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\(a.\left(-15\right)^5:\left(-15\right)^3\)
\(=\left(-15\right)^{5-3}\)
\(=\left(-15\right)^2\)
\(b.\left(\frac{5}{8}\right)^{10}:\left(\frac{5}{8}\right)^3\)
\(=\left(\frac{5}{8}\right)^{10-3}\)
\(=\left(\frac{5}{8}\right)^7\)
\(c.x^{10}:x^7\)
\(=x^{10-7}\)
\(=x^3\)
\(d.y^5:y^4\)
\(=y^{5-4}\)
\(=y^1=y\)
\(e.\left(-x.y^2.z\right)^3:\left(-x.y^2.z\right)^3\)
\(=\left(-x.y^2.z\right)^{3-3}\)
\(=\left(-x.y^2.z\right)^0=1\)
a/\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{54}{2y}\)
\(\Rightarrow2y\cdot y=54\cdot3\Rightarrow2y^2=162\Rightarrow y^2=\dfrac{162}{2}=81\)
Mà y > 0 (gt) => \(y=\sqrt{81}=9\Rightarrow x=\dfrac{54}{9}=6\)
Vậy..............
b/ \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{4}\cdot25=\dfrac{25}{4}\\y^2=\dfrac{1}{4}\cdot9=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\sqrt{\dfrac{25}{4}}=\pm\dfrac{5}{2}\\y=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\end{matrix}\right.\)
Vậy.............
c/ x/2 = y/3 => x/10 = y/15
y/5 = z/7 => y/15 = z/21
=> x/10 = y/15 = z/21
Áp dụng t/c của dãy tỉ số = nhau là ra....
6) Ta có
\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+2xz+yz+2xy+zx+2yz}\)
\(\Leftrightarrow A\ge\frac{1}{3\left(xy+yz+zx\right)}\ge\frac{1}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)