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\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{100}{100}-\frac{1}{x+1}=\frac{99}{100}\)
\(\frac{1}{x+1}=\frac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=99\)
ta có
x+y+y+z+z+x=\(\frac{13}{12}\)
2(x+y+z)=\(\frac{13}{12}\)
=>x+y+z=\(\frac{13}{24}\)
z=(x+y+z)-(x+y)
y=y+z-z
x=x+Y-y
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
a) \(x-\frac{1}{12}+x-\frac{1}{20}+x-\frac{1}{30}+x-\frac{1}{42}+x-\frac{1}{56}+x-\frac{1}{72}=224\)
\(\left(x+x+x+x+x+x\right)-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(6x-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=224\)
\(6x-\left(\frac{1}{3}-\frac{1}{9}\right)=224\)
\(6x-\frac{2}{9}=224\)
\(6x=224+\frac{2}{9}\)
\(6x=\frac{2018}{9}\)
\(\Rightarrow x=\frac{2018}{9}:6=\frac{1009}{27}\)
b) ( 2x - 1 ) 2 = \(\frac{1}{4}\)
( 2x - 1 ) 2 = \(\left(\frac{1}{2}\right)^2\)
\(\Rightarrow\)2x - 1 = \(\frac{1}{2}\)
\(\Rightarrow\)2x = \(\frac{1}{2}+1\)
\(\Rightarrow\)2x = \(\frac{3}{2}\)
\(\Rightarrow\)x = \(\frac{3}{2}:2\)
\(\Rightarrow\)x = \(\frac{3}{4}\)
2.
Ta có : 3300 = ( 33 ) 100 = 27100
5200 = ( 52 ) 100 = 25100
Vì 27100 > 25100 nên 3300 > 5200
3.
150 - ( 100 - 99 + 98 - 97 + 96 - 95 + ... + 4 - 3 + 2 - 1 )
= 150 - [ (100 - 99 ) + ( 98 - 97 ) + ( 96 - 95 ) + ... + ( 4 - 3 ) + ( 2 - 1 ) ]
= 150 - ( 1 + 1 + 1 + ... + 1 + 1 )
= 150 - 50
= 100
4.
ta có :
9x + 5y + 4 .( 2x + 3y )
= 9x + 5y + 8x + 12y
= ( 9x + 8x ) + ( 5y + 12y )
= 17x + 17y
= 17 ( x + y ) \(⋮\)17
Vì 9x + 5y \(⋮\)17 \(\Rightarrow\)4 . ( 2x + 3y ) \(⋮\)17
Mà ( 4 ; 17 ) = 1
\(\Rightarrow\)2x + 3y \(⋮\)17
bài 1
a) \(\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=224\)
\(\left(x-1\right).\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=224\)
\(\left(x-1\right).\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=224\)
\(\left(x-1\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=24\)
\(\left(x-1\right).\left(\frac{1}{3}-\frac{1}{9}\right)=\left(x-1\right)\cdot\frac{2}{9}=224\)
\(\Rightarrow\left(x-1\right)=224:\frac{2}{9}=1008\Rightarrow x=1008+1=1009\)
\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)......\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)
\(=\left(-\dfrac{1}{2}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{3}{4}\right)......\left(-\dfrac{98}{99}\right)\left(-\dfrac{99}{100}\right)\)
\(=\dfrac{1}{100}\) ( do có 100 số hạng nên tích ra 1 số dương nhé ! ( chẵn mà )
Nhi có 10 viên bi,Nhi cho Oanh 5 viên bi.Hỏi Nhi còn lại bao nhiêu viên bi?