Lấy \(A\left(2;3\right)\)) có điểm đối xứng qua trục \(Ox\) là \(A'\left(2;-3\right)\)

\(A'\left(2;-3\right);B\left(2;-4\right)\Rightarrow\left(A'B\right):x=2\) (Có cùng hoành độ \(x=2\))

\(\left(Ox\right):y=0\Rightarrow C=\left(A'B\right)\cap\left(Ox\right)=\left(2;0\right)\)

\(CA+CB=CA'+CB\ge A'B\left(CA=CA'\right)\)

\(\Rightarrow\left(CA+CB\right)_{max}=A'B\) khi \(C\left(2;0\right)\)

Chiều cao trung bình của trẻ = \(0,75+0,05.\left(t-1\right)\)  \(\left(m\right)\)

\(\Rightarrow\) Chiều cao trung bình của trẻ = \(0,75.100+0,05.100.\left(t-1\right)\) \(\left(cm\right)\)

\(=75+5\left(t-1\right)\left(cm\right)\Rightarrow Chọn.B\)

\(A=\dfrac{\sqrt{x}}{x+9}\left(x\ge0\right)\)

Áp dụng bất đẳng thức AM-GM :

\(\sqrt{9.x}\le\dfrac{x+9}{2}\)

\(\Rightarrow3\sqrt{x}\le\dfrac{x+9}{2}\)

\(\Rightarrow A=\dfrac{\sqrt{x}}{x+9}\le\dfrac{1}{6}\)

Dấu '=' xảy ra khi \(x=9\)

Vậy \(A\left(max\right)=\dfrac{\sqrt{x}}{x+9}=\dfrac{1}{6}\left(tại.x=9\right)\)

\(P\left(\overline{A}\cup B\right)=1-P\left(A\cap\overline{B}\right)\)

\(\Rightarrow P\left(A\cap\overline{B}\right)=1-P\left(\overline{A}\cup B\right)=1-0,5=0,5\)

\(P\left(A\right)=P\left(A\cap B\right)+P\left(A\cap\overline{B}\right)\)

\(\Rightarrow P\left(A\cap B\right)=P\left(A\right)-P\left(A\cap\overline{B}\right)=0,7-0,5=0,2\) (Chọn \(0,2\))

\(P\left(\overline{B}\right)=1-P\left(B\right)=1-0,3=0,7\)

\(P\left(A|\overline{B}\right)=\dfrac{P\left(A\cap\overline{B}\right)}{P\left(\overline{B}\right)}=\dfrac{0,2}{0,7}=\dfrac{2}{7}\) (Chọn \(\dfrac{2}{7}\))

\(y'=\dfrac{2x^2-4mx+4m^2-2m-4}{4\left(x-m\right)^2}=\dfrac{x^2-2mx+2m^2-m-2}{2\left(x-m\right)^2}\)

\(y'=0\Leftrightarrow x^2-2mx+2m^2-m-2=0\left(1\right)\)

Để hàm số có cực đại và cực tiểu, phương trình \(\left(1\right)\) phải có hai nghiệm phân biệt

\(\Leftrightarrow\Delta'=m^2-2m^2+m+2>0\)

\(\Leftrightarrow m^2-m-2< 0\)

\(\Leftrightarrow-1< m< 2\)

\(\Rightarrow Chọn.B\)

Chọn hệ trục tọa độ \(Oxy\) sao cho: \(B\left(0;0\right)A\left(0;6\right);C\left(6;0\right);D\left(6;6\right)\)

\(I\) là trung điểm \(BC\Rightarrow I\left(\dfrac{0+6}{2};\dfrac{0+0}{2}\right)=\left(3;0\right)\)

\(A\left(0;6\right)\in\left(P_1\right)\Rightarrow\left(P_1\right):y=ax^2+6\)

\(I\left(3;0\right)\in\left(P_1\right)\Rightarrow0=9a+6\Rightarrow a=-\dfrac{2}{3}\)

\(\Rightarrow\left(P_1\right):y=-\dfrac{2}{3}x^2+6\)

Tương tự \(\left(P_2\right):y=-\dfrac{2}{3}\left(x-6\right)^2+6\)

\(M\left(a;6\right)\left(0\le a\le6\right);N\left(6-a;6\right)\)

\(Q\left(a;-\dfrac{2}{3}a^2+6\right)\in\left(P_1\right);P\left(6-a;-\dfrac{2}{3}a^2+6\right)\in\left(P_2\right)\)

\(MN=6-2a;MQ=\dfrac{2}{3}a^2\)

