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a)đặt B=1/2.3+1/3.4+...+1/99.100
=1/1.2+1/2.3+1/3.4+...+1/99.100
=1-1/2+1/2-1/3+...+1/99-1/100
=1-1/100<1 (1)
Mà 1<2(2)
A =1/1+1/2.2+1/3.3+...+1/100.100<1-1/2+1/2-1/3+...+1/99-1/100 (3)
từ (1),(2),(3) =>A<2
b,c tự làm
Câu 8( Mình không viết đè nữa nha)
a) 2-1/1.2 + 3-2/2.3 + 4-3/3.4 +…..+ 100-99/99.100
= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…..+ 1/99 – 1/100
= 1 – 1/100 < 1
= 99/100 < 1
Vậy A< 1
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
...
\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)
Cộng vế theo vế
\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)
\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)
Lại có \(\frac{7}{8}< 1\)
Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)
\(\Rightarrow B< 1\left(đpcm\right)\)
Làm bài hình thôi nhé.
Hình b tự vẽ.
a/ Ta có: góc xOy + góc yOz = 180 độ (kề bù)
=> 120 + góc yOz = 180
=> góc yOz = 180 - 120 = 60 độ
b/ Vì Om là pgiác góc yOz => góc yOm = góc zOm = góc yOz : 2 = 60 : 2 = 30 độ
Ta có: góc xOm = góc xOy + góc yOm = 120 + 30 = 150 độ
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
Bài 1 :
\(x\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)
\(\Rightarrow x\cdot\frac{24}{50}=1\)
\(\Rightarrow x=1\div\frac{24}{50}=\frac{25}{12}\)
#Louis
\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)
\(\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)x=1\)
\(\left(\frac{1}{2}-\frac{1}{50}\right)x=1\)
\(\frac{12}{25}x=1\)
Đến đây dễ rồi :)))
Bn tự tính típ nha
Đặt A = \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}\)
Ta có \(A=\frac{1}{3\cdot3}+\frac{1}{4\cdot4}+\frac{1}{5\cdot5}+...+\frac{1}{100\cdot100}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\left(đpcm\right)\)
\(\frac{1}{3^2}\)\(< \frac{1}{2.3}\)
\(\frac{1}{4^2}\)\(< \frac{1}{3.4}\)
\(...................\)
\(\frac{1}{100^2}\)\(< \frac{1}{99.100}\)
\(=>\)\(\frac{1}{3^2}\)\(+\frac{1}{4^2}\)\(+...+\frac{1}{100^2}\)\(< \frac{1}{2.3}\)\(+\frac{1}{3.4}\)\(+...+\frac{1}{99.100}\)
\(=>\)\(\frac{1}{3^2}\)\(+\frac{1}{4^2}\)\(+...+\frac{1}{100^2}\)\(< \frac{1}{2}\)\(-\frac{1}{3}\)\(+\frac{1}{3}\)\(-\frac{1}{4}\)\(+...+\frac{1}{99}\)\(-\frac{1}{100}\)
\(=>\)\(\frac{1}{3^2}\)\(+\frac{1}{4^2}\)\(+...+\frac{1}{100^2}\)\(< \frac{1}{2}\)\(-\frac{1}{100}\)\(< \frac{1}{2}\)
Vậy: \(\frac{1}{3^2}\)\(+\frac{1}{4^2}\)\(+...+\frac{1}{100^2}\)\(< \frac{1}{2}\)