Ta có : (2x + 3)2 - (2x + 1)(2x - 1) = 22
=> 4x2 + 12x + 9 - 4x2 + 1 = 22
=> 12x + 10 = 22
=> 12x = 12
=> x = 1
Vậy x = 1
\(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\)
\(\Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
\(\Leftrightarrow3.\left(4x+3\right)=21\)
\(\Leftrightarrow4x+3=7\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - [ ( 2x )2 - 1 ] = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
\(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(\left(4x^2+12x+9\right)-\left[\left(2x\right)^2-1^2\right]=22\)
\(\left(4x^2+12x+9\right)-\left(4x-1\right)=22\)
\(4x^2+12x+9-4x+1=22\)
\(\left(4x^2-4x^2\right)+12x+\left(9+1\right)=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(=12\div12\)
\(x=1\)
\(4x^2+12x+9-\left[\left(2x\right)^2-1^2\right]=22\)
\(4x^2+12x+9-\left(4x^2-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x=22-9-1\)
\(12x=12\)
\(x=1\)
Bài làm :
Ta có :
\(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
\(\Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\)
\(\Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\)
\(\Leftrightarrow3.\left(4x+3\right)=21\)
\(\Leftrightarrow4x+3=7\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)
Vậy x=1