\(S_{MNPQ}=S\left(a\right)=MN.MQ=\left(6-2a\right).\dfrac{2}{3}a^2=\dfrac{4}{3}\left(3a^2-a^3\right)\)

\(S'\left(a\right)=\dfrac{4}{3}\left(6a-3a^2\right)=4a\left(2-a\right)\)

\(S'\left(a\right)=0\Leftrightarrow a=0\cup a=2\)

Lập BTT ta thấy \(S\left(a\right)_{max}=S\left(2\right)=\dfrac{4}{3}\left(12-8\right)=\dfrac{16}{3}\approx5,3\left(đvdt\right)\)

Vậy \(S_{MNPQ}\left(max\right)=5,3\left(đvdt\right)\) thỏa đề bài

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow1+a\left(\dfrac{b+c}{bc}\right)=0\Rightarrow a\left(b+c\right)=-bc\left(1\right)\)

\(\left(a-b\right)\left(a-c\right)=a^2-a\left(b+c\right)+bc=a^2+bc+bc=a^2+2bc\left(do.\left(1\right)\right)\)

\(\Rightarrowđpcm\)

Tương tự ta cũng có :

\(\left(b-a\right)\left(b-c\right)=b^2+2ac\)

\(\left(c-b\right)\left(c-a\right)=c^2+2ab\)

\(\Rightarrow\dfrac{bc}{a^2+2bc}+\dfrac{ca}{b^2+2ac}+\dfrac{ab}{c^2+2ab}=\dfrac{bc}{\left(a-b\right)\left(a-c\right)}+\dfrac{ca}{\left(b-c\right)\left(b-a\right)}+\dfrac{ab}{\left(c-a\right)\left(c-b\right)}\)

\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left[-bc\left(b-c\right)-ca\left(a-c\right)-ab\left(a-b\right)\right]\) \(\)

\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\left[a\left(b^2-c^2\right)+b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\right]\) \(\left(do.\left(1\right)\right)\)

\(=\dfrac{1}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}.\left(a-b\right)\left(b-c\right)\left(c-a\right)=1\)

\(\Rightarrowđpcm\)

a) \(75\%-\dfrac{3}{4}:x^2=-2\dfrac{1}{4}\)

\(\Rightarrow\dfrac{3}{4}-\dfrac{3}{4}:x^2=-\dfrac{9}{4}\)

\(\Rightarrow\dfrac{3}{4}:x^2=\dfrac{3}{4}+\dfrac{9}{4}=3\)

\(\Rightarrow x^2=\dfrac{3}{4}:3=\dfrac{3}{4}.\dfrac{1}{3}=\dfrac{1}{4}=\left(\dfrac{1}{2}\right)^2\)

\(\Rightarrow x=-\dfrac{1}{2};x=\dfrac{1}{2}\) hay \(x\in\left\{-\dfrac{1}{2};\dfrac{1}{2}\right\}\)

b) \(8x^4+x=0\)

\(\Rightarrow x\left(8x^3+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\8x^3+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^3=-\dfrac{1}{8}=\left(-\dfrac{1}{2}\right)^3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{0;-\dfrac{1}{2}\right\}\)

\(A=\dfrac{3x}{x+3}=\dfrac{3x+9-9}{x+3}=\dfrac{3\left(x+3\right)-9}{x+3}=3-\dfrac{9}{x+3}\)

Ta có \(x\in N\Rightarrow x+3\ge3\)

\(\Rightarrow\dfrac{9}{x+3}\le3\)

\(\Rightarrow-\dfrac{9}{x+3}\ge-3\)

\(\Rightarrow A=3-\dfrac{9}{x+3}\ge3-3=0\)

Vậy \(A\left(min\right)=0\left(tại.x=0\right)\)

\(h=35768\left(km\right);R=6400\left(km\right)\)

Khoảng cách từ vệ tinh đến tâm Trái Đất là:

\(d=R+h=42168\left(km\right)\)

Góc tạo bởi bán kính trái đất và đường thẳng nối vệ tinh với tiếp điểm \(A\left(B\right)\) trên mặt đất:

\(sin\alpha=\dfrac{R}{d}=\dfrac{6400}{42168}\approx0,152\)

\(\Rightarrow\alpha=8,73^o\)

Nửa độ dài \(\stackrel\frown{AB}:R.\alpha=6400.0,152=972,8\left(km\right)\)

Độ dài cung \(AB\) là \(2.972,8=1945,6\left(km\right)\approx2000\left(km\right)\)