Gọi thời gian ô tô đuổi kịp xe máy là x (h) (x>0)

Thời gian xe máy đi là: x+1 (h)

Quãng đường xe máy đi là: 36(x+1) (km)

Quãng đường ô tô đi là: 48x (km)

Theo bài ra ta có:

36(x+1)=48x

48x-36x=36

12x=36

x=3 (h)

............

\(x^2+4x-7=\left(x+4\right)\sqrt{x^2-7}\)     \(\left(ĐKXĐ:\left[{}\begin{matrix}x\ge\sqrt{7}\\x\le-\sqrt{7}\end{matrix}\right.\right)\)

\(\Leftrightarrow x^2-7-\left(x+4\right)\sqrt{x^2-7}+4x=0\)

Đặt \(\sqrt{x^2-7}=t\) \(\left(t\ge7\right)\)

\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)

\(\Delta=x^2+8x+16-16x=x^2-8x+16=\left(x-4\right)^2\)

\(\left[{}\begin{matrix}t=\dfrac{x+4-x+4}{2}=4\\t=\dfrac{x+4+x-4}{2}=x\end{matrix}\right.\)

Với t=4 thì

\(\sqrt{x^2-7}=4\)

\(x^2-7=16\)

\(x=\pm\sqrt{23}\)

Với t=x thì

\(\sqrt{x^2-7}=x\)

\(x^2-7=x^2\left(vn\right)\)

Vậy.........

\(8\times8\times8\times8:8\times8:8\times8:[\left(8+8\right):8]\)

\(\left(x+3\right)^{2022}+\left(\sqrt{y-2}-1\right)^{2023}=0\)    \(\left(ĐKXĐ: y\ge2\right)\)

Xét \(\left(x+3\right)^{2022}\ge0\forall x\)

\(\Rightarrow\left(\sqrt{y-2}-1\right)^{2023}\le0\)

\(\Leftrightarrow\sqrt{y-2}-1\le0\)

\(\Leftrightarrow\sqrt{y-2}\le1\) 

\(\Leftrightarrow y-2\le1\)

\(\Rightarrow y\le3\)

\(\Rightarrow2\le y\le3\) mà \(y\in Z\)

\(\Rightarrow\left\{{}\begin{matrix}y=2\Leftrightarrow x=-2\\y=3\Leftrightarrow x=-3\end{matrix}\right.\)

\(a^2+b^2+c^2\ge ab+bc+ac\)

\(2a^2+2b^2+2c^2\ge2ab+2bc+2ac\)

\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(lđ\right)\)

\(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\)

\(a^2+b^2\ge\dfrac{a^2+2ab+b^2}{2}\)

\(2a^2+2b^2\ge a^2+2ab+b^2\)

\(a^2-2ab+b^2\ge0\)

\(\left(a-b\right)^2\ge0\left(lđ\right)\)

Em nghĩ Max là 17/4

Thầy thêm 4 nhưng ko triệt tiêu

\(\sqrt[3]{x+2}=\sqrt{x-2}\)  \(\left(DK:x\ge2\right)\)

\(\left(\sqrt[3]{x+2}\right)^6=\left(\sqrt{x-2}\right)^6\)

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

\(x^2+4x+4=x^3-6x^2+12x-8\)

\(x^3-7x^2+8x-12=0\)

\(\left(x-6\right)\left(x^2-x+2\right)=0\)

=> x=6 (tm)

\(\dfrac{tan^215^o-1}{cot75^o-1}-tan15^o\)

\(=\dfrac{cot^275^o-1}{cot75^o-1}-cot75^o\)

\(=\dfrac{\left(cot75^o+1\right)\left(cot75^o-1\right)}{cot75^o-1}-cot75^o\)

\(=cot75^o+1-cot75^o\)

=1

\(E=\dfrac{10}{9}\sqrt{x}+\sqrt{x}-1\)

     \(=\dfrac{19}{9}\sqrt{x}-1\)

Vì  \(\dfrac{19}{9}\sqrt{x}\ge0\)  với mọi x

=> \(E\ge0-1=-1\)

Dấu"=" xảy ra khi \(\dfrac{19}{9}\sqrt{x}=0\)

                              \(\sqrt{x}=0\)

                                x=0

Vậy \(Min_E=-1\)  tại x=0