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\(\begin{array}{l}A + B = \left( {5{x^2}y + 5x - 3} \right) + \left( {xy - 4{x^2}y + 5x - 1} \right)\\ = 5{x^2}y + 5x - 3 + xy - 4{x^2}y + 5x - 1\\ = \left( {5{x^2}y - 4{x^2}y} \right) + xy + \left( {5x + 5x} \right) + \left( { - 3 - 1} \right)\\ = {x^2}y + xy + 10x - 4\end{array}\)
a) \(\left(4x^4-8x^2y^2+12x^5y\right):\left(-4x^2\right)\)
\(=4x^4:-4x^2-8x^2y^2:-4x^2+12x^4y:-4x^2\)
\(=-x^2+2y^2-3x^2y\)
b) \(x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\)
\(=x^3-x^2y^2-xy+x^2y^2-x^3\)
\(=-xy\)
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
`a)`
`4x^3 * (-6x^3y)`
`= 4*(-6) * (x^3*x^3) * y`
`= -24x^6y`
`b)`
`(-2y)*(-5xy^2)`
`= (-2)*(-5)*x*(y*y^2)`
`= 10xy^3`
`c)`
`(-2a)^3 * (2ab)^2`
`= (-8a^3) * (4a^2b^2)`
`= (-8*4)*(a^3*a^2)*b^2`
`= -32a^5b^2`
a) \(4x^3\cdot\left(-6x^3y\right)\)
\(=\left(4\cdot-6\right)\cdot\left(x^3\cdot x^3\right)\cdot y\)
\(=-24x^6y\)
b) \(\left(-2y\right)\cdot\left(-5xy^2\right)\)
\(=\left(-2\cdot-5\right)\cdot\left(y\cdot y^2\right)\cdot x\)
\(=10xy^3\)
c) \(\left(-2a\right)^3\cdot\left(2ab\right)^2\)
\(=-8a^3\cdot4a^2b^2\)
\(=\left(-8\cdot4\right)\cdot\left(a^3\cdot a^2\right)\cdot b^2\)
\(=-32a^5b^2\)
TL:
\(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+x^2y+xy^2-yx^2-xy^2-y^3\)
\(=x^3-y^3\)
\(\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x\left(x^2+xy+y^2\right)-y\left(x^2+xy+y^2\right)\)
\(=\left(x^3+x^2y+xy^2\right)-\left(x^2y+xy^2+y^3\right)\)
\(=x^3+x^2y+xy^2-x^2y-xy^2-y^3\)
\(=x^3-y^3\)
B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)
b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)
\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)
B2:
a)\(2x^2-4x+2=2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)
b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv
\(\begin{array}{l}A - B = \left( {5{x^2}y + 5x - 3} \right) - \left( {xy - 4{x^2}y + 5x - 1} \right)\\ = 5{x^2}y + 5x - 3 - xy + 4{x^2}y - 5x + 1\\ = \left( {5{x^2}y + 4{x^2}y} \right) - xy + \left( {5x + 5x} \right) + \left( { - 3 + 1} \right)\\ = 9{x^2}y - xy + 10x - 2\end{array}\)