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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. \(a.\left(-15\right)^5:\left(-15\right)^3\) \(=\left(-15\right)^{5-3}\) \(=\left(-15\right)^2\) \(b.\left(\frac{5}{8}\right)^{10}:\left(\frac{5}{8}\right)^3\) \(=\left(\frac{5}{8}\right)^{10-3}\) \(=\left(\frac{5}{8}\right)^7\) \(c.x^{10}:x^7\) \(=x^{10-7}\) \(=x^3\) \(d.y^5:y^4\) \(=y^{5-4}\) \(=y^1=y\) \(e.\left(-x.y^2.z\right)^3:\left(-x.y^2.z\right)^3\) \(=\left(-x.y^2.z\right)^{3-3}\) \(=\left(-x.y^2.z\right)^0=1\) 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 \(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\) \(=\frac{x^4}{xy+2xz}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\) \(\ge\frac{\left(x^2+y^2+z^2\right)^2}{xy+2xz+yz+2xy+zx+2yz}\) \(\Leftrightarrow A\ge\frac{1}{3\left(xy+yz+zx\right)}\ge\frac{1}{3\left(x^2+y^2+z^2\right)}=\frac{1}{3}\)
