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a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
\(ĐKXĐ:x\ne\pm\frac{3}{2};x\ne1;x\ne0\)
\(A=\left(\frac{2+3x}{2-3x}-\frac{36x^2}{9x^2-4}-\frac{2-3x}{2+3x}\right):\frac{x^2-x}{2x^2-3x^3}\)
\(=\left[\frac{\left(2+3x\right)^2}{\left(2+3x\right)\left(2-3x\right)}+\frac{36x^2}{\left(2-3x\right)\left(2+3x\right)}-\frac{\left(2-3x\right)^2}{\left(2-3x\right)\left(2+3x\right)}\right]:\frac{x\left(x-1\right)}{x^2\left(2-3x\right)}\)
\(=\frac{4+12x+9x^2+36x^2-4+12x-9x^2}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{36x^2+24x}{\left(2+3x\right)\left(2-3x\right)}\cdot\frac{x\left(2-3x\right)}{x-1}\)
\(=\frac{12x\left(3x+2\right)}{2+3x}\cdot\frac{x}{x-1}\)
\(=\frac{12x^2}{x-1}\)
Để A nguyên dương hay \(\frac{12x^2}{x-1}\) nguyên dương
Mà \(12x^2\ge0\Rightarrow x-1>0\Rightarrow x>1\)
Vậy để A nguyên dương thì x là số nguyên dương lớn hơn 1.
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a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
Bài 1:
a) Sửa đề \(x\left(x+y\right)-3y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3y\right)\)
b) \(x^2+2019x-xy-2019y\)
\(=x\left(x+2019\right)-y\left(x+2019\right)\)
\(=\left(x+2019\right)\left(x-y\right)\)
c) \(x^2-9y^2-4x+4\)
\(=\left(x^2-4x+4\right)-9y^2\)
\(=\left(x-2\right)^2-\left(3y\right)^2\)
\(=\left(x-2-3y\right)\left(x-2+3y\right)\)
d) \(3x^2-5x+2\)
\(=3x^2-3x-2x+2\)
\(=3x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-2\right)\)
Bài 2:
a) \(\left(6x^3y^3-27xy^2\right):\left(3x^2y\right)-2xy^2\)
\(=6x^3y^3:3x^2y-27xy^2:3x^2y-2xy^2\)
\(=2xy^2-\dfrac{9y}{x}-2xy^2\)
\(=-\dfrac{9y}{x}\)
b) \(\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}+\dfrac{3x+2}{4-x^2}\)
\(=\dfrac{2}{x-2}+\dfrac{1-2x}{x+2}-\dfrac{3x+2}{x^2-4}\)
\(=\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(1-2x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2\left(x+2\right)+\left(1-2x\right)\left(x-2\right)-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x+4+x-2-2x^2+4x-3x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-2x}{x+2}\)
Bài 3:
a) \(3x\left(2x-3\right)-x\left(6x+4\right)=7-12x\)
\(\Rightarrow6x^2-9x-6x^2-4x=7-12x\)
\(\Rightarrow-13x=7-12x\)
\(\Rightarrow-13x+12x-7=0\)
\(\Rightarrow-x-7=0\)
\(\Rightarrow-x=7\)
\(\Rightarrow x=-7\)
b) \(3\left(x-5\right)-2x^2+10x=0\)
\(\Rightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
a, Sửa đề:
\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)
b,
\(x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\)
c,
\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)
d,
\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)