\(3x\left(x+5\right)-\left(x+2\right)^2=2x^2+7\)
\(\Leftrightarrow3x^2+15x-x^2-4x-4=2x^2+7\)
\(\Leftrightarrow3x^2-2x^2-x^2+15x-4x=7+4\)
\(\Leftrightarrow11x=11\)
\(\Leftrightarrow x=1\)
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đăng kí hộ
3x(x + 5) - (x + 2)2 = 2x2 + 7
<=> 3x2 + 15x - x2 - 4x - 4 = 2x2 + 7
<=> 2x2 + 11x - 4 = 2x2 + 7
<=> 2x2 + 11x - 4 - 2x2 = 7
<=> 11x - 4 = 7
<=> 11x = 7 + 4
<=> 11x = 11
<=> x = 1
3x(x+5)-(x+2)^2=2x^2+7
=> 3x^2 + 15x - x^2 - 4x - 4 = 2x^2 + 7
=> 2x^2 + 11x - 4 = 2x^2 + 7
=> 11x = 7 + 4
=> 11x = 11
=> x = 1
\(3x\left(x+5\right)-\left(x+2\right)^2=2x^2+7.\)
\(3x^2+15x-x^2-4x-4=2x^2+7\)
\(2x^2+11x-4=2x^2+7\)
\(11x-4=7\)
\(11x=7+4\)
\(11x=11\)
\(x=1\)
dấu \(\Leftrightarrow\)dùng phải đúng trường hợp chứ nhiều bn làm vậy cho ngầu hả
3.x .( x + 5 ) - ( x + 2)2 = 2.x2 + 7
<=> 3.x2 + 15.x - x2-4.x-4=2.x2+7
<=> 3.x2-2.x2-x2 + 15.x - 4.x = 7 + 4
<=>11 .x = 11
<=> x = 11 : 11
<=> x = 1