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Bài 1.
Ta có : B = ( x + 2 )2 + ( x - 2 )2 - 2( x + 2 )( x - 2 )
= [ ( x + 2 ) - ( x - 2 ) ]2
= ( x + 2 - x + 2 )2
= 42 = 16
=> B không phụ thuộc vào x
Vậy với x = -4 thì B vẫn bằng 16
Bài 2.
4x2 - 4x + 1 = ( 2x )2 - 2.2x.1 + 12 = ( 2x - 1 )2
Bài 3.
Ta có : \(A=\frac{3}{2}x^2+2x+3\)
\(=\frac{3}{2}\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{7}{3}\)
\(=\frac{3}{2}\left(x+\frac{2}{3}\right)^2+\frac{7}{3}\ge\frac{7}{3}\forall x\)
Dấu "=" xảy ra khi x = -2/3
=> MinA = 7/3 <=> x = -2/3
a, \(A=\left(\frac{4}{2x+1}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4\left(x^2+1\right)}{\left(2x+1\right)\left(x^2+1\right)}+\frac{4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\left(\frac{4x^2+4+4x-3}{\left(x^2+1\right)\left(2x+1\right)}\right)\frac{x^2+1}{x^2+2}\)
\(=\frac{\left(2x+1\right)^2}{\left(x^2+1\right)\left(2x+1\right)}\frac{x^2+1}{x^2+2}=\frac{2x+1}{x^2+2}\)
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
