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AH
Akai Haruma
Giáo viên
19 tháng 3 2019

1.

\((\frac{1}{3}xy)^2.x^3+\frac{3}{2}(2x)^3(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)

\(=(\frac{1}{9}x^2y^2)x^3+\frac{3}{2}(8x^3)(-\frac{7}{4}x^2y^2)-\frac{2}{3}x^5y^2\)

\(=\frac{1}{9}(x^2.x^3)y^2+(\frac{3}{2}.8.\frac{-7}{4})(x^3.x^2).y^2-\frac{2}{3}x^5y^2\)

\(=\frac{1}{9}x^5y^2-21x^5y^2-\frac{2}{3}x^5y^2=\frac{-194}{9}x^5y^2\)

2.

\(\frac{-2}{5}x^2y(-y^6)+\frac{3}{2}xy(\frac{-1}{15}xy^6)+(-2xy)^2y^5\)

\(=\frac{2}{5}x^2(y.y^6)+(\frac{3}{2}.\frac{-1}{15})(x.x).(y.y^6)+4x^2(y^2.y^5)\)

\(=\frac{2}{5}x^2y^7-\frac{1}{10}x^2y^7+4x^2y^7=\frac{43}{10}x^2y^7\)

AH
Akai Haruma
Giáo viên
19 tháng 3 2019

3.

\(\frac{3}{7}xy^2z+\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2-\frac{3}{7}xy^2z\)

\(=(\frac{3}{7}xy^2z-\frac{3}{7}xy^2z)+(\frac{1}{2}x^3y^2+\frac{1}{3}x^3y^2)\)

\(=\frac{5}{6}x^3y^2\)

4.

\(\frac{2}{3}xy^2-\frac{5}{2}yz+\frac{1}{2}xy^2-\frac{2}{3}yz\)

\(=(\frac{2}{3}xy^2+\frac{1}{2}xy^2)-(\frac{5}{2}yz+\frac{2}{3}yz)\)

\(=\frac{7}{6}xy^2+\frac{19}{6}yz\)

5.

\(\frac{3}{2}xy^2z^5-\frac{5}{4}xyz^2+\frac{4}{3}xy^2z^5+\frac{1}{2}xyz^2\)

\(=(\frac{3}{2}xy^2z^5+\frac{4}{3}xy^2z^5)+(\frac{-5}{4}xyz^2+\frac{1}{2}xyz^2)\)

\(=\frac{17}{6}xy^2z^5-\frac{3}{4}xyz^2\)

15 tháng 3 2019

a,-200 x10 t10z3

b,\(\frac{-5}{4}\)x11 y5 z4

c,\(\frac{2}{15}\)x6 y6 z9

d,\(\frac{1}{7}\)x10 y6 z7

e,-4z6 y10 z6

AH
Akai Haruma
Giáo viên
15 tháng 3 2019

Lời giải:
1.

\((-2x^4y^3z^7)^2(\frac{1}{4}xy^5)(-3x^2yz)^3(\frac{-1}{27}x^3yz^2)\)

\(=(4x^8y^6z^{14})(\frac{1}{4}xy^5)(-27x^6y^3z^3)(-\frac{1}{27}x^3yz^2)\)

\(=(4.\frac{1}{4}.-27.\frac{-1}{27})(x^8.x.x^6.x^3)(y^6.y^5.y^3.y)(z^{14}.z^3.z^2)\)

\(=x^{18}.y^{15}.z^{19}\)

2.

\(=(\frac{-1}{3}.\frac{4}{5}.\frac{-27}{10})(x.x^5.x^2)(y^2.y^6.y)(z.z.z^4)\)

\(=\frac{18}{25}.x^8.y^9.z^6\)

3.

\(=(49.x^{10}y^2z^4)(\frac{-1}{4}.x^3yz^7)(\frac{8}{21}x^5z^4)\)

\(=(49.\frac{-1}{4}.\frac{8}{21})(x^{10}.x^3.x^5)(y^2.y)(z^4.z^7.z^4)\)

\(=\frac{-14}{3}.x^{18}.y^3.z^{15}\)

4.

