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Bài 1 :
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\left(\frac{2017}{1}+1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{\frac{2018}{1}+\frac{2018}{2}+\frac{2018}{3}+....+\frac{2018}{2017}+\frac{2018}{2018}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2017}+\frac{1}{2018}\right)}\)
\(=\frac{1}{2018}\)
B=\(\frac{\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}}{\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}}\)
\(\)TA CÓ E=\(\frac{1}{101.99}+\frac{1}{103.97}+...+\frac{1}{149.51}\)
\(200E=\frac{200}{101.99}+\frac{200}{103.97}+..+\frac{200}{149.51}\)
\(200E=\frac{101+99}{101.99}+\frac{103+97}{103.97}+...+\frac{149+51}{149.51}\)
\(200E=\frac{1}{99}+\frac{1}{101}+\frac{1}{97}+\frac{1}{103}+...+\frac{1}{51}+\frac{1}{149}\)
\(200E=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\)
\(E=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right):200\)\(=\left(\frac{1}{51}+\frac{1}{53}+...+\frac{1}{147}+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{51}+\frac{1}{53}+...+\frac{1}{149}\)/\(\left(\frac{1}{51}+\frac{1}{53}+..+\frac{1}{149}\right).\frac{1}{200}\)
\(\Rightarrow B=\frac{1}{\frac{1}{200}}=200\)
VẬY B=200
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
đặt A=1/2+(1/2)^2+(1/2)^3+...+(1/2)^98+(1/2)^99+(1/2)^99
=>A=1/2+12/22+13/23+...+198/298+199/299+199/299
=>A=1/2+1/22+1/23+...+1/298+1/299+1/299
=>2A-1/299=1+1/2+1/22+...+1/298
=>(2A-1/299)-(A-1/299)=(1+1/2+1/22+...+1/298)-(1/2+1/22+1/23+...+1/298+1/299)
=>(2A-1/299)-(A-1/299)=1-1/299
=>A=1-1/299 +1/299=1
vậy A=1
chắc thế
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
\(B=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(B=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(B=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(B=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(B=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
Ta có: \(63.1,2-21.3,6=0,9.7.10.1,2-21.3,6\)
\(=6,3.12-21.3,6\)
\(=0,9.7.4.3-7.3.0,9.4\)
\(=6,3.12-6,3.12\)
\(=0\)
\(\Rightarrow\frac{\left(1+2+...+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)0}{1-2+3-4+...+99-100}=0\)
Vậy \(\frac{\left(1+2+...+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}=0\)
Ta có:Xét tử số
TS có 99 tổng,1 có mặt trong 99 tổng,2 có mặt trong 98 tổng,3 có mặt trong 97 tổng,...,99 có mặt trong 1 tổng
Vì thế ta được:\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1.99+2.98+3.97+...+1.99}\)
\(=\frac{1.99+2.98+3.97+...+99.1}{1.99+2.98+3.97+...+99.1}=1\)