22+...+251)-(1+2+22+...+250)
S=251-1<251
=>S<251
đang cần gấp xin mn giúp mik
#) Sửa đề bỏ cái 2^51 nhé.
\(A=1+2^2+2^4+.....+2^{50}\)
\(2^2A=2^2\left(1+2^2+2^4+...+2^{50}\right)\)
\(4A=2^2+2^4+2^6+...+2^{52}\)
\(4A-A=2^{52}-1\)
\(3A=2^{52}-1\)
\(A=\dfrac{2^{52}-1}{3}\)
~ Hok tốt ~