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\(a)\) Ta có :
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=1-\frac{1}{2^{100}}< 1\)
Vậy \(A< 1\)
Chúc bạn học tốt ~
a) \(\frac{6^{10}.27^5}{4^5.81^6}=\frac{\left(2.3\right)^{10}.\left(3^3\right)^5}{\left(2^2\right)^5.\left(3^4\right)^6}=\frac{2^{10}.3^{10}.3^{15}}{2^{10}.3^{24}}=\frac{2^{10}.3^{25}}{2^{10}.3^{24}}=\frac{3^{25}}{3^{24}}=3\)
b) \(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=\frac{2^{11}}{2^8}=2^3=8\)
c) \(\frac{27^4.2^3-3^{10}.4^3}{6^4.9^3.4}=\frac{\left(3^3\right)^4.2^3-3^{10}.\left(2^2\right)^3}{\left(2.3\right)^4.\left(3^2\right)^3.2^2}=\frac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6.2^2}\)
= \(\frac{3^{12}.2^3-3^{10}.2^6}{2^6.3^{10}}=\frac{3^{12}.2^3}{2^6.3^{10}}-\frac{3^{10}.2^6}{2^6.3^{10}}=\frac{3^2}{2^3}-1=\frac{9}{8}-1=\frac{1}{8}\)
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
\(A\frac{27^4.8^{17}}{9^6.32^3}=\frac{\left(3^3\right)^4.\left(2^3\right)^{17}}{\left(3^2\right)^6.\left(2^5\right)^3}=\frac{3^{12}.2^{51}}{3^{12}.2^{15}}=\frac{3^{12}.2^{15}.2^{36}}{3^{12}.2^{15}}=2^{36}\)
\(B=\frac{72^3.54^3:8^3}{108^5:4^5}=\frac{\left(72.54:8\right)^3}{\left(108:4\right)^5}=\frac{486^3}{27^5}=\frac{\left(3^5.2\right)^3}{\left(3^3\right)^5}=\frac{3^{15}.2^3}{3^{15}}=2^3=8\)
Bài 2
A = 2 +22 + 23 + 24 + ....+ 2100
A = ( 2+22 ) + (23 + 24 ) + ....+ (299 + 2100 )
A = 2(1+2 ) + 23 (1+2 ) + ...+ 299(1+2)
A = 2.3 + 23.3 + ....+ 299 .3
A = 3(2+23 + ...+ 299 )
=> A \(⋮\) 3 ( đpcm )
Bài 3
a, 2.3x = 312 .34 + 20 .274
2.3x = 312 . 34 + 20 . (33 ) 4
2.3x = 312 .34 + 20 .312
2.3x = 312(34+20 )
2.3x = 312 . 54
2.3x = 312 . 27 .2
2.3x = 312 . 33 .2
2.3x = 315 .2
=> x=15
b , (2x +1 ) 2 + 3.(22 + 1 ) = 22 .10
(2x +1 ) 2 + 3.(4+1 ) = 4.10
(2x +1 ) 2 + 3.5 = 40
(2x +1 ) 2 + 15 = 40
(2x +1 ) 2 = 40-15
(2x +1 ) 2 = 25
(2x +1 ) 2 = 52
=> 2x + 1 = 5
2x = 5-1
2x = 4
2x = 22
=> x=2