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3/4 x 8/9 x 15/16 x ... x 99/100 x 120/121 = 3 x 8 x 15 x 99 x 120/ 4 x 9 x 16 x 100 x 121
= ( 1 x 3 ) x ( 2 x 4 ) x ( 3 x 5 ) x ... x ( 9 x 11 ) x ( 10 x 12 ) / ( 2 x 2 ) x ( 3 x 3 ) x ( 4 x 4 ) x ... x ( 10 x 10 ) x ( 11 x 11 )
= ( 1 x 2 x 3 x ... x 10 ) x ( 3 x 4 x 5 x ... x 12 ) / ( 2 x 3 x ... x 11 ) x ( 2 x 3 x ... x 11 ) = 12/11x2 = 6/11
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\)
\(=\frac{1.2........2016}{2.3.............2017}\)
\(=\frac{1}{2017}\)
a) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(x-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right).\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{2015}{2016}.\frac{2016}{2017}=\frac{1}{2017}\)
<=>x.1+x.3/2+x.7/2=252
x.(1+3/2+7/2)=252
x.6=252
x=252:6
=>x=42
\(x+x\times\frac{1}{3}\div\frac{2}{9}+x\div\frac{2}{7}=252\)
\(\Leftrightarrow x+x\times\frac{1}{3}\times\frac{9}{2}+x\times\frac{7}{2}=252\)
\(\Leftrightarrow x\left(1+\frac{1}{3}\times\frac{9}{2}+\frac{7}{2}\right)=252\)
\(\Leftrightarrow x\left(\frac{6}{6}+\frac{9}{6}+\frac{21}{6}\right)=252\)
\(\Leftrightarrow x\times\frac{36}{6}=252\)
\(\Leftrightarrow x\times6=252\)
\(\Leftrightarrow x=252\div6\)
\(\Leftrightarrow x=43\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)
a) Ta có :
\(\frac{7}{12}< \frac{x}{24}< \frac{2}{3}\)
\(\Rightarrow\frac{14}{24}< \frac{x}{24}< \frac{16}{24}\)
\(\Rightarrow14< x< 16\)
\(\Rightarrow x=15\)
Vậy x = 15
b) \(\left(x+1\right)+\left(x+4\right)+\left(x+7\right)+...+\left(x+28\right)=195\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+4+7+...+28\right)=195\)
\(\Rightarrow10x+145=195\)
\(\Rightarrow10x=195-145\)
\(\Rightarrow10x=50\)
\(\Rightarrow x=50:10\)
\(\Rightarrow x=5\)
Vậy x = 5
c) \(\left(x+0,5\right)+\left(x+1,5\right)+\left(x+2,5\right)=33\)
\(\Rightarrow\left(x+x+x\right)+\left(0,5+1,5+2,5\right)=33\)
\(\Rightarrow3x+4,5=33\)
\(\Rightarrow3x=33-4,5\)
\(\Rightarrow3x=28,5\)
\(\Rightarrow x=28,5:3\)
\(\Rightarrow x=9,5\)
Vậy x = 9,5
_Chúc bạn học tốt_
a, \(\frac{7}{12}\)\(< \)\(\frac{x}{24}\)\(< \)\(\frac{2}{3}\)
\(\frac{14}{24}\)\(< \)\(\frac{x}{24}\)\(< \)\(\frac{16}{24}\)
Số lớn hơn 14 và nỏ hơn 16 là : 15
\(\Rightarrow\)Vậy \(x\)= 15
\(x+x\times\frac{1}{3}:\frac{2}{9}+x:\frac{2}{7}=252\)
\(x+x\times\frac{3}{2}+x\times\frac{7}{2}=252\)
\(x\times\left(1+\frac{3}{2}+\frac{7}{2}\right)=252\)
\(x\times6=252\)
\(x=252:6\)
\(x=42\)
X + X x \(\frac{1}{3}\)x\(\frac{9}{2}\)+ X x\(\frac{7}{2}\)= 252
X + X x \(\frac{3}{2}\)+ X x \(\frac{7}{2}\)= 252
X x (\(\frac{3}{2}\)+ \(\frac{7}{2}\)+1) = 252
X x (\(\frac{3}{2}\)+ \(\frac{7}{2}\)+ \(\frac{2}{2}\)) = 252
X x 6 = 252
X = 252 : 6
X= 42
Chúc bạn học tốt! ^^
\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)
\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)
\(\Leftrightarrow x\cdot18=252\)
hay x=14