Ta có : 1 + 3 + 5 + … + x = ( x + 1 )[ ( x – 1 ) : 2 + 1 ] : 2 Mà : 1 + 3 + 5 + … + x = 1600 => ( x + 1 )[ ( x – 1 ) : 2 + 1 ] : 2 = 1600 <=> ( x + 1 )[ ( x – 1 ) : 2 + 1 ] = 3200 <=> ( x + 1 ) ( x2−12 + 1 ) = 3200 <=> ( x + 1 ) ( x2+12 ) = 3200 <=> ( x + 1 ) [ ( x + 1 ) : 2 ] = 3200 =>( x + 1 )2 : 2 = 3200 ( x + 1 )2 = 6400 => ( x + 1 )2 = (±80)2 => x + 1 = ±80 => x = { 79 ; – 81 } Mf theo quy luật thì x phải > 0 => x = 79 Vậy x = 79 Bình luận
a) \((4x+5):3-121:11=4\) \((4x+5):3-11=4\) \((4x+5):3=4+11\) \((4x+5):3=15\) \((4x+5)=15\times3\) \(4x+5=45\) \(4x=45-5\) \(4x=40\) \(x=40:4\) \(x=4\) a’) \((4x+5):3-\dfrac{122}{11}=4\) \((4x+5):3=4+\dfrac{122}{11}\) \((4x+5):3=\dfrac{44}{11}+\dfrac{122}{11}\) \((4x+5):3=\dfrac{44+122}{11}\) \((4x+5):3=\dfrac{166}{11}\) \((4x+5)=\dfrac{166}{11}\times3\) \(4x+5=\dfrac{498}{11}\) \(4x=\dfrac{498}{11}-5\) \(4x=\dfrac{498}{11}-\dfrac{55}{11}\) \(4x=\dfrac{498-55}{11}\) \(4x=\dfrac{443}{11}\) \(x=\dfrac{443}{11}:4\) \(x=\dfrac{443}{11}.\dfrac{1}{4}\) \(x=\dfrac{443}{44}\). b) \(1+3+5+….+x=1600\) \(\Rightarrow\left({\dfrac{x-1}{2}+1}\right)\left({\dfrac{x+1}{2}}\right)=1600\) \(\Rightarrow\left({\dfrac{x-1}{2}+\dfrac{2}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\) \(\Rightarrow\left({\dfrac{x-1+2}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\) \(\Rightarrow\left({\dfrac{x+1}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\) \(\Rightarrow\left({\dfrac{x+1}{2}}\right)=40\) \(\Rightarrow x+1=40.2\) \(\Rightarrow x+1=80\) \(\Rightarrow \) \(x=80-1\) \(\Rightarrow\) \( x=79\). Bình luận
Ta có : 1 + 3 + 5 + … + x = ( x + 1 )[ ( x – 1 ) : 2 + 1 ] : 2
Mà : 1 + 3 + 5 + … + x = 1600
=> ( x + 1 )[ ( x – 1 ) : 2 + 1 ] : 2 = 1600
<=> ( x + 1 )[ ( x – 1 ) : 2 + 1 ] = 3200
<=> ( x + 1 ) ( x2−12 + 1 ) = 3200
<=> ( x + 1 ) ( x2+12 ) = 3200
<=> ( x + 1 ) [ ( x + 1 ) : 2 ] = 3200
=>( x + 1 )2 : 2 = 3200
( x + 1 )2 = 6400
=> ( x + 1 )2 = (±80)2
=> x + 1 = ±80
=> x = { 79 ; – 81 }
Mf theo quy luật thì x phải > 0 => x = 79
Vậy x = 79
a) \((4x+5):3-121:11=4\)
\((4x+5):3-11=4\)
\((4x+5):3=4+11\)
\((4x+5):3=15\)
\((4x+5)=15\times3\)
\(4x+5=45\)
\(4x=45-5\)
\(4x=40\)
\(x=40:4\)
\(x=4\)
a’) \((4x+5):3-\dfrac{122}{11}=4\)
\((4x+5):3=4+\dfrac{122}{11}\)
\((4x+5):3=\dfrac{44}{11}+\dfrac{122}{11}\)
\((4x+5):3=\dfrac{44+122}{11}\)
\((4x+5):3=\dfrac{166}{11}\)
\((4x+5)=\dfrac{166}{11}\times3\)
\(4x+5=\dfrac{498}{11}\)
\(4x=\dfrac{498}{11}-5\)
\(4x=\dfrac{498}{11}-\dfrac{55}{11}\)
\(4x=\dfrac{498-55}{11}\)
\(4x=\dfrac{443}{11}\)
\(x=\dfrac{443}{11}:4\)
\(x=\dfrac{443}{11}.\dfrac{1}{4}\)
\(x=\dfrac{443}{44}\).
b) \(1+3+5+….+x=1600\)
\(\Rightarrow\left({\dfrac{x-1}{2}+1}\right)\left({\dfrac{x+1}{2}}\right)=1600\)
\(\Rightarrow\left({\dfrac{x-1}{2}+\dfrac{2}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\)
\(\Rightarrow\left({\dfrac{x-1+2}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\)
\(\Rightarrow\left({\dfrac{x+1}{2}}\right)\left({\dfrac{x+1}{2}}\right)=1600\)
\(\Rightarrow\left({\dfrac{x+1}{2}}\right)=40\)
\(\Rightarrow x+1=40.2\)
\(\Rightarrow x+1=80\)
\(\Rightarrow \) \(x=80-1\)
\(\Rightarrow\) \( x=79\).