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a: \(A=-4x^5y^3-2x^2y^3z^2-2y^4\)
b: \(B=-4x^5y^3-2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3=-4x^5y^3+\dfrac{1}{5}x^4y^3-\dfrac{8}{3}y^4\)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
\(B=\dfrac{3}{4}xy^2-\dfrac{1}{3}x^2y-\dfrac{5}{6}xy^2+2x^2y=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y\)
Bậc:3
Thay x=-1, y=1 vào B ta có:
\(B=-\dfrac{1}{12}xy^2+\dfrac{5}{3}x^2y=-\dfrac{1}{12}.\left(-1\right).1^2+\dfrac{5}{3}.\left(-1\right)^2.1=\dfrac{1}{12}+\dfrac{5}{3}=\dfrac{7}{4}\)
a, \(A=-4x^5y^3+x^4y^3-3x^2y^3z^2+4x^5y^3-x^4y^3+x^2y^3z^2-2y^4\)
\(=2x^2y^3z^2-2y^4\)
Bậc của đa thức A là 7
Vậy...
b, Ta có: \(B-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=A\)
\(\Rightarrow B-2x^2y^3z^2+\dfrac{2}{3}y^4-\dfrac{1}{5}x^4y^3=2x^2y^3z^2-2y^4\)
\(\Rightarrow B=2x^2y^3z^2-2y^4+2x^2y^3z^2-\dfrac{2}{3}y^4+\dfrac{1}{5}x^4y^3\)
\(=4x^2y^3z^2-\dfrac{8}{3}y^4+\dfrac{1}{5}x^4y^3\)
Vậy...
a) \(\left(3x-2\right)^2=\left(3x\right)^2-2.3x.2+2^2=9x^2-12x+4\)
b) \(\left(\dfrac{x}{3}+y^3\right)^2=\left(\dfrac{x}{3}\right)^2+2\dfrac{x}{3}y^3+\left(y^3\right)^2=\dfrac{x^2}{9}+\dfrac{2}{3}xy^3+y^6\)
c) \(9x^2-225=\left(3x\right)^2-\left(15\right)^2=\left(3x-15\right)\left(3x+15\right)\)
d) \(\left(2x-3y\right)^3=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^2-\left(3y\right)^3=8x^3-3.4x^2.3y+6x.9y^2-27y^3=8x^3-36x^2y+54xy^2-27y^3\)
e) \(\left(2x^2+\dfrac{3}{2}\right)^3=\left(2x^2\right)^3+3\left(2x^2\right)^2\dfrac{3}{2}+3.2x^2\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3=8x^6+3.4x^4.\dfrac{3}{2}+6x^2.\dfrac{9}{4}+\dfrac{27}{8}=8x^6+18x^4+\dfrac{27}{2}x^2+\dfrac{27}{8}\)
f) \(\left(-2xy^2+\dfrac{1}{2}x^3y\right)^3=\left(-2xy^2\right)+3\left(-2xy^2\right)^2\dfrac{1}{2}x^3y+3\left(-2xy^2\right)\left(\dfrac{1}{2}x^3y\right)^2+\left(\dfrac{1}{2}x^3y\right)^3=-8x^3y^6+3.4x^2y^4.\dfrac{1}{2}x^3y-6xy^2.\dfrac{1}{4}x^6y^2+\dfrac{1}{8}x^9y^3=-8x^3y^6+6x^5y^5-\dfrac{3}{2}x^7y^4+\dfrac{1}{8}x^9y^3\)
a) \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)
\(=\dfrac{\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5}{\dfrac{3}{5}ax^3}\)
\(=\dfrac{\dfrac{3}{5}ax^3\left(a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\right)}{\dfrac{3}{5}ax^3}\)
\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\)
b) \(\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)\cdot y^2\)
\(=\dfrac{3x^2y\left(3y^2-5x^2y^3\right)}{3x^2y}-2y^2+3x^2y^3\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
c) \(\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=\dfrac{6x^2-xy}{x}+\dfrac{2x^3y+3xy^2}{xy}-x\left(2x-1\right)\)
\(=\dfrac{x\left(6x-y\right)}{x}+\dfrac{xy\left(2x^2+3y\right)}{xy}-2x^2+x\)
\(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d) \(\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^2\right):\dfrac{3}{2}x^2y^3\)
\(=\dfrac{x^2-xy}{x}+\dfrac{6x^2y^5-9x^3y^4+15x^4y^2}{\dfrac{3}{2}x^2y^3}\)
\(=\dfrac{x\left(x-y\right)}{x}+\dfrac{\dfrac{3}{2}x^2y^2\left(4y^3-6xy^2+10x^2\right)}{\dfrac{3}{2}x^2y^3}\)
\(=x-y+\dfrac{4y^3-6xy^2+10x^2}{y}\)
a: \(A=3x^2y^3-5x^2+3x^3y^2\)
\(B=x^2y^3+\dfrac{5}{2}x^5y-5x^2y\)
b: \(A+B=4x^2y^3+5x^2+\dfrac{5}{2}x^5y+3x^3y^2-5x^2y\)
\(A-B=2x^2y^3-5x^2+3x^3y^2-\dfrac{5}{2}x^5y+5x^2y\)
c: Khi x=-1 và y=-1/3 thì \(A=3\cdot\left(-1\right)^2\cdot\dfrac{-1}{27}-5\cdot\left(-1\right)^2+3\cdot\left(-1\right)^3\cdot\dfrac{1}{9}\)
\(=-\dfrac{1}{9}-5-\dfrac{1}{3}=\dfrac{-49}{9}\)
a: \(A=31x^2y^3-2xy^3+\dfrac{1}{4}x^2y^2+2\)
\(B=2xy^3+\dfrac{3}{4}x^2y^2-31x^2y^3-x^2-5\)
P=\(A+B=x^2y^2-x^2-3\)
\(A-B=62x^2y^3-4xy^3-\dfrac{1}{2}x^2y^2+x^2+7\)
b: Khi x=6 và y=-1/3 thì \(P=\left(6\cdot\dfrac{-1}{3}\right)^2-6^2-3=4-36-3=1-36=-35\)
Lời giải:
a.
$C=-x^3y^3+x^2y+xy^2$
Bậc: $3+3=6$
b.
$D=3x^2y^3+3x^3y^2+7y^2-12x^2$
Bậc: $2+3=5$
c.
$E=\frac{5}{2}x^5y+\frac{7}{3}xy^4-\frac{1}{4}x^2y^3
Bậc: $5+1=6$