Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x.3/10+x.1/4+89=100
x.(3/10+1/4)+89=100
x.11/20+89=100
x.11/20=100-89
x.11/20=11
x=11:11/20
x=20
X x 3/10 + X x 1/4 + 89 = 100
X x ( 3/10 + 1/4 ) + 89 = 100
X x 11/20 + 89 = 100
X x 11/20 = 100 - 89
X x 11/20 = 11
X = 11 : 11/20
X = 20
a = 1/2 - 1/3 + 2/5 : 1/2
= 1/6 + 4/5
= 29/30
b X x 3/10 + X x 1/4 + 89 = 100
X x ( 3/10 + 1/4 ) + 89 = 100
X x 11/20 + 89 = 100
X x 11/20 = 100 - 89
X x 11/20 = 11
X = 11 : 11/20
X = 20
Ta có: \(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\cdot100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\Leftrightarrow100\left(\dfrac{1}{1}-\dfrac{1}{10}\right)-\left[\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)\right]\cdot2=89\)
\(\Leftrightarrow\dfrac{5}{2}:\left(x+\dfrac{103}{50}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow x+\dfrac{103}{50}=5\)
hay \(x=\dfrac{147}{50}\)
Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}=\frac{9}{10}\)
\(\Rightarrow\frac{9}{10}.100-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Leftrightarrow90-\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{206}{100}\right):\frac{1}{2}=90-89=1\)
\(\Leftrightarrow x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}:1=5\)
\(\Rightarrow x=5-\frac{206}{100}=\frac{147}{50}\)
Vậy \(x=\frac{147}{50}.\)