Có gì khó đâu bạn -..-
( 2x + 5 )( 2x - 7 ) - ( -4x - 3 )2 = 16
<=> 2x( 2x - 7 ) + 5( 2x - 7 ) - [ (-4x)2 - 2.3.(-4x) + 32 ] = 16
<=> 4x2 - 14x + 10x - 35 - [ 16x2 + 24x + 9 ] = 16
<=> 4x2 - 4x - 35 - 16x2 - 24x - 9 = 16
<=> -12x2 - 28x - 44 - 16 = 0
<=> -12x2 - 28x - 60 = 0
<=> -4( 3x2 + 7x + 15 ) = 0
<=> 3x2 + 7x + 15 = 0
Ta có : 3x2 + 7x + 15 = 3( x2 + 7/3x + 49/36 ) + 131/12 = 3( x + 7/6 )2 + 131/12 ≥ 131/12 > 0 ∀ x
=> Vô nghiệm
\(4x^2-14x+10x-35-\left(16x^2+24x+9\right)=16\)
\(4x^2-4x-35-16x^2-24x-9-16=0\)
\(-12x^2-28x-60=0\)
\(-4\left(3x^2+7x+15\right)=0\)
\(3x^2+7x+15=0\)
\(3\left(x^2+\frac{7}{3}x+5\right)=0\)
\(x^2+\frac{7}{3}x+5=0\)
\(x^2+2\cdot x\cdot\frac{7}{6}+\left(\frac{7}{6}\right)^2-\left(\frac{7}{6}\right)^2+5=0\)
\(\left(x+\frac{7}{6}\right)^2+\frac{131}{36}=0\)
\(\left(x+\frac{7}{6}\right)^2=-\frac{131}{36}\) ( vô lí vì \(\left(x+\frac{7}{6}\right)^2\ge0\forall x\) )
Vậy phương trình vô nghiệm
\(\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\)
\(\Leftrightarrow4x^2-14x+10x-35-16x^2-24x-9-16=0\)
\(\Leftrightarrow-12x^2-28x-60=0\)
\(\Leftrightarrow12x^2+28x+60=0\)
\(\Leftrightarrow12\left(x^2+\frac{7}{3}x+\frac{49}{36}\right)+\frac{131}{3}=0\)
\(\Leftrightarrow12\left(x+\frac{7}{6}\right)^2+\frac{131}{3}=0\)
Mà \(12\left(x+\frac{7}{6}\right)^2+\frac{131}{3}\ge\frac{131}{3}\)
=> Pt vô nghiệm
(2x+5)(2x-7)-(-4x-3)2= 16
4x2-14x+10x-35-16x2-24x-9=16
-12x2-28x-60=0
-4(3x2+7x+15)=0
3x2+7x+15=0
3(x2+7/3x+49/36)+131/36
3(x+7/6)2+131/36>=131/36>0 với mọi x (vô nghiệm)
\(\left(2x+5\right)\left(2x-7\right)-\left(-4x-3\right)^2=16\)
\(\Leftrightarrow4x-14x+10x-35-16x^2-24x-9=16\)
\(\Leftrightarrow3x^2+7x+15=0\)( vô nghiệm )
