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\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
a: Ta có: \(x^2-4-\left(x+2\right)^2\)
\(=x^2-4-x^2-4x-4\)
=-4x-8
b: Ta có: \(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-x^2+2x+3\)
=2x-1
c: ta có: \(\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(x+5\right)\)
\(=\left(x-2\right)\left(x+2-x-5\right)\)
\(=-3x+6\)
d: Ta có: \(\left(6x+1\right)^2-2\left(6x+1\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(=\left(6x+1-6x+1\right)^2\)
=4
e: ta có: \(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)
\(=21a^2-35a+8a^2+2a-12a-3-\left(36a^2-24a+4\right)\)
\(=29a^2-45a-3-36a^2+24a-4\)
\(=-7a^2-21a-7\)
g: ta có: \(\left(5y-3\right)\left(5y+3\right)-\left(5y-4\right)^2\)
\(=25y^2-9-25y^2+40y-16\)
=40y-25
h: Ta có: \(\left(3x+1\right)^3-\left(1-2x\right)^3\)
\(=27x^3+27x^2+9x+1-1+6x-12x^2+8x^3\)
\(=35x^3+15x^2+15x\)
i: Ta có: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=16x^2\)
a, = 3x2y3 : x2y2 - 5x2y2 : x2y2 +6x4y7 :x2y2 -9x5y4 :x2y2
= 3y -5+ x2y5 -9x3y2
b., = a2.(6a-3):a2 + 3a(4a+3):3a = 6a - 3+ 4a +3= 10a
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
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
ĐKXĐ: \(a\ne1\)
a. \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3+\left(1-a\right).\left(a-1\right)-2.\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-a^2+2a-1-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a+1}{\left(a-1\right).\left(a^2+a+1\right)}\)
\(=-\dfrac{1}{a^2+a+1}\)
a) Ta có: \(\dfrac{3a^2-a+3}{a^3-1}+\dfrac{1-a}{a^2+a+1}+\dfrac{2}{1-a}\)
\(=\dfrac{3a^2-a+3}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{\left(a-1\right)^2}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{2\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{3a^2-a+3-\left(a^2-2a+1\right)-2a^2-2a-2}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{a^2-3a+1-a^2+2a-1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b) Ta có: \(x-\dfrac{xy}{x+y}-\dfrac{x^3}{x^2y^2}\)
\(=x-\dfrac{xy}{x+y}-\dfrac{x}{y^2}\)
\(=\dfrac{xy^2\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}+\dfrac{y^2\cdot xy}{y^2\cdot\left(x+y\right)}-\dfrac{x\cdot\left(x+y\right)}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+xy^3+xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
\(=\dfrac{x^2y^2+2xy^3-x^2-xy}{y^2\cdot\left(x+y\right)}\)
Trả lời:
a, \(\left(x-\frac{3}{4}\right)^2=x^2-2.x.\frac{3}{4}+\left(\frac{3}{4}\right)^2=x^2-\frac{3}{2}x+\frac{9}{16}\)
b, \(\left(3a+1\right)^2=\left(3a\right)^2+2.3a.1+1^2=9a^2+6a+1\)
c, \(\left(3a+\frac{1}{3}\right)\left(\frac{1}{3}-3a\right)=\left(\frac{1}{3}\right)^2-\left(3a\right)^2=\frac{1}{9}-9a^2\)
d, \(\left(x^2-2\right)^2=\left(x^2\right)^2-2.x^2.2+2^2=x^4-4x^2+4\)
e, \(\left(9x+2y\right)^2=\left(9x\right)^2+2.9x.2y+\left(2y\right)^2=81x^2+36xy+4y^2\)
g, \(\left(x-5y\right)\left(x+5y\right)=x^2-\left(5y\right)^2=x^2-25y^2\)
h, \(\left(5-x\right)^2=5^2-2.5.x+x^2=25-10x+x\)