Đk:\(x\ge1;x\le-2\)

Đặt \(t=\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}\)

\(\Rightarrow t^2=\left(x-1\right)\left(x+2\right)\)

Pttt: \(t^2+4t=12\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-6\end{matrix}\right.\)

TH1: \(t=2\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1>0\\\left(x-1\right)\left(x+2\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>1\\x^2+x-6=0\end{matrix}\right.\)\(\Rightarrow x=2\) (thỏa mãn)

TH2:\(t=-6\Rightarrow\left(x-1\right)\sqrt{\dfrac{x+2}{x-1}}=-6\)\(\Leftrightarrow\left\{{}\begin{matrix}x-1< 0\\\left(x-1\right)\left(x+2\right)=36\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x< 1\\x^2+x-38=0\end{matrix}\right.\)\(\Rightarrow x=\dfrac{-1-3\sqrt{17}}{2}\) (thỏa mãn)

Vậy...

Chỗ khoanh thứ nhất là nhân cả tử và mẫu với \(\sqrt{ab+2c^2}\)

Chỗ khoanh thứ hai: Áp dụng AM-GM có:

\(\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}\le\dfrac{a^2+b^2+ab+ab+c^2+c^2}{2}=\dfrac{a^2+b^2+2ab+2c^2}{2}\)

\(\Rightarrow\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)

\(x^2\left(x^2+2\right)=4-x\sqrt{2x^2+4}\)

Đặt \(t=x\sqrt{2x^2+4}\)

Pttt: \(\dfrac{t^2}{2}=4-t\)

\(\Leftrightarrow t^2+2t-8=0\) \(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-4\end{matrix}\right.\)

TH1: \(t=2\Rightarrow x\sqrt{2x^2+4}=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2\left(2x^2+4\right)=4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^4+2x^2-2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x^2=-1+\sqrt{3}\end{matrix}\right.\)(do \(x^2\ge0\)) \(\Rightarrow x=\sqrt{-1+\sqrt{3}}\)

TH2: \(t=-4\Rightarrow x\sqrt{2x^2+4}=-4\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^2\left(2x^2+4\right)=16\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\x^4+2x^2-8=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x^2=2\end{matrix}\right.\)(do \(x^2\ge0\))\(\Rightarrow x=-\sqrt{2}\)

Vậy...

\(\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}=\sqrt{\dfrac{ab+2c^2}{a^2+b^2+ab}}\)\(=\dfrac{ab+2c^2}{\sqrt{\left(a^2+b^2+ab\right)\left(ab+c^2+c^2\right)}}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{a^2+b^2+2ab+2c^2}\)\(\ge\dfrac{2\left(ab+2c^2\right)}{2\left(a^2+b^2\right)+2c^2}\)\(=\dfrac{ab+2c^2}{a^2+b^2+c^2}\)

\(\Rightarrow\sqrt{\dfrac{ab+2c^2}{1+ab-c^2}}\ge ab+2c^2\)

Tương tự: \(\sqrt{\dfrac{bc+2a^2}{1+bc-a^2}}\ge bc+2a^2\); \(\sqrt{\dfrac{ac+2b^2}{1+ac-b^2}}\ge ac+2b^2\)

Cộng vế với vế \(\Rightarrow VT\ge2a^2+2b^2+2c^2+ab+bc+ac=2+ab+bc+ac\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)

Tóm tắt:

\(v_0=0\)

\(a=0,5m/s\)²

a) Quãng đường xe đi được sau hai phút là:

\(S=v_0t+\dfrac{1}{2}at^2=\dfrac{1}{2}.0,5.60^2=900\left(m\right)\)

b) Vân tốc xe sau khi đi được 100m là:

\(v=\sqrt{2as+v_0^2}=\sqrt{2.0,5.100}=10\) (m/s)

c) \(v_1=36km/h=10m/s\)

Xe đạt 10m/s sau:

\(t=\dfrac{v_1-v_0}{a}=\dfrac{10}{0,5}=20\left(s\right)\)

Đỗ Thanh Hải bằng tuổi

a) Có \(\widehat{DAC}+\widehat{C}=140^0+40^0=180^0\)

mà hai góc nằm ở vị trí hai góc so le trong

\(\Rightarrow\)\(AD//CF\)

b) Có \(\widehat{DAB}=360^0-\widehat{BAC}-\widehat{DAC}=130^0\)

\(\Rightarrow\)\(\widehat{DAB}+\widehat{B}=130^0+50^0=180^0\)

 mà hai góc nằm ở vị trí hai góc so le trong

\(\Rightarrow\)\(AD//EB\)