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a )\(-x^2yz+12x^2yz-10x^2yz+x^2yz\)
\(=\left(-1+12-10+1\right)x^2yz\)
\(=2x^2yz\)
b ) \(11xy^2z^3-6xy^2z+20xy^2z^3\)
\(=\left(11xy^2z^3+20xy^2z^3\right)-6xy^2z\)
\(=31xy^2z^3-6xy^2z\)
c ) \(\left(92x^3y+51x^3y\right)-\left(105x^3y-7x^3y\right)\)
\(=143x^3y-98x^3y\)
\(=45x^3y\)
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.
\(=(\frac{1}{16}.x^8.y^4z^2)(-8xyz^2).(-\frac{1}{2}x^4yz)\)
\(=(\frac{1}{16}.-8.\frac{-1}{2})(x^8.x.x^4)(y^4.y.y)(z^2.z^2.z)\)
\(=\frac{1}{4}.x^{13}.y^6.z^5\)
\(\Leftrightarrow2+4+6+...+2k=72\)
Số số hạng là (2k-2):2+1=k-1+1=k(số)
Tổng là \(\dfrac{\left(2k+2\right)\cdot k}{2}=k\left(k+1\right)\)
Theo đề, ta có: k(k+1)=72
=>k=8
Bài 1:
\(\Leftrightarrow3x^{n+4}y^{14}=3x^{25}y^{14}\)
=>n+4=25
=>x=21
D = x6 - x4yz + x3 yz2 - x3y2z + x3y2z - z6 + 2018
=> D = -z6 + x3 yz2 + ( - x4) yz + x6 + 2018
=> D = - ( z6 - x3 yz2 + x4yz - x6 - 2018 )
.... :)
Mình ko ghi đề nữa nha!
a) \(=\left(5+3+1\right)xy^2\)
\(=9xy^2\)
b)\(=\left[\dfrac{1}{4}+\dfrac{2}{3}+\dfrac{-1}{2}\right]xyz\)
\(=\dfrac{5}{12}xyz\)
c)\(=\left[\dfrac{-1}{2}+\dfrac{5}{8}\right]yz^3\)
\(=\dfrac{1}{8}yz^3\)
d)\(=\left(-3-0,5+2,5\right)x^2\)
\(=-1x^2\)
\(=-x^2\)
1 kick đúng nhé
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