1/2x1/3x1/4x...X1/10
Tính tổng
1/1x1/2x1/3x1/4x...x99/100
Tính tổng
1/1x1/2x1/3x1/4x...x99/100
a)[1-1/2]x[1-1/3]x[1-1/4]x..........[1-1/18]x[1-1/19]x[1-1/20]
b)1/1/2x1/1/3x1/1/4x1/1/5........x1/1/2005x1/1/2006x1/1/2007
a,
=1/2x2/3x3/4x...x17/18x18/19x19/20
=1/20
b,
a)2/9:2/3x1/2. b)2+1/4x4/3
c)3x1/2x1/4
a) 2/9 : 2/3 x 1/2
= 2/9 x 3/2 x 1/2
= 6/18 x 1/2
= 6/36
b) 2 + 1/4 x 4/3
= 2 + 4/7
= 2/1(7) + 4/7
= 14/7 + 4/7
= 18/7
c) 3 x 1/2 x 1/4
= 3/1 x 1/2 x 1/4
= 3/2 x 1/4
= 3/8
#Hok_tốt
Thông báo,Te2qAMq.png (1280×800)
Các bạn vô link trên sẽ xem đc độ dell có văn hóa của con aTRẦN LÊ KIM MAI,link nick https://olm.vn/thanhvien/kimmai123az
b1:tính:1/2x1/3+1/3x1/4+...+1/8x1/9
b2: tìm x
1+1/3+1/10+...+1/x(x+1):2
1/1x1/2+1/2x1/3+1/3x1/4+...+1/79x1/80
\(\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{79}\cdot\frac{1}{80}\)
\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{79\cdot80}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{79}-\frac{1}{80}\)
\(=1-\frac{1}{80}\)
\(=\frac{79}{80}\)
(1/1x1/2) + (1/2x1/3) + (1/3x1/4) + (1/4x1/5) +....+ (1/72x1/73) =
`@Fù`
`=1/(1×2)+1/(2×3)+1/(3×4)+...+1/(72×73)`
`=1/1-1/2+1/2-1/3+...+1/72-1/73`
`=1-1/73`
`=72/73`
1/1x1/2+1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6
= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5 x 6
= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6)
= 1 - 1/6
= 5/6
nha,mơn nhìu
\(\frac{1}{1}x\frac{1}{2}+\frac{1}{2}x\frac{1}{3}+\frac{1}{3}x\frac{1}{4}+\frac{1}{4}x\frac{1}{5}+\frac{1}{5}x\frac{1}{6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=1-\frac{1}{6}=\frac{5}{6}\)
c4
cho pt ẩn x: \(x^2-2x-m^2-4=0\)(1)
a/ giải pt đã cho khi m=\(\dfrac{1}{2}\)
b/ chứng minh pt luôn có 2 nghiệm phân biệt vs mọi m
c/ tính giá trị của m để pt (1) có 2 nghiệm x1,x2 sao cho 2x1,x2(2-3x1)=2
a: Khi m=1/2 thì \(x^2-2x-\dfrac{1}{4}-4=0\)
\(\Leftrightarrow x^2-2x-\dfrac{17}{4}=0\)
\(\Leftrightarrow4x^2-8x-17=0\)
\(\Leftrightarrow\left(2x-2\right)^2=21\)
hay \(x\in\left\{\dfrac{\sqrt{21}+2}{2};\dfrac{-\sqrt{21}+2}{2}\right\}\)
b: \(\text{Δ}=\left(-2\right)^2-4\left(-m^2-4\right)\)
\(=4+4m^2+16=4m^2+20>0\)
Do đó: Phương trình luôn có hai nghiệm phân biệt