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a, \(2x^2+3\left(x+1\right)\left(x-1\right)-5x\left(x+1\right)\)
\(=2x^2+3\left(x^2-1\right)-5x^2-5x\)
\(=2x^2+3x^2-3-5x^2-5x\)
\(=\left(2x^2+3x^2-5x^2\right)-3-5x\)
\(=-\left(5x+3\right)\)
b, \(\left(4x+3y\right)\left(2x-5y\right)-\left(2x+6y\right)\left(3x-5y\right)\)
\(=8x^2-20xy+6xy-\left(15y^2-6x^2-10xy-18xy-30y^2\right)\)
\(=8x^2-20xy+6xy-15y^2+6x^2+10xy+18xy+30y^2\)
\(=\left(8x^2+6x^2\right)+\left(-20xy+6xy+10xy+18xy\right)+\left(-15y^2+30y^2\right)\)
\(=14x^2+14xy+15y^2\)
\(=14x.\left(x+y\right)+15y^2\)
Chúc bạn học tốt!!!
a) \(=-10x^6y^7+10x^5y^6+5x^3y^5\)
b) \(=-8x^5y^3+16x^7y^2-12x^3y^4\)
\(a,A=4x^{n+1}y^2;B=3x^3y^{n-1}\)
Để \(A⋮B\) thì:
\(\left\{{}\begin{matrix}n+1\ge3\\n-1\le2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n\ge2\\n\le3\end{matrix}\right.\Leftrightarrow2\le n\le3\)
Vậy....
\(b,A=7x^{n-1}y^5-5x^3y^4;B=5x^3y^n\)
Để \(A⋮Bthì:\)
\(\left\{{}\begin{matrix}n-1\ge2\\\\\\n\le4\end{matrix}\right.\)
\(\Leftrightarrow....\Leftrightarrow3\le n\le4\)
Vậy....
De ma bn
De \(A⋮B\)thi cac so mu o B phai nho hon hoac bang so mu o A
the la tim dc n
a) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-12xy-\frac{12}{5}y^2+12xy=3x^2-\frac{12}{5}y^2\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=24-11x\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=6y^2-x^2y\)
c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)
\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)
\(=4y^3+y^2+6xy^2\)
\(=\dfrac{2x+y}{2\left(x+y\right)}-\dfrac{x+2y}{x-y}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-2xy+xy-y^2}{2\left(x+y\right)\left(x-y\right)}-\dfrac{2\left(x+2y\right)\left(x-y\right)}{2\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{2x^2-xy-y^2-2\left(x^2+xy-2y^2\right)}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{2x^2-xy-y^2-2x^2-2xy+4y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-3xy+3y^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{4x}{3\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9xy+9y^2-8x}{6\left(x-y\right)\left(x+y\right)}+\dfrac{5}{x}\)
\(=\dfrac{-9x^2y+9xy^2-8x^2+30\left(x^2-y^2\right)}{6x\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{-9x^2y+9xy^2+22x^2-30y^2}{6x\cdot\left(x-y\right)\left(x+y\right)}\)
a) \(\left(5x+3y\right)\left(5x-3y\right)+\left(4x-3y\right)^2\)
\(=25x^2-9y^2+16x^2-24xy+9y^2\)
\(=41x^2-24xy\)
b) \(\left(2x-3y\right)^3-\left(3x+2y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27xy^2-27x^3-54x^2y-36xy^2-8y^3\)
\(=-19x^3-90x^2y+18xy^2-35y^3\)