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a,
\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)+\sqrt{2}\cdot\dfrac{\sqrt{2^5}}{1-\sqrt{9}}\)
\(=2^2-\left(\sqrt{3}\right)^2+\dfrac{\sqrt{2}\cdot\sqrt{2^5}}{1-3}=4-3+\dfrac{\sqrt{2^6}}{-2}=1+\dfrac{8}{-2}=1+\left(-4\right)=-3\)
b,
\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{49\cdot50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}\)
\(=\left(1-\dfrac{1}{50}\right)\cdot\dfrac{49}{50}=\dfrac{49}{50}\cdot\dfrac{49}{50}=\dfrac{49^2}{50^2}=\dfrac{2401}{2500}\)
Ta có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A>\dfrac{1}{1.2}+\dfrac{1}{3.4}=\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{14}{24}=\dfrac{7}{12}\)\(\left(1\right)\)
Lại có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-\left(\dfrac{1}{6}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)\(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
Vậy \(\dfrac{7}{12}< A< \dfrac{5}{6}\) ( Điều phải chứng minh ).
Ta có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}\right)+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\dfrac{7}{12}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}>\dfrac{7}{12}\left(1\right)\\ \Rightarrow A>\dfrac{7}{12}\left(1\right)\)
Ta lại có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ A=\dfrac{5}{6}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}< \dfrac{5}{6}\\ \Rightarrow A=< \dfrac{5}{6}\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{7}{12}< A< \dfrac{5}{6}\left(dpcm\right)\)
1.
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\)\(\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
2.
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\)\(\dfrac{1}{100!}\)
Ta có:
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+...+\)\(\dfrac{99.100}{100!}-\dfrac{1}{100}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+...+\dfrac{99.100}{100!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
Lời giải:
$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$
$=1$
`# \text {DNamNgV}`
\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}\)
\(\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)=\dfrac{1}{99}\)
\(\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}\)
\(\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}\)
\(\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}\)
\(\Rightarrow x=\dfrac{99}{99}\)
\(\Rightarrow x=1\)
Vậy, `x = 1.`
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
1, \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)...........\left(1-\dfrac{1}{n+1}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...........\left(\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}..............\dfrac{n}{n+1}\)
\(=\dfrac{1.2.3........n}{2.3.......\left(n+1\right)}\)
\(=\dfrac{1}{n+1}\)
2, \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}\)
\(=\dfrac{99}{100}\)
C=\(-66\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{11}\right)+124.\left(-37\right)+63.\left(-124\right)\)
=\(-66.\left(\dfrac{5}{66}\right)+124\left(-37-63\right)=-5+124.\left(-100\right)\)
=-12405
1) Ta có :
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
Vậy \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\rightarrowđpcm\)
2) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+............+\dfrac{1}{99.100}\)
\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow A=1-\dfrac{1}{100}\)
\(\Leftrightarrow A=\dfrac{99}{100}\)
Cảm ơn bạn nhưng cho mình hỏi đpcm là gì thế?