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\(B=1-\left(\dfrac{1}{2.6}+\dfrac{1}{4.9}+\dfrac{1}{6.12}+...+\dfrac{1}{35.67}+\dfrac{1}{38.60}\right)\left(1\right)\)
Đặt \(S=\dfrac{1}{2.6}+\dfrac{1}{4.9}+\dfrac{1}{6.12}+...+\dfrac{1}{35.67}+\dfrac{1}{38.60}\)
\(S=\dfrac{1}{2.3.\left(1.2\right)}+\dfrac{1}{2.3.\left(2.3\right)}+\dfrac{1}{2.3.\left(3.4\right)}+...+\dfrac{1}{2.3.\left(18.19\right)}+\dfrac{1}{2.3.\left(19.20\right)}\)
\(S=\dfrac{1}{6}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)
\(S=\dfrac{1}{6}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(S=\dfrac{1}{6}.\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}.\dfrac{19}{20}=\dfrac{19}{120}\)
\(\left(1\right)\Rightarrow B=1-\dfrac{19}{120}=\dfrac{101}{120}\)
Đạ biểu thức trong dấu ngoặc đơn là A
\(A=\dfrac{1}{2.1.3.2}+\dfrac{1}{2.2.3.3}+\dfrac{1}{2.3.3.4}+\dfrac{1}{2.4.3.5}+...+\dfrac{1}{2.18.3.19}+\dfrac{1}{2.19.3.20}=\)
\(=\dfrac{1}{2.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)=\)
Đặt biểu thức trong dấu ngoặc đơn là C
\(C=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{20-19}{19.20}=\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(\Rightarrow B=1-\dfrac{1}{6}.C=1-\dfrac{1}{6}.\dfrac{19}{20}=\dfrac{101}{120}\)
2a/ Ta có: \(\left|x+1\right|\ge0\Rightarrow A=\left|x+1\right|+5\ge5\)
Đẳng thức xảy ra khi: |x + 1| = 0 => x = -1
Vậy giá trị nhỏ nhất của A là 5 khi x = -1
6.30 túi là tớ đi học rùi.cứu tớ nhé.huhu\(\frac{1}{2.6}\)+ \(\frac{1}{4.9}\)+ \(\frac{1}{6.12}\)+ ... + \(\frac{1}{198.300}\)
giúp mình đi ạ!
ta có \(\frac{1+5y}{5x}\)=\(\frac{1+7y}{4x}\)
=> 4x(1+5y)=5x(1+7y)
=> 4x+20xy=5x+35xy
=> 4x-5x =35xy-20xy
=> -x =15xy
=> -1 =15y
=> y =\(\frac{-1}{15}\)
có y roi thi có thể dễ dàng tìm được x=-2
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)
\(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có :
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)
\(B=\frac{1-3-5-7-....-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)
Tổng của \(3+5+7+...+49\)là:
\(\frac{\left(3+49\right).24}{2}=624\)
\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)
\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
Đặt BT là A
\(\Rightarrow A=2016-\left(\frac{1}{1.2.6}+\frac{1}{2.3.6}+\frac{1}{3.4.6}+....+\frac{1}{19.20.6}\right)\)
\(\Rightarrow A=2016-\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{19}-\frac{1}{20}\right)\)
\(\Rightarrow A=2016-\frac{1}{6}\left(1-\frac{1}{20}\right)\)
\(A=2016-\frac{1}{6}.\frac{19}{20}=2016-\frac{19}{120}=\frac{241901}{120}\)
đúng ko bn