\(AB=\sqrt{\left(4-2\right)^2+\left(1+1^2\right)+\left(3-1\right)^2}=2\sqrt3\)

\(R=\frac{AB}{2}=\frac{2\sqrt3}{2}=\sqrt3\)

Tâm \(I\) là trung điểm \(AB\) nên \(I\left(\frac{2+4}{2};\frac{-1+1}{2};\frac{1+3}{2}\right)=\left(3;0;2\right)\)

\(d\left(I;P\right)=\frac{\left\vert2.3+2.0+1.2+1\right\vert}{\sqrt{2^2+2^2+1^2}}=3\) không bằng \(R\)

Vậy \(\left(P\right)\) không tiếp xúc với mặt cầu đường kính \(AB\)

\(A:B=5x^3+x^2-3x-4+\frac{6}{2x-3}\)

Để \(A\) chia hết \(B\) khi \(2x-3\in U\left(6\right)=\left\lbrace\pm1;\pm2;\pm3;\pm6\right\rbrace\)

\(\rArr x\in\left\lbrace0;1;2;3\right\rbrace\) thỏa đề bài

3b) \(2c=8\left(m\right)\Rightarrow c=4\left(m\right)\)

\(2a=10\left(m\right)\Rightarrow a=5\left(m\right)\)

\(a^2=b^2+c^2\Rightarrow b=\sqrt{a^2-c^2}=\sqrt{25-16}=3\left(m\right)\)

\(\Rightarrow\left(E\right):\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\)

Điểm cách tâm đường hầm \(x=2\left(m\right)\)

\(\Rightarrow\dfrac{2^2}{25}+\dfrac{y^2}{9}=1\Rightarrow y^2=9\left(1-\dfrac{4}{25}\right)\Rightarrow h=y=\dfrac{3\sqrt{21}}{5}=\dfrac{a\sqrt{b}}{c}\)

\(\Rightarrow a=3;b=21;c=5\)

\(T=a+b+2c=3+21+2.5=34\)

Gọi \(a>0\left(m\right)\) là chiều dài mảnh đất HCN

Chiều rộng mảnh đất HCN \(b=\dfrac{726}{a}\left(m\right)\)

Chiều dài mảnh đất HCN làm trang trại \(a-2.1,5=a-3\left(m\right)\)

Chiều rộng mảnh đất HCN làm trang trại \(\dfrac{726}{a}-2\left(m\right)\)

Diện tích trang trại \(S\left(x\right)=\left(a-3\right)\left(\dfrac{726}{a}-2\right)\left(m^2\right)\)

\(\Rightarrow S\left(x\right)=732-\left(2a+\dfrac{2178}{a}\right)\le732-2\sqrt{2a.\dfrac{2178}{a}}=600\left(m^2\right)\left(Bđt.Cauchy\right)\)

Dấu '=' xảy ra khi và chỉ khi \(2a=\dfrac{2178}{a}\Rightarrow a=33\left(m\right)\)

\(\Rightarrow b=\dfrac{726}{33}=22\left(m\right)\)

Vậy, để diện tích phần trang trại còn lại là lớn nhất, bác Minh nên chọn mảnh đất có kích thước \(33mx22m\). Diện tích phần trang trại còn lại lớn nhất là \(600\left(m^2\right)\)

Câu 1 :

a) \(n\left(\Omega\right)=C^3_{18}=\dfrac{18!}{15!.3!}=\dfrac{18.17.16}{3.2.1}=816\RightarrowĐúng\)

b) \(n\left(3Đ\right)=C^3_6=\dfrac{6!}{3!.3!}=\dfrac{6.5.4}{3.2.1}=20\)

\(P\left(3Đ\right)=\dfrac{n\left(3Đ\right)}{n\left(\Omega\right)}=\dfrac{20}{816}=\dfrac{5}{204}\ne\dfrac{1}{272}\Rightarrow Sai\)

c) \(n\left(3M\right)=C^1_5.C^1_6.C^1_7=5.6.7=210\)

\(P\left(3M\right)=\dfrac{n\left(3M\right)}{n\left(\Omega\right)}=\dfrac{210}{816}=\dfrac{35}{136}\RightarrowĐúng\)

d) \(n\left(\overline{D}\right)=C^3_5=\dfrac{5!}{2!.3!}=\dfrac{5.4}{2.1}=10\)

\(\Rightarrow n\left(D\right)=816-10=808\)

\(P\left(D\right)=\dfrac{n\left(D\right)}{n\left(\Omega\right)}=\dfrac{806}{816}=\dfrac{403}{408}\RightarrowĐúng\)

