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V
24 tháng 10 2019
x3 + x + 2
\(=x^3+x^2-x^2-x+2x+2\)
\(=x^2\left(x+1\right)-x\left(x+1\right)+2\left(x+1\right)\)
\(\left(x+1\right)\left(x^2-x+2\right)\)
c) x3 + 32x - 4
\(=x^3-x^2+4x^2-4x+4x-4\)
\(=x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+4x+4\right)\)
\(=\left(x-1\right)\left(x+2^2\right)\)
d)x3y3 + x2y2 + 4
\(=x^3y^3-4xy+x^2y^2-4xy+4\)
\(=xy\left(x^2y^2-4\right)+\left(xy+2\right)^2\)
\(=xy\left(xy-2\right)\left(xy+2\right)+\left(xy+2\right)^2\)
\(=\left(xy+2\right)\left(xy\left(xy-2\right)+xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-2xy+xy+2\right)\)
\(=\left(xy+2\right)\left(x^2y^2-xy+2\right)\)
e) x3 + 3x2y - 9xy2 + 5y3
\(=x^3-3x^2y+3xy^2-y^3+6x^2y-12xy^2+6y^3\)
\(=\left(x-y\right)^3\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\left(x-y\right)^2=\left(x-y\right)^2\left(x-y-1\right)\)
TN
0
VN
5 tháng 8 2018
A = \(4\left(x-5\right)-x^2\left(x+1\right)-x^3\left(x-3\right)-\left(x-4+x^2\right)\)
A = \(4x-20-x^3-x^2-x^4+3x^3-x+4-x^2\)
A = \(-x^3-3x^3-x^2+x^2-x^4+4x-x-20+4\)
A = \(-4x^3-x^4+4x-16\)
B = \(-3\left(x^2-x+1\right)-2\left(4-x^2\right)-6\left(x+1\right)-x^4-x^3\)
B = \(-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
B = \(-x^4-x^3-3x^2-2x^2+3x-6x-3-8-6\)
B = \(-x^4-x^3-5x^2-3x-17\)
C = \(-\left(x^4+3x^2-2\right)-x^2\left(5-x\right)+3\left(x-1\right)\)
C = \(-x^4-3x^2+2-5x^2+x^3+3x-3\)
C = \(-x^4+x^3-3x^2+5x^2+3x+2-3\)
C = \(-x^4+x^3-2x^2+3x-1\)
#Yiin
PN
5 tháng 8 2018
\(A=4x-20-x^3-x-x^4+3x^3-x+4-x^2\)
\(=-x^4+2x^3-x^2+2x-16\)
\(B=-3x^2+3x-3-8+2x^2-6x-6-x^4-x^3\)
\(=-x^4-x^3-x^2-3x-17\)
\(C=-x^4-3x^2+2-5x^2+x^3+3x-3\)
\(=-x^4+x^3-8x^2+3x-1\)
Từ đó có:
\(A-B=-x^4+2x^3-x^2+2x-16-\left(-x^4-x^3-x^2-3x-17\right)\)
\(=-x^4+2x^3-x^2+2x-16+x^4+x^3+x^2+3x+17\)\(=3x^3+5x+1\)
\(B-C=-x^4-x^3-x^2-3x-17-\left(-x^4+x^3-8x^2+3x-1\right)\)
\(=-x^4-x^3-x^2-3x-17+x^4-x^3+8x^2-3x+1\)
\(=-2x^3+7x^2-6x-16\)
\(C-A=-x^4+x^3-8x^2+3x-1-\left(-x^4+2x^3-x^2-2x-16\right)\)
\(=-x^4+x^3-8x^2+3x-1+x^4-2x^3+x^2+2x+16\)
\(=-x^3-7x^2+5x+15\)
NQ
2
A
18 tháng 10 2020
Chỉ biết cách châu bò này :#
\(\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}=\frac{x}{3^2}+\frac{x}{3^3}+\frac{x}{3^4}\)
\(\Leftrightarrow\frac{x}{4}+\frac{x}{8}+\frac{x}{16}=\frac{x}{9}+\frac{x}{27}+\frac{x}{81}\)
\(\Leftrightarrow\frac{4x}{16}+\frac{2x}{16}+\frac{x}{16}=\frac{9x}{81}+\frac{3x}{81}+\frac{x}{81}\)
\(\Leftrightarrow\frac{7x}{16}=\frac{13x}{81}\Leftrightarrow567x=208x\Leftrightarrow x=\frac{1}{359}\)
\(|x^2|x+\dfrac{3}{4}||=x^2\)
=>\(x^2\cdot\left|x+\dfrac{3}{4}\right|=x^2\)
=>\(\left|x+\dfrac{3}{4}\right|=1\)
=>\(\left[{}\begin{matrix}x+\dfrac{3}{4}=1\\x+\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
|\(x^2\).|\(x+\dfrac{3}{4}\)| |= \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| = \(x^2\)
\(x^2\).|\(x+\dfrac{3}{4}\)| - \(x^2\) = 0
\(x^2\).(|\(x+\dfrac{3}{4}\)| - 1) = 0
\(\left[{}\begin{matrix}x=0\\\left|x+\dfrac{3}{4}\right|=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x+\dfrac{3}{4}=-1\\x+\dfrac{3}{4}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=-\dfrac{7}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\) \(\in\) { - \(\dfrac{7}{4}\); 0; \(\dfrac{1}{4}\)}