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\(\frac{1}{2^2}< \frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2}-\frac{1}{3}\)
\(.......\)
\(\frac{1}{50^2}< \frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< \frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Mà \(\frac{49}{50}< 2\)
\(\Rightarrow A< 2\)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(\frac{1}{1^2}=1\)
Ta có :
\(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4}\)
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{49\cdot50}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1+1-\frac{1}{50}\)
\(=2-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\left(dpcm\right)\)
CHO \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}.\)CHỨNG MINH A<2
\(\frac{1}{2^2}< \frac{1}{1.2}\)
...................\(\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(\Rightarrow A< 1-\frac{1}{50}< \frac{49}{50}< 1< 2\)
1/2^2<1/1*2;1/3^2<1/2*3;1/4^2<1/3*4;1/50^2<1/49*50
ta có:
=> 1/1^2+1/2*3+1/3*4+...+1/49*50
<=> 1/1-1/2+1/2-1/3+1/3-1/4+...+1/49-1/50
<=> 1-1/50 < 2
=> A < 2
\(A=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
Khi đó \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{49.50}=B\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{49}-\frac{1}{50}< 1\)
Vì \(A< 1+B\)mà \(B< 1\)nên \(B+1< 2\)do đó \(A< 2\)
Vậy \(A< 2\)
1/12+1/22+....+1/502<1/1+1/1x2+1/2x3+....+1/49x50=1-1/50=49/50<2
=>A<2(đpcm)
Ta có
\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3}......\frac{1}{50^2}<\frac{1}{49.50}\)
\(=>A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=> A<2-1/50
=> A < 2
=> đpcm
Ta có: A = \(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
A < \(\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=> A < 1 +( \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\))
A< 1 +1 -\(\frac{1}{50}\)
A< 2 - \(\frac{1}{50}\)
Vậy A< 2
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
=>A< \(\frac{1}{1^2}+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
=\(1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
=\(1+\left(1-\frac{1}{50}\right)\)
=\(1+\frac{49}{50}\) =\(1\frac{49}{50}<2\)
Vậy A<2
A=1/1^2+1/2^2+1/3^2+.....+1/50^2
A<B=1+1/1.2+1/2.3+1/3.4+........+1/49.50
=1+(1-1/2+1/2-1/3+1/3-1/4+......+1/49-1/50)
=1+(1-1/50)
=1+ 49/50
=99/100<50/100 SUY RA 99/100<50/100 DO A<B<2
SUY RA A<2
ỦNG HỘ CHO MÌNH NHÉ
Bạn xem lời giải ở đường link sau nhé:
Câu hỏi của nguyenducminh - Toán lớp 6 - Học toán với OnlineMath
A=\(\frac{1}{1^2}\)\(+\frac{1}{2^2}\)\(+\frac{1}{3^2}\)\(+...+\frac{1}{50^2}\)
A<1\(+\frac{1}{1.2}\)\(+\frac{1}{2.3}\)\(+...\frac{1}{49.50}\)
=1+1-\(-\frac{1}{2}\)\(+\frac{1}{2}\)\(-\frac{1}{3}\)\(+...+\frac{1}{49}\)\(-\frac{1}{50}\)
=\(1+1-\frac{1}{50}\)
=\(2-\frac{1}{50}\)\(< 2\)
\(\Rightarrow A< 2\)
\(A<1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1+\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A<1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}<2\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};....;\frac{1}{50^2}<\frac{1}{49.50}\)
=>\(A<1+\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{49.50}\)
=>\(A<1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}\)
=>\(A<2-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}<2\)
=>A<2 (đpcm)
\(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(A=1+\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)=1+B\)( B là biểu thức trong ngoặc )
Xét B
\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\)
\(B<\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{49.50}\)
\(B<\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(B<\frac{1}{1}-\frac{1}{50}\)
\(B<\frac{49}{50}<1\)
Vậy B < 1
\(\Rightarrow A=1+B<1+1=2\)
\(\Rightarrow A<2\)
A\(<\frac{1}{1}+\frac{1}{1.2}+\frac{1}{2.3}+...........+\frac{1}{49.50}\)
=\(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..............+\frac{1}{49}-\frac{1}{50}\)
=\(2-\frac{1}{50}<2\)
\(\Rightarrow A<2\left(đpcm\right)\)
Cho góc bẹt xOy. Vẽ tia Oz sao cho góc yOz = 800.
a) Tính góc xOz ?
b) Vẽ Om, On lần lượt là tia phân giác của góc xOz và góc yOz. Hỏi hai góc và có phụ nhau không ? Tại sao?