\(DK:x\in\left[-2;2\right]\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{4-x^2}\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2-x\right)+\sqrt{\left(2+x\right)\left(2-x\right)}\left(3x-1\right)=0\)

\(\Leftrightarrow\sqrt{2-x}\left[\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(1\right)\\\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\left(2\right)\end{cases}}\)

Xet PT(2)

\(\sqrt{2-x}+\sqrt{2+x}\left(3x-1\right)=0\)

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

\(\Leftrightarrow\left(2+x\right)\left(9x^2-6x+1\right)=2-x\)

\(\Leftrightarrow9x^3+12x^2-10x=0\)

\(\Leftrightarrow x\left(9x^2+12x-10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(n\right)\\9x^2+12x-10=0\end{cases}}\)

Xet PT 

\(9x^2+12x-10=0\)

Ta co:

\(\Delta^`=6^2-9.\left(-10\right)=126>0\)

\(\Rightarrow\hept{\begin{cases}x_1=\frac{-4+\sqrt{14}}{6}\left(n\right)\\x_2=\frac{-4-\sqrt{14}}{6}\left(n\right)\end{cases}}\)

Vay tap nghiem cua PT la \(S=\left\{\frac{-4-\sqrt{14}}{6};\frac{-4+\sqrt{14}}{6};0;2\right\}\)

\(\sqrt{x^2+7}\)có nghĩa

\(\Leftrightarrow x^2+7\ge0\)

Mà \(x^2\ge0\Rightarrow x^2+7\ge7>0\)

Vậy bt có nghĩa với mọi x

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 \(x\in\left[4;9\right]\)