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Đặt \(A=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
Ta có:
\(3A=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+\dfrac{1}{3.4.5.6}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3A=\dfrac{1}{6}-\dfrac{1}{24360}\)
\(\Rightarrow3A=\dfrac{1353}{8120}\)
\(\Rightarrow A=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(A=\dfrac{451}{8120}\)
P=1/1.2.3.4 +1/2.3.4.5 +1/3.4.5.6 +...+1/97.98.99.100
3P=3/1.2.3.4 +3/2.3.4.5 +3/3.4.5.6 +...+3/97.98.99.100
3P=1/1.2.3-1/2.3.4+1/2.3.4-1/3.4.5+................+1/97.98.99-1/98.99.100
3P = 1/1.2.3 - 1/98.99.100
3P =( 98.99.100-1.2.3)/1.2.3.98.99.100
P=( 98.99.100-1.2.3)/1.2.3.98.99.100.3
P=(98.33.50-1)/98.99.100.3
P= 161699/2910600
Nhận xét: 1/1.2.3 - 1/2.3.4 = 3/1.2.3.4, 1/2.3.4 - 1/3.4.5 =3/2.3.4.5,...,1/27.28.29 - 1/28.29.30
Gọi biểu thức phải tính bằng A,ta tính được:
3A=1/2.3 - 1/28.29.30 = 4059/28.29.30
vậy A = 1353/8120
a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6
a) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{27.28}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{28.29}\right)\)
\(=\frac{1}{2}.\frac{405}{812}=\frac{405}{1624}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{27.28.29}=\frac{405}{1624}\)
b) \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+\frac{1}{3.4.5}-\frac{1}{4.5.6}+....+\frac{1}{27.28.29}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{1.2.3}-\frac{1}{28.29.30}\right)\)
\(=\frac{1}{3}\cdot\frac{1353}{8120}=\frac{451}{8120}\)
Vậy giá trị của biểu thức \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}=\frac{451}{8120}\)
Ta có\(A=1.2.3....2019-1.2.3...2018-1.2.3...2017.2018^2\)
\(=1.2.3...2018.\left(2019-1-2018\right)\)
\(=1.2.3....2018.0\)
\(=0\)
2A=1-1/2+1/2^2-...+1/2^98-1/2^99
=>3A=1-1/2^100
=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)
\(c=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(c=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)\(=\)\(\dfrac{1}{6}-\dfrac{1}{24360}\)
\(C=\) \(\dfrac{4059}{24360}\)
Ta có:
\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1353}{8120}\)
\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(C=\dfrac{451}{8120}\)