Cho mình hỏi bài này sử dụng bđt cauchy trực tiếp luôn có được không?

Đặt \(\frac{a}{b}=t\)

Ta có:\(t^2+\frac{1}{t^2}+4\ge3\left(t+\frac{1}{t}\right)\)

\(\Leftrightarrow t^2+\frac{1}{t^2}+4-3t-\frac{3}{t}\ge0\)

\(\Leftrightarrow\left(t^2-2t+1\right)+\left(\frac{1}{t^2}-\frac{3}{t}+1\right)+2-t-\frac{1}{t}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2+\left(\frac{1}{t}-1\right)^2+1-t-\frac{1}{t}+t\cdot\frac{1}{t}\ge0\)

\(\Leftrightarrow\left(t-1\right)^2+\left(\frac{1}{t}-1\right)^2+\left(t-1\right)\left(\frac{1}{t}-1\right)\ge0\)

Đặt \(\left(t-1;\frac{1}{t}-1\right)\rightarrow\left(p,q\right)\)

Ta có:

\(p^2+q^2+pq\ge0\)

\(\Leftrightarrow\left(p^2+pq+\frac{q^2}{4}\right)+\frac{3q^2}{4}\ge0\)

\(\Leftrightarrow\left(p+\frac{q}{2}\right)^2+\frac{3q^2}{4}\ge0\) *luôn đúng*

a+b=-1; a.b=\(\frac{-1}{4}\)  => a²+b²=(a+b)²-2ab=1+\(\frac{1}{2}=\frac{3}{2}\)

a⁷+b⁷=(a³+b³)(a⁴+b⁴)-a³b³(a+b)   (1)

a³+b³=(a+b)((a+b)²-ab))=-1(1+\(\frac{1}{4}\))=\(\frac{-5}{4}\)

a⁴+b⁴=(a²+b²)²-2a²b²=\(\left(\frac{3}{2}\right)^2-2.\left(\frac{-1}{4}\right)^2=\frac{17}{8}\)

Thay vào (1) P=\(\frac{-5}{4}.\frac{17}{8}-\left(\frac{-1}{4}\right)^3.\left(-1\right)=\frac{-171}{64}\)

mầy câu 1;3;;4;5 cách làm nhu nhau(nhân liên hop hoac bình phuong lên)

1.

\(DK:x\in\left[-4;5\right]\)

\(\Leftrightarrow\sqrt{x-5}+\left(\sqrt{x+4}-3\right)=0\)

\(\Leftrightarrow\sqrt{x-5}+\frac{x-5}{\sqrt{x+4}+3}=0\)

\(\Leftrightarrow\sqrt{x-5}\left(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}\right)=0\)

Vi \(1+\frac{\sqrt{x-5}}{\sqrt{x+4}+3}>0\)

\(\Rightarrow\sqrt{x-5}=0\)

\(x=5\left(n\right)\)

Vay nghiem cua PT la \(x=5\)

2.

\(DK:x\ge0\)

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

\(\Leftrightarrow|\sqrt{x}-2|+|\sqrt{x}-3|=1\)

Ta co:

\(|\sqrt{x}-2|+|\sqrt{x}-3|=|\sqrt{x}-2|+|3-\sqrt{x}|\ge|\sqrt{x}-2+3-\sqrt{x}|=1\)

Dau '=' xay ra khi \(\left(\sqrt{x}-2\right)\left(3-\sqrt{x}\right)\ge0\)

TH1:

\(\hept{\begin{cases}\sqrt{x}-2\ge0\\3-\sqrt{x}\ge0\end{cases}\Leftrightarrow4\le x\le9\left(n\right)}\)

TH2:(loai)

Vay nghiem cua PT la \(x\in\left[4;9\right]\)

Tham khảo:

\(\sqrt{3b\left(a+2b\right)}\le\frac{3b+\left(a+2b\right)}{2}\); \(\sqrt{3a\left(b+2a\right)}\le\frac{3a+\left(b+2a\right)}{2}\)

=> M\(\le a\frac{a+5b}{2}+b\frac{5a+b}{2}\)=\(\frac{a^2+b^2+10ab}{2}\)\(\le\frac{6\left(a^2+b^2\right)}{2}\)( áp dụng 2ab\(\le a^2+b^2\))=3(a²+b²)\(\le\)6

dấu = khi a =b =1