\(=(\frac{-1}{64}.x^8.y^9.z^{12})(4x^2y^2z^4)(\frac{-5}{3}x^4yz)\)

\(=(\frac{-1}{64}.4.\frac{-5}{3})(x^8.x^2.x^4)(y^9.y^2.y)(z^{12}.z^4.z)\)

\(=\frac{5}{48}.x^{14}.y^{12}.z^{17}\)

5.

\(=(\frac{1}{16}.x^8.y^4z^2)(-8xyz^2).(-\frac{1}{2}x^4yz)\)

\(=(\frac{1}{16}.-8.\frac{-1}{2})(x^8.x.x^4)(y^4.y.y)(z^2.z^2.z)\)

\(=\frac{1}{4}.x^{13}.y^6.z^5\)

7 tháng 5 2019

1, \(\left(xy\right)^2-\frac{1}{2}x^2y^2+3xy^2.\left(-\frac{1}{3}x\right)\)

\(=x^2y^2-\frac{1}{2}x^2y^2-x^2y^2\)

\(=-\frac{1}{2}x^2y^2\)

2, \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)

\(=x^2+\frac{3}{2}x^2+\frac{1}{3}x^2\)

\(=\frac{17}{6}x^2\)

3, \(-4.\left(2x\right)^2y^3+\frac{1}{2}xy.\left(-2xy^2\right)+\frac{1}{4}x^2y^3\)

\(=-16x^2y^3-x^2y^3+\frac{1}{4}x^2y^3\)

\(=-\frac{67}{4}x^2y^3\)

4, \(\frac{1}{3}x^4y-\frac{5}{3}x^3.\left(\frac{5}{2}xy\right)+\frac{3}{4}x^4y\)

\(=\frac{1}{3}x^4y-\frac{25}{6}x^4y+\frac{3}{5}x^4y\)

\(=-\frac{97}{30}x^4y\)

5, \(\left(-2x^3y^4\right)^2-5x^2y.\left(\frac{3}{4}x^4y^7\right)-\frac{2}{3}x^6y^8\)

\(=4x^6y^8-\frac{15}{4}x^6y^8-\frac{2}{3}x^6y^8\)

\(=-\frac{5}{12}x^6y^8\)

AH
Akai Haruma
Giáo viên
19 tháng 3 2019

1.

\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)

\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)

\(=2x^5y^4-4x^2y^3\)

2.

\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)

\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)

\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)

3.

\(5x-7xy^2+3x-\frac{1}{2}xy^2\)

\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)

\(=8x-\frac{15}{2}xy^2\)

AH
Akai Haruma
Giáo viên
19 tháng 3 2019

4.

\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)

\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)

\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)

5.

\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)

\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)

\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)

6.

\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)

\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)

m: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)

Do đó: x=8; y=10; z=7

n: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Do đó: x=18; y=16; z=15

1 tháng 8 2020

a) (-xy)10 : (-xy)5 = (-xy)2 

b) (5x2y4) : 10x2y = y3/2

c) (15x4y3z2) : (5x2y2z2) = 3x2y

d) \(\frac{3}{4}x^3y^3:\left(-\frac{1}{2}x^2y^2\right)=\frac{\frac{3}{4}x^3y^2}{-\frac{1}{2}x^2y^2}=-\frac{3}{2}x\)

e) 6x3y5 : 12x3y2 = y3/2

f) (25x5 - 5x4 + 10x2) : (5x2

= 5x3 - x2 + 2

g) \(\frac{2}{3}xy\cdot\left(2x^2y-3xy+y^2\right)=\frac{4}{3}x^3y^2-2x^2y^2+\frac{2}{3}xy^3\)

h) \(\frac{3}{4}x^3y^5z:\left(5x^2y^2z\right)=\frac{\frac{3}{4}x^3y^5z}{5x^2y^2z}=\frac{3}{20}xy^3\)

1 tháng 8 2020

Cảm ơn a CTV ạ!

12 tháng 3 2019

2.a.\(A=6x^2y-\frac{2}{3}x^2y-\frac{4}{3}x^2y=4x^2y\)

b. Thay x=-2; y=\(\frac{1}{8}\):

\(A=4\left(-2\right)^2.\frac{1}{8}=2\)