\(A=\frac{a}{b}+\sqrt{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}=\frac{a}{b}+\frac{1}{2}.\sqrt{\frac{b}{c}}+\frac{1}{2}\sqrt{\frac{b}{c}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}+\frac{1}{3}\sqrt[3]{\frac{c}{a}}.\)

Áp dụng BĐT Cauchy cho 6 số dương

\(A\ge6\sqrt[6]{\frac{1}{2^2}.\frac{1}{3^3}.\frac{a}{b}.\sqrt{\frac{b}{c}}.\sqrt{\frac{b}{c}}.\sqrt[3]{\frac{c}{a}}.\sqrt[3]{\frac{c}{a}}.\sqrt[3]{\frac{c}{a}}}=6.\sqrt[6]{\frac{1}{108}}>6.\sqrt[6]{\frac{5^6}{12^6}}=\frac{6.5}{12}=\frac{5}{2}\)

x = -1.96

ko biết

• Ngô Hồ Quỳnh Hân ko biết thì trả lời lm j??

​mk cũng ko biết nha :))    > mk chỉ nhắc Ngô Hồ Quỳnh Hân thui :3

i am in class 6

\(A=1-\sqrt{1-6x+9x^2}+\left(3x-1\right)^2\)

\(A=1-\sqrt{\left(3x-1\right)^2}+\left(3x-1\right)^2\)

\(A=1-\left(3x-1\right)+\left(3x-1\right)^2\)

\(A=1-3x+1+9x^2-6x+1\)

\(A=9x^2-9x+3\)

\(A=\left(3x\right)^2-2.3x.\frac{9}{6}+\frac{81}{36}-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{3}{4}\ge0\forall x\)

Dấu = xảy ra khi:

\(3x-\frac{9}{6}=0\Leftrightarrow3x=\frac{9}{6}\Leftrightarrow x=0,5\)

Vậy A_min = -3/4 tại x = 0,5

A=1-\(\sqrt{\left(3x-1\right)^2}\)+(3x-1)^2

A=1-/3x-1/+(3x-1)^2

đặt t=/3x-1/ với t>=0

khi đó A=t^2-t+1

A=t^2-t+1/4+3/4

A=(t-1/2)^2+3/4

khi đó A>=3/4

dấu bằng xảy ra khi t=1/2 hay x=1/2

Chúc bạn học tốt!

Lời giải :

\(A=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)

\(A=\left(5-2\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(A=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(A=2\left(9-5\right)\)

\(A=8\)