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\(a,=2\left(\dfrac{1}{4}x^2-y^2\right)=2\left(\dfrac{1}{2}x-y\right)\left(\dfrac{1}{2}x+y\right)\\ b,=\dfrac{1}{3}x\left(y+3xz+3z\right)\\ c,=2x\left(9x^2-\dfrac{4}{25}\right)=2x\left(3x-\dfrac{2}{5}\right)\left(3x+\dfrac{2}{5}\right)\)
\(d,=x^2\left(\dfrac{2}{5}+5x+y\right)\\ e,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ f,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ g,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)
Ta có \(4x-5\sqrt{x}-3\) = (\(4x-\frac{2×2×5\sqrt{x}}{2×2}+\frac{25}{16}\)) - \(\frac{73}{16}\)
= (\(2\sqrt{x}-\frac{5}{4}\))2 - \(\frac{73}{16}\)
= (\(2\sqrt{x}-\frac{5}{4}-\frac{\sqrt{73}}{4}\))(\(2\sqrt{x}-\frac{5}{4}+\frac{\sqrt{73}}{4}\))
a ) x 2 - 3 = x 2 - ( √ 3 ) 2 = ( x - √ 3 ) ( x + √ 3 ) b ) x 2 - 6 = x 2 - ( √ 6 ) 2 = ( x - √ 6 ) ( x + √ 6 ) c ) x 2 + 2 √ 3 x + 3 = x 2 + 2 √ 3 x + ( √ 3 ) 2 = ( x + √ 3 ) 2 d ) x 2 - 2 √ 5 x + 5 = x 2 - 2 √ 5 x + ( √ 5 ) 2 = ( x - √ 5 ) 2
a) x8+x4+1 = (x8+x7+x6) +(-x7-x6-x5)+(x5+x4+x3)+(-x3-x2-x)+(x2+x+1) = (x2+x+1)(x6-x5+x3-x+1)
b) x5+x4+1 = x5 +x4+x3-x3-x2-x+x2+x+1=(x2+x+1)(x3-x+1)
tương tự thì c) và d) cx có nhân tử x2+x+1
e) = x3-x2-5x2+5x+6x+6 = (x-1)(x2-5x+6) = (x-1)(x2-2x-3x+6) = (x-1)(x-2)(x-3)
a) Ta có: \(x^8+x^4+1=\left(x^4\right)^2+2.x^4.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\) Không phân tích được
câu b sai r
\(\dfrac{1}{3}xy+x^2z+xz=3x\left(\dfrac{1}{9}y+\dfrac{1}{3}xz+\dfrac{1}{3}z\right)\)
Lời giải:
a.
$=\frac{1}{2}(x^2-4y^2)=\frac{1}{2}[x^2-(2y)^2]=\frac{1}{2}(x-2y)(x+2y)$
b.
$=\frac{1}{3}x(y+3xz+3z)$
c.
$=\frac{2}{25}x(225x^2-4)=\frac{2}{25}(15x-2)(15x+2)$
d.
$=\frac{1}{5}x^2(2+25x+5y)$
Chọn đáp án C.