Câu 8 : \(a=\dfrac{1}{2x};b=-2x;n=5\)

Hệ số chứa \(x^k\) là \(C^k_5.\left(\dfrac{1}{2x}\right)^{5-k}.\left(-2\right)^k.x^k=C^k_5.\left(\dfrac{1}{2}\right)^{5-k}.\left(-2\right)^k.x^{2k-5}\)

\(2k-5=5\Rightarrow k=5\Rightarrow\) Hệ số chứa \(x^5\) là \(C^5_5.\left(\dfrac{1}{2}\right)^{5-5}.\left(-2\right)^5=-32\)

\(2k-5=-5\Rightarrow k=0\Rightarrow\) Hệ số chứa \(\dfrac{1}{x^5}=x^{-5}\) là \(C^0_5.\left(\dfrac{1}{2}\right)^{5-0}.\left(-2\right)^0=\dfrac{1}{32}\)

\(\Rightarrow\) Tích hệ số của \(x^5\&\dfrac{1}{x^5}\) lầ \(-32.\dfrac{1}{32}=-1\Rightarrow Chọn.B\)

3) Sửa lại đề bài \(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\)

Áp dụng Bất đẳng thức Cauchy-Schwarz :

\(\left(\dfrac{x}{\sqrt{a}}.\sqrt{a}+\dfrac{y}{\sqrt{b}}.\sqrt{b}+\dfrac{z}{\sqrt{c}}.\sqrt{c}\right)^2\le\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right).\left(a+b+c\right)\)

\(\Rightarrow\left(x+y+z\right)^2\le\left(\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\right).\left(a+b+c\right)\)

\(\Rightarrow\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}\ge\dfrac{\left(x+y+z\right)^2}{a+b+c}\left(đpcm\right)\)

Dấu "=" xảy ra khi và chỉ khi \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

2) \(A=5x^2+2y^2+4xy-2x+4y+2015\)

\(\Rightarrow A=\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+2010\)

\(\Rightarrow A=\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+2010\ge2010\)

Dấu "=' xảy ra khi và chỉ khi

\(\Leftrightarrow\left\{{}\begin{matrix}2x+y=0\\x-1=0\\y+2=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Vậy \(A\left(min\right)=2010\) tại \(\left(x;y\right)=\left(1;-2\right)\)

1) \(x^2+xy-2022x-2023y-2024=0\)

\(\Leftrightarrow y\left(x-2023\right)=-x^2+2022x+2024\)

\(\Leftrightarrow y=\dfrac{-x^2+2022x+2024}{x-2023}=\dfrac{\left(-x^2+2023x\right)-x+2023+1}{x-2023}\)

\(\Leftrightarrow y=\dfrac{-x\left(x-2023\right)-\left(x-2023\right)+1}{x-2023}=\dfrac{\left(x-2023\right)\left(-x-1\right)+1}{x-2023}\)

\(\Leftrightarrow y=-x-1+\dfrac{1}{x-2023}\in Z\)

\(\Leftrightarrow x-2023\in U\left(1\right)=\left\{-1;1\right\}\)

\(\Leftrightarrow x\in\left\{2022;2024\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(2022;-2024\right);\left(2024;-2024\right)\right\}\)

a) \(...P=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\)

\(Đkxđ\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+3\ne0\left(đúng\right)\\\sqrt{x}-2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\Rightarrow Sai\)

b) \(P=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(\Rightarrow P=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(\Rightarrow P=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\Rightarrow Sai\)

c) \(P+4=\sqrt{x}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}+4=\sqrt{x}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-4+4\sqrt{x}-8}{\sqrt{x}-2}=\dfrac{x-2\sqrt{x}}{\sqrt{x}-2}\)

\(\Leftrightarrow5\sqrt{x}-12=x-2\sqrt{x}\)

\(\Leftrightarrow x-7\sqrt{x}+12=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=9\end{matrix}\right.\) (thỏa) \(\Rightarrow\) Có \(2\) giá trị \(x\RightarrowĐúng\)

d) \(P< 1\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}< 1\)

\(\Leftrightarrow\dfrac{\sqrt{x}-4}{\sqrt{x}-2}-1< 0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-4-\sqrt{x}+2}{\sqrt{x}-2}< 0\)

\(\Leftrightarrow\dfrac{-2}{\sqrt{x}-2}< 0\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)

\(\Rightarrow\) Số chính phương của \(x\) là \(9;16;25...\Rightarrow x_{min}=9\RightarrowĐúng\)