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Bài 3:
Vì x,y,z tỉ lệ với 2;3;4 nên x/2=y/3=z/4
Đặt x/2=y/3=z/4=k
=>x=2k; y=3k; z=4k
\(M=\dfrac{5x+2y+z}{x+4y-3z}=\dfrac{10k+6k+4k}{2k+12k-12k}=10\)
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.
\(=(\frac{1}{16}.x^8.y^4z^2)(-8xyz^2).(-\frac{1}{2}x^4yz)\)
\(=(\frac{1}{16}.-8.\frac{-1}{2})(x^8.x.x^4)(y^4.y.y)(z^2.z^2.z)\)
\(=\frac{1}{4}.x^{13}.y^6.z^5\)
a) x - y + z = 0
<=> (x - y + z)2 = 0
<=> (x - y + z).x - (x - y + z).y + (x - y + z).z = 0
<=> x2 - xy + xz - xy + y2 - zy + xz - zy + z2 = 0
=> x2 + y2 + z2 - 2xy + 2xz - 2zy = 0
=> x2 + y2 + z2 = 2xy - 2xz + 2zy = 2.(xy - xz + yz)
Vì \(x^2+y^2+z^2\ge0\) nên \(2.\left(xy-xz+yz\right)\ge0\)
\(\Leftrightarrow xy-xz+yz\ge0\left(đpcm\right)\)
b) ĐK: x ϵ N
\(8.2^n+2^{n+1}=8.2^n+2^n.2=2^n.\left(8+2\right)=2^n.10⋮10\)
a mik ko biết
b. 8.2^n +2^(n+1)
A= 8. 2^n + 2^n +2
=2^n(8+2)
=2^n.10
vậy A chia hết cho 10 (đpcm)