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\(=24x^2y^{6-3n}-\dfrac{24}{7}x^{3-2m}y+8x^3y^6-24x^{3-2m}y^{8-3n}\)
a: \(=24x^{2m-1+3-2m}y^{6-3m}-\dfrac{24}{7}y^{3n-7+6-3n}\cdot x^{3-2m}+8x^{3-2m+2m}\cdot y^{6-3n+3m}-24x^{3-2m}y^{6-2n+2}\)
\(=24x^2y^{6-3m}-\dfrac{24}{7}x^{3-2m}\cdot y^{-1}+8x^3y^{-3n+3m+6}-24x^{3-2m}y^{-2n+8}\)
b: \(=2x^{2n+1-2n}-6x^{2n+2-2n}+3x^{2n-1+1-2n}-9x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
\(=8x^{3-2m}\cdot y^{6-3n}\cdot3x^{2m-1}-8x^{3-2m}\cdot y^{6-3n}\cdot\dfrac{3}{7}y^{3n-5}+8x^{3-2m}\cdot y^{6-3n}\cdot x^{2m}y^{3n}-8x^{3-2m}\cdot y^{6-3n}\cdot3y^2\)
\(=24x^2y^{6-3n}-\dfrac{24}{7}x^{3-2m}\cdot y+8x^3y^6-24x^{3-2m}\cdot y^{8-3n}\)
1 ) Ta có : \(2n^2+3n+3\)
\(=2n^2-n+4n-2+5\)
\(=n\left(2n-1\right)+2\left(2n-1\right)+5\)
\(=\left(n+2\right)\left(2n-1\right)+5\)
Để \(2n^2+3n+3⋮2n-1\)
\(\Leftrightarrow\left(n+2\right)\left(2n-1\right)+5⋮2n-1\)
\(\Leftrightarrow5⋮2n-1\)
Do \(n\in Z\Rightarrow2n-1\in Z\)
\(\Rightarrow2n-1\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow2n\in\left\{2;0;6;-4\right\}\)
\(\Rightarrow n\in\left\{1;0;3;-2\right\}\)
Vậy ...
Bài 2 :
a ) \(3x^2-3y^2+4x-4y=3\left(x^2-y^2\right)+4\left(x-y\right)=3\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)
\(=\left(x-y\right)\left[3\left(x+y\right)+4\right]=\left(x-y\right)\left(3x+3y+4\right)\)
b ) \(12x^2-3xy+8x-2y\)
\(=3x\left(4x-y\right)+2\left(4x-y\right)\)
\(=\left(3x+2\right)\left(4x-y\right)\)
c ) \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=\left(x^2-xz\right)\left(x+y\right)\)
\(=x\left(x-z\right)\left(x+y\right)\)
d ) \(xy+y-2x-2\)
\(=y\left(x+1\right)-2\left(x+1\right)\)
\(=\left(y-2\right)\left(x+1\right)\)
e ) \(x^3-3x^2+3x-9\)
\(=x^2\left(x-3\right)+3\left(x-3\right)\)
\(=\left(x^2+3\right)\left(x-3\right)\)
1.\(\left(2n^2+3n+3\right):\left(2n-1\right)=n+2\) dư 5 (đoạn này bạn tự chia nha)
Muốn \(2n^2+3n+3\)\(⋮2n-1\) thì \(5⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(5\right)\)
\(\Rightarrow2n-1=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow2n=\left\{-4;0;2;6\right\}\)
\(\Rightarrow n=\left\{-2;0;1;3\right\}\)
bài 2 nhờ Nguyễn Thanh Hằng hay Mysterious Person giải nha
tui đi học rồi
câu a nè = (4x-1)(2x-3)
câu f = (x+y+z) ( x^ 2 + y^2 + z^2 +xy + yz + zx)
c) \(\frac{x+9}{x^2-9}+\frac{1}{x+3}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}+\frac{x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+9+x-3}{\left(x-3\right)\left(x+3\right)}=\frac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3}\)
d) \(\frac{8m+8}{11n^2}.\frac{22n^2}{m^2+2m+1}\)
\(=\frac{8\left(m+1\right)}{11n^2}.\frac{22n^2}{\left(m+1\right)^2}\)
\(=\frac{8.2}{m+1}=\frac{16}{m+1}\)
e) \(\frac{5x+3}{4xy^3}:\frac{10x+6}{x^2y}\)
\(=\frac{5x+3}{4xy^3}.\frac{x^2y}{2.\left(5x+3\right)}\)
\(=\frac{x}{8y^2}\)
Lời giải:
\((3x^{2m-1}-\frac{3}{7}y^{3n-5}+x^{2m}y^{3n}-3y^2).8x^{3-2m}y^{6-3n}\)
\(=3x^{2m-1}.8x^{3-2m}.y^{6-3n}-\frac{3}{7}y^{3n-5}.8x^{3-2m}y^{6-3n}+x^{2m}.y^{3n}.8x^{3-2m}y^{6-3n}-3y^2.8x^{3-2m}y^{6-3n}\)
\(=24x^{2m-1+3-2m}.y^{6-3n}-\frac{24}{7}y^{3n-5+6-3n}x^{3-2m}+8x^{2m+3-2m}y^{3n+6-3n}-24y^{2+6-3n}x^{3-2m}\)
\(=24x^2y^{6-3n}-\frac{24}{7}yx^{3-2m}+8x^3y^6-24y^{8-3n}x^{3-2m}\)