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a, \(\dfrac{3a^2b-4ab^2}{5ab}=\dfrac{ab\left(3a-4b\right)}{5ab}=\dfrac{3a-4b}{5}\)
b, \(\dfrac{3x^3y^2-5x^2y^3+4x^3y^3}{x^2y^2}=\dfrac{x^2y^2\left(3x-5y+4xy\right)}{x^2y^2}\)
\(=3x-5y+4xy\)
c, \(\dfrac{2a^5b^4+3a^4b^3}{-3a^4b^5}=\dfrac{a^4b^3\left(2ab+3\right)}{-3a^4b^5}=\dfrac{2ab+3}{-3b^2}\)
d, \(\dfrac{-a^5b^4+3a^6b^2}{4a^4b^2}=\dfrac{-a^4b^2\left(ab^2+3a^2\right)}{4a^4b^2}=\dfrac{-\left(ab^2+3a^2\right)}{4}\)
Chúc bạn học tốt!!!
a. \(\left(3a^2b-4ab^3\right):5ab=3a^2b:5ab-4ab^3:5ab=\dfrac{3}{5}a-\dfrac{4}{5}b^2\)
b. \(\left(3x^3y^2-5x^2y^3+4x^3y^3\right):x^2y^2=3x^3y^2:x^2y^2-5x^2y^3:x^2y^2+4x^3y^3:x^2y^2=3x-5y+4xy\)
c. \(\left(2a^5b^4+3a^4b^3\right):\left(-3a^4b^5\right)=2a^5b^4:\left(-3a^4b^5\right)+3a^4b^3:\left(-3a^4b^5\right)=-\dfrac{2a}{3b}-\dfrac{1}{b^2}\)
d. \(\left(-a^5b^4+3a^6b^2\right):4a^4b^2=\left(-a^5b^4\right):4a^4b^2+3a^6b^2:4a^4b^2=-\dfrac{1ab^2}{4}+\dfrac{3a^2}{4}\)
1: =(4x-1)^2-3(4x-1)
=(4x-1)(4x-1-3)
=4(x-1)(4x-1)
2: =-8x^4y^5(2y+3x)
3: =(a-5)^2-4b^2
=(a-5-2b)(a-5+2b)
5: =x^2-mx-nx+mn
=x(x-m)-n(x-m)
=(x-m)(x-n)
6: =(4a^2-3a-18-4a^2-3a)(4a^2-3a-18+4a^2+3a)
=(-6a-18)(8a^2-18)
=-6(2a-3)(2x+3)(a+3)
a)\(\left(3x^2+2xy\right).\left(5xy^2-4x+\frac{1}{3}y^2\right)\)
\(=15x^3y^2-12x^3+x^2y^3+10x^2y^3-8x^2y+\frac{2}{3}xy^4\)
\(=15x^3y^2-12x^3+11x^2y^3-8x^2y+\frac{2}{3}xy^4\)
b)\(\left(x^3-x^2-7x+3\right):\left(x-3\right)\)
\(=\left(x^3-3x^2+2x^2-6x-x+3\right):\left(x-3\right)\)
\(=\left[x^2\left(x-3\right)+2x\left(x-3\right)-\left(x-3\right)\right]:\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+2x-1\right):\left(x-3\right)\)
\(=x^2+2x-1\)
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
a, 2x3 / xn+1
= 2/x3-n+1
= 2/x2-n
Để 2x3 chia hết cho xn+1 thì 2-n \(\ge\)0 <=> n \(\le\)2
b, ( 5x3 - 7x2 + x ) / 3xn
= 5/33-n - 7/32-n + 1/31-n
Để ( 5x3 - 7x2 + x ) chia hết cho 3xn thì 3 - n \(\ge\)0
2 - n \(\ge\)0
1 - n \(\ge\)0
<=> n \(\le\)3
n \(\le\) 2
n \(\le\) 1
<=> n \(\le\)1
Còn lại tương tự nha!
c, ( 13x4y3 - 5x3y3 + 6x2y2 ) / 5xnyn
d, ( 5x3 - 7x2 + x ) / 5xnyn
e, ( 13x4y3 - 5x3y3 + 6x2y2 ) / 5xnyn
1) 4x\(^2\).(5x3+2x-1)
= 20x\(^5\)+8x\(^3\)-4x\(^2\).
2) 4x\(^3\): x2
= 4x
3) ( 15x2y3-10x3y3+6xy): 5xy
= 3xy2-2x2y2+\(\dfrac{6}{5}\)
4) (5x3+14x2+12x+8 ): (x+2)
= 5x2+4x+4
5)\(\dfrac{7}{2x}\)+\(\dfrac{11}{3y^2}\)
=\(\dfrac{7.3y^2+11.2x}{6xy^2}\) =\(\dfrac{21y^2+22x}{6xy^2}\) = \(\dfrac{21+22}{6}\) =\(\dfrac{43}{6}\)
6) \(\dfrac{x}{x+2}\) +\(\dfrac{3}{\left(x+2\right)\left(4x-7\right)}\)
7)\(\dfrac{3}{x-y}\)-\(\dfrac{2x^2}{x+y}\)
= \(\dfrac{3\left(x+y\right)-2\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{3x+3y-2x-2y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{1}{x-y}\).
8)\(\dfrac{1}{2}\)x2y2.(2x+y)(2x-y)
= \(\dfrac{1}{2}\)x2y2.(4x2-2xy+2xy-y2)
= \(\dfrac{1}{2}\)x2y2.(4x2-y2)
= 2x4y2-\(\dfrac{1}{2}\)x2y4
9) (x-\(\dfrac{1}{2}\)).(x+\(\dfrac{1}{2}\)).(4x-1)
= x2.(4x-1)
= 4x3-x2
10)\(\dfrac{3x}{2x+6}\)+\(\dfrac{6-x}{2x^2+6x}\)
= \(\dfrac{3x}{2\left(x+3\right)}\)+\(\dfrac{6-x}{2x\left(x+3\right)}\)= \(\dfrac{3x^2+6-x}{2x\left(x+3\right)}\)=\(\dfrac{3-x}{3}\)= -x
11) x2-\(\dfrac{1}{2x-2}\)+3x+\(\dfrac{3}{1-x^2}\)
12)\(\dfrac{x^2}{x^2-y^2}\)-\(\dfrac{x-y}{x^2-y^2}\)
= \(\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}\)=\(\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\)= \(\dfrac{x}{x+y}\)
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A = \(\frac{3}{5}\)a - \(\frac{4}{5}\) b2
B = 3x - 5y + 4x
TỚ nhầm. B ko thực hiện đc nhé