4x^3+4xy^2+8xy^2-16x
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\(a.\)
\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)
\(b.\)
\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)
a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)
\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)
\(=\dfrac{4x+1}{4x-1}\)
b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)
\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)
\(=\dfrac{y-2x}{y+2x}\)
4x3y - 4xy3 - 8xy2 - 4xy
= 4xy.(x2 - y2 - 2y - 1)
= 4xy.[(x2 - 1) - (y2 + 2y)]
= 4xy.[(x+1).(x-1) - y.(y+2)]
\(C=\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\ge-\dfrac{1}{2}\)
\(C_{min}=-\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)
\(D=\left(16x^2+2x+\dfrac{1}{16}\right)-\dfrac{1}{16}=\left(4x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\ge-\dfrac{1}{16}\)
\(D_{min}=-\dfrac{1}{16}\) khi \(x=-\dfrac{1}{16}\)
\(E=\left(x^2-4xy+4y^2\right)+\left(4x^2-4x+1\right)+2\)
\(E=\left(x-2y\right)^2+\left(2x-1\right)^2+2\ge2\)
\(E_{min}=2\) khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{4}\right)\)
a) Ta có: \(4x^2-28xy+49y^2\)
\(=\left(2x\right)^2-2\cdot2x\cdot7y+\left(7y\right)^2\)
\(=\left(2x-7y\right)^2\)
b) Ta có: \(x^2+8xy+16y^2\)
\(=x^2+2\cdot x\cdot4y+\left(4y\right)^2\)
\(=\left(x+4y\right)^2\)
c) Ta có: \(x^2-12x+36\)
\(=x^2-2\cdot x\cdot6+6^2\)
\(=\left(x-6\right)^2\)
a.16/4 . x = 16/2
x = 16/2 : 16/4
x = 16/2 . 4/16
x = 1/2.
b, b.2xy3+ 4x2y+8xy3−1x2y
= 2xy3 + 8xy3 + 4x2y - 1x2y
= ( 2 + 8 )xy3 + ( 4 - 1 )x2y
= 10xy3 + 3x2y.
a: x=16/2:16/4=2
b: \(=10xy^3+3x^2y\)
c: \(15x-14x+3xyz^3+2xyz^3-4xy^2+xy^2\)
\(=x+5xyz^3-3xy^2\)
\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
b)\(x^2-7x+6=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
Câu a khó hiểu quá
\(4x^3+4xy^2+8xy^2-16x\)
\(=4x.x^2+4x.y^2+4x.2y^2-4x.4\)
\(=4x.\left(x^2+y^2+2y^2-4\right)\)
\(=4x.\left(x^2+3y^2-4\right)\)