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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!!
\(\dfrac{4x^2+4x+1}{4x^2-1}-\dfrac{2}{2x-1}-3\)
\(=\dfrac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{2}{2x-1}-3\)
\(=\dfrac{2x+1}{2x-1}-\dfrac{2}{2x-1}-\dfrac{3\left(2x-1\right)}{2x-1}\)
\(=\dfrac{2x+1-2-6x+3}{2x-1}\)
\(=\dfrac{-4x+2}{2x-1}\)
\(=\dfrac{-2\left(2x-1\right)}{2x-1}=-2\)
Đề bài ko chính xác
Biểu thức này chỉ có GTLN, không có GTNN
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
\(p=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+2010\)\(=\left(x^3+1\right)+x-\left(x^3-1\right)+2010=x^3+1+x-x^3+1+2010=x+2012\)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 \(x=\dfrac{1}{5}\Rightarrow q=-80.\dfrac{1}{5}-1=-17\)
\(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{\left(1+2x\right)\cdot3}{\left(x+4\right)\cdot2}\)
ii) (4x + 1)2 + (4x – 1)2 – 2(4x + 1)(4x – 1)
= [(4x + 1) - (4x - 1)]2
= (4x + 1 - 4x + 1)2
= 22 = 4
16x (4x²-5) + 17 = (4x + 1) (16x²-4x + 1)
64x³-80x + 17 = 64x³ + 1
64x³-64x³ = 1-17-80x
-80x = -16
x = -16 / ( -80)
x = 0,2
\(\left(4x+1\right)\left(1-4x+16x^2\right)-16x\left(4x^2-5\right)=17\)
\(\Leftrightarrow4x-16x^2+64x^2+1-4x+16x^2-64x^2+80x-17=0\)
\(\Leftrightarrow\left(-16x^2+16x^2\right)+\left(64x^2-64x^2\right)+\left(4x-4x\right)+80x+1-17=0\)
\(\Leftrightarrow80x=16\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
Thứ nhất: Làm chi tiết ra k dc ạ?
Thứ 2: Kết quả sai. Xem lại.
bạn thử phẩn tích mẫu thành nhân tử xem băng phuowg pháp tách hoạch nhẩm nghiệm cx đc
\(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\\ \)