đề?

c1 x²-3x-1

c2 3x²-5x-2

minh camon bn !

\(\left\{{}\begin{matrix}4x^2-12x+9\\x^2+10x+25\end{matrix}\right.\)

=2x² - 2.2x.3 + 3²

=4x²-12x+ 9

b) =x²+ 2.x.5+5²

=x²+ 10x+25

ii) (4x + 1)2 + (4x – 1)2 – 2(4x + 1)(4x – 1)

= [(4x + 1) - (4x - 1)]2

= (4x + 1 - 4x + 1)2

= 22 = 4

1) ( 4x + 1 )² + ( 4x - 1 )² - 2( 4x + 1 ).( 4x - 1 )

= ( 4x + 1 - 4x - 1 )²

= 2²

= 4

2) 4x² - 9 + ( 2x + 3 )

= ( 2x )² - 3² + ( 2x + 3 )

= ( 2x + 3 ).( 2x - 3 ) + ( 2x + 3 )

= ( 2x + 3 ). ( 2x - 3 + 1 )

= ( 2x + 3 ) .( 2x - 2 )

= 2.( 2x + 3 ) .( x - 1 )

1, (4x+1)^2 + (4x-1)^2 - 2(4x+1)(4x-1)

=[(4x+1)-(4x-1)]^2

=(4x+1-4x+1)^2

=2^2

=4

2, 4x^2 - 9 +(2x+3)

=(4x^2 - 9)+(2x+3)

=(2x+3)(2x-3)+(2x+3)

=(2x+3)(2x-3+1)

=(2x+3)(2x-2)

=2(x-1)(2x+3)

=.= hok tốt!!

\(\dfrac{4x^2+4x+1}{4x^2-1}-\dfrac{2}{2x-1}-3\)

\(=\dfrac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2}{2x-1}-3\)

\(=\dfrac{2x+1}{2x-1}-\dfrac{2}{2x-1}-\dfrac{3\left(2x-1\right)}{2x-1}\)

\(=\dfrac{2x+1-2-6x+3}{2x-1}\)

\(=\dfrac{-4x+2}{2x-1}\)

\(=\dfrac{-2\left(2x-1\right)}{2x-1}=-2\)

\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)Với \(x=-2010\Rightarrow p=-2010+2012=2\)

\(q=16x\left(4x^2-5\right)-\left(4x+1\right)\left(16x^2-4x+1\right)=64x^3-80x-64x^3-1=-80x-1\)Với \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)

Đề bài ko chính xác

Biểu thức này chỉ có GTLN, không có GTNN

Haizzzzzzzzzzz!

ĐKXĐ: \(x\ne0;\dfrac{-1}{2};\dfrac{1}{2}\)

\(\left(\dfrac{1+x}{x}+\dfrac{1}{4x^2}\right)\left(\dfrac{1-2x}{1+2x}-\dfrac{1}{1-4x^2}.\dfrac{1-4x+4x^2}{1+2x}\right)-\dfrac{1}{2x}\)

=

\(\dfrac{4x\left(x+1\right)+1}{4x^2}.\left[\dfrac{\left(1-2x\right)\left(1+2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{\left(1-2x\right)\left(1+2x\right)}.\dfrac{\left(1-2x\right)^2}{1+2x}\right]\)\(-\dfrac{1}{2x}\)

= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\left(\dfrac{1-4x^2}{\left(2x+1\right)^2}-\dfrac{1-2x}{\left(2x+1\right)^2}\right)-\dfrac{1}{2x}\)

= \(\dfrac{\left(2x+1\right)^2}{4x^2}.\dfrac{2x\left(1-2x\right)}{\left(2x+1\right)^2}-\dfrac{1}{2x}\)

= \(\dfrac{1-2x}{2x}-\dfrac{1}{2x}=\dfrac{-2x}{2x}=1\)

Sửa cho tui đoạn kết quả nhé: = -1