Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b: \(=\left(12\sqrt[3]{2}+2\sqrt[3]{2}-2\sqrt[3]{2}\right)\cdot\left(5\sqrt[3]{4}-3\sqrt[3]{\dfrac{1}{2}}\right)\)
\(=12\sqrt[3]{2}\cdot5\sqrt[3]{4}-12\sqrt[3]{2}\cdot3\sqrt[3]{\dfrac{1}{2}}\)
\(=12\cdot5\cdot2-12\cdot3=120-36=84\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{a-1-\left(a-4\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}:\frac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\left(\sqrt{a}-1\right)\sqrt{a}}.\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
\(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right).\)
\(A=\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(A=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)\)
\(A=\frac{\sqrt{a}-2}{\sqrt{a}}\)
\(a,A=\frac{1-\sqrt{a^3}}{a-1}=-\frac{\sqrt{a^3}-1}{a-1}.\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{a+\sqrt{a}+1}{\sqrt{a}+1}\)
\(b,B=3\sqrt{\frac{12\left(a-2\right)^2}{27}}=\sqrt{9}.\sqrt{\frac{12\left(a-2\right)^2}{27}}\)
\(=\sqrt{\frac{9.3.4.\left(x-2\right)^2}{27}}=2\sqrt{\left(x-2\right)^2}=2.|x-2|\)
\(c,C=\left(a-b\right)\sqrt{\frac{ab}{\left(a-b\right)^2}}=\sqrt{\frac{\left(a-b\right)^2ab}{\left(a-b\right)^2}}=\sqrt{ab}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}\right)^2-4\left(a+\frac{1}{a}\right)^2+12}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}\right)^2-4\left(a^2+2a.\frac{1}{a}+\frac{1}{a^2}\right)+12}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}\right)^2-4\left(a^2+\frac{1}{a^2}+2\right)+12}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}\right)^2-4\left(a^2+\frac{1}{a^2}\right)-8+12}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}\right)^2-4\left(a^2+\frac{1}{a^2}\right)+4}\)
\(A=\sqrt{\left(a^2+\frac{1}{a^2}-2\right)^2}\)
\(A=\left|a^2+\frac{1}{a^2}-2\right|\)
Ta có \(a^2>0\)nên \(\frac{1}{a^2}>0\)(không có dấu bằng xảy ra vì \(a^2\)nằm dưới mẫu)
Áp dụng BĐT Cô-si cho 2 số dương \(a^2\)và \(\frac{1}{a^2}\), ta có:
\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2.\frac{1}{a^2}}=2\)\(\Leftrightarrow a^2+\frac{1}{a^2}-2\ge0\)
Chính vì vậy \(A=a^2+\frac{1}{a^2}-2\)