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a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
a: \(\dfrac{5}{2x+6}=\dfrac{5\left(x-3\right)}{2\left(x+3\right)\left(x-3\right)}\)
3/x^2-9=6/2(x+3)(x-3)
b: \(\dfrac{2x}{x^2-8x+16}=\dfrac{2x}{\left(x-4\right)^2}=\dfrac{6x^2}{3x\left(x-4\right)^2}\)
\(\dfrac{x}{3x^2-12x}=\dfrac{x}{3x\left(x-4\right)}=\dfrac{x\left(x-4\right)}{3x\left(x-4\right)^2}\)
c: \(\dfrac{x+y}{x}=\dfrac{\left(x+y\right)\cdot\left(x-y\right)}{x\left(x-y\right)}\)
x/x-y=x^2/x(x-y)
e: \(\dfrac{1}{x+2}=\dfrac{2x-x^2}{x\left(x+2\right)\left(2-x\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8\left(x+2\right)}{x\left(2-x\right)\left(2+x\right)}\)
ĐKXĐ bạn tự xét nha ^^
a) \(\dfrac{3}{2y+4}-\dfrac{1}{3y+6}=\dfrac{3}{2\left(y+2\right)}-\dfrac{1}{3\left(y+2\right)}\)
\(=\dfrac{9}{6\left(y+2\right)}-\dfrac{2}{6\left(y+2\right)}=\dfrac{9-2}{6\left(y+2\right)}\)
\(=\dfrac{7}{6\left(y+2\right)}\)
b) \(\dfrac{1}{2x-3}-\dfrac{1}{2x+3}=\dfrac{2x+3}{\left(2x-3\right)\left(2x+3\right)}-\dfrac{2x-3}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{2x+3-2x+3}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{6}{4x^2-9}\)
c) \(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y}{xy\left(y-x\right)}-\dfrac{x}{xy\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}\)
\(=\dfrac{1}{xy}\)
d) \(\dfrac{x+1}{x+4}-\dfrac{x^2-4}{x^2-16}=\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+4\right)\left(x-4\right)}\)
\(=\dfrac{x^2-3x-4-x^2+4}{\left(x+4\right)\left(x-4\right)}=\dfrac{-3x}{x^2-16}\)
Bài giải:
a) 5x2y4 : 10x2y = 510510x2 – 2. y4 – 1 = 1212y3
b) 3434x3y3 : (- 1212x2y2) = 3434 . (-2) . x3 – 2 . y3 – 2 = -3232xy
c) (-xy)10 : (-xy)5 = (-xy)10 – 5 = (-xy)5 = -x5y5.
a,\(x^2+2y^2+z^2-2xy-2y+2z+2=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2+2x+1\right)=0\)\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z+1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z+1\right)^1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-y=0\\y-1=0\\z+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=1\\z=-1\end{matrix}\right.\)
a/
\(\Leftrightarrow A=\dfrac{3}{8}xy^2+B-\dfrac{5}{6}x^2y+\dfrac{3}{4}x^2y-\dfrac{5}{8}xy^2\\ \Leftrightarrow A-B=-\dfrac{1}{12}x^2y-\dfrac{1}{4}xy^2\)
b/
\(\Leftrightarrow A-B=5xy^3-\dfrac{5}{8}yx^3-\dfrac{21}{4}xy^3+\dfrac{3}{7}x^3y\\ \Leftrightarrow A-B=-\dfrac{1}{4}xy^3-\dfrac{11}{56}x^3y\)