B=5/1.2+5/2.3+..........+5/99.100. Tính B
Giải gấp giùm mik
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tính giá trị của biểu thức
C= 5/1.2 + 5/2.3 + 5/3.4 +...+ 5/99.100
giải chi tiết giùm mình
cảm ơn nhìu
C=5/1.2+5/2.3+5/3.4+...+5/99.100
C=5.(1/1.2+1/2.3+1/3.4+...+1/99.100)
C=5.(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)
C=5.(1-1/100)
C=5.99/100
C=99/20
K cho mik nha các bạn
\(C=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\right)\)
\(=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=5.\left(1-\frac{1}{100}\right)\)
\(=5.\frac{99}{100}=\frac{495}{100}\)
C=5/1.2+5/2.3+5/3.4+...+5/99.100
C=5.(1/1.2+1/2.3+1/3.4+...+1/99.100)
C=5.(1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100)
C=5.(1-1/100)
C=5.(100/100-1/100)
C=5.9/100
C=9/20
Chúc bạn học tốt nha, Lan Anh
\(C=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+....+\frac{5}{99.100}\)
\(C=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{99.100}\right)\)
\(C=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=5.\left(\frac{1}{1}-\frac{1}{100}\right)=5.\frac{99}{100}=\frac{99}{20}\)
Vậy C=99/20
A =\(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3\cdot4}+...+\frac{5}{99.100}\)
A = 5 x (\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\) )
A = 5 x \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
A = 5 x \(\left(1-\frac{1}{100}\right)\)
A = 5 x \(\frac{99}{100}\)
A = \(\frac{495}{100}\)
A= \(\frac{99}{20}\)
Ta co : A =5.(1/1.2+1/2.3+1/3.4+....+1/99.100)
A= 5.(1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100)
Rut gon tung so ta co :A=5.(1-1/100)
A=5.99/100
A=1.99/50=99/50
\(=5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=5.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=5\left(1-\dfrac{1}{100}\right)\)
\(=5.\dfrac{99}{100}=\dfrac{99}{20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{100}{100}-\frac{1}{100}\)
\(=\frac{99}{100}\)
A = 5/1.2 + 5/2.3 +...+ 5/99.100
2A = 10/1.2 + 10/2.3 +...+ 10/99.100
2 A = 5/1-5/2+5/2-5/3+5/3-5/4+...+5/99-5/100
2A=5/1-5/100
2A=9/2 => A=9/2:2=9/4
cho 1 đ-ú-n-g nha bạn!!
99/20 đảm bảo
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c) Đặt \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)
\(\Leftrightarrow3A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(\Leftrightarrow3\cdot A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+98\cdot99\cdot100-98\cdot99\cdot100+99\cdot100\cdot101\)
\(\Leftrightarrow3\cdot A=99\cdot100\cdot101\)
\(\Leftrightarrow A=33\cdot100\cdot101=333300\)
b) Ta có: \(1+2-3-4+...+97+98-99-100\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(97+98-99-100\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=-4\cdot25=-100\)
A = 5 .(1/1.2 +1/2.3 + 1/3.4 +.....1/99.100)
A= 5 . (1/1 - 1/2 +1/2 -1/3 +1/3 -1/4 ........ +1/99 - 1/100)
A= 5 . (1/1 - 1/100)
A= 5.99/100
A= 99/20
A=5.(1-1/2+1/2-1/3 + .... + 1/99 - 1/100)
= 5.(1-1/100)
=5 . 99/100
= 99/20
1/5 .B=1/1×2+1/2×3+.....+1/99×100
1/5 .B=2-1/1×2+3-2/2×3+......+100-99/99×100
1/5 .B=1/1-1/2+1/2-1/3+......+1/99-1/100
1/5.B=1-1/100
1/5. B=99/100
B=99/100÷1/5
B=99/20
\(B=\frac{5}{1.2}+\frac{5}{2.3}+.....+\frac{5}{99.100}\)
\(B=5-\frac{5}{2}+\frac{5}{2}-\frac{5}{3}+.....+\frac{5}{99}-\frac{5}{100}\)
\(B=5-\frac{5}{100}\)
\(B=\frac{500}{100}-\frac{5}{100}\)
\(B=\frac{495}{100}=\frac{99}{20}\)