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Câu 4

Gọi chiều rộng ban đầu của mảnh vườn là \(x\left(m\right)\), chiều dài ban đầu là \(y\left(m\right)\)

Ta có : \(xy=396\left(m^2\right)\Rightarrow y=\dfrac{396}{x}\left(m\right)\)

Diện tích phần trồng rau là:

\(S\left(rau\right)=\left(x-6\right)\left(y-3\right)\)

\(\Rightarrow S\left(x\right)=S\left(rau\right)=\left(x-6\right)\left(\dfrac{396}{x}-3\right)\)

\(\Rightarrow S\left(x\right)=396-3x-\dfrac{2376}{x}+18=-3x-\dfrac{2376}{x}+414\)

\(\Rightarrow S'\left(x\right)=-3-+\dfrac{2376}{x^2}\left(x\ne0\right)\)

\(S'\left(x\right)=0\Leftrightarrow x^2=792\Leftrightarrow x=28,14\left(m\right)\Rightarrow y=14,1\left(m\right)\)

\(\Rightarrow S_{max}\left(x=28,14;y=14,1\right)=\left(28,14-6\right)\left(14,1-3\right)\approx246\left(m^2\right)\)

Ta có : \(h\left(x\right)=1+2018\int\limits^x_0f\left(t\right)dt=f^2\left(x\right)\)

\(\Rightarrow h'\left(x\right)=2018f\left(x\right)=2f\left(x\right).f'\left(x\right)\)

\(\Rightarrow f'\left(x\right)=\dfrac{2018}{2}=1009\)

\(\Rightarrow f\left(x\right)=\int\limits f'\left(x\right)dx=1009x+C\)

\(\Rightarrow h\left(x\right)=\left(1009x+C\right)^2\)

mà \(h\left(0\right)=1\)

\(\Rightarrow C=1\)

\(\Rightarrow\int\limits^1_0f\left(x\right)dx=\int\limits^1_0\left(1009x+1\right)dx=\dfrac{1009}{2}x^2+x|^1_0=\dfrac{1009}{2}+1=\dfrac{1011}{2}\)

\(\Rightarrow\) Chọn D

\(\overrightarrow{AM}=\left(a+1;b-2\right)\)

\(\overrightarrow{BM}=\left(a+2;b\right)\)

\(\overrightarrow{CM}=\left(a;b-5\right)\)

\(\overrightarrow{AM}+2\overrightarrow{BM}+3\overrightarrow{CM}=\overrightarrow{0}\)

\(\Rightarrow\left\{{}\begin{matrix}a+1+2\left(a+2\right)+3a=0\\b-2+2b+3\left(b-5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}6a=-5\\6b=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-\dfrac{5}{6}\\b=\dfrac{17}{6}\end{matrix}\right.\)

\(\Rightarrow2a+6b=2.\left(-\dfrac{5}{6}\right)+6.\dfrac{17}{6}=-\dfrac{5}{3}+17=\dfrac{46}{3}\)

\(y=ax^2+bx+3\left(P\right)\Rightarrow c=3\)

\(\left(P\right)\) có trục đối xứng \(x=2\)

\(\Rightarrow\) Hoành độ điểm cực trị là \(x=2\)

\(\Rightarrow-\dfrac{b}{2a}=2\Leftrightarrow4a+b=0\left(1\right)\)

\(A\left(1;-2\right)\in\left(P\right)\Leftrightarrow a+b+3=-2\Leftrightarrow a+b=-5\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow3a=5\Leftrightarrow a=\dfrac{5}{3}\)

\(\left(2\right)\Rightarrow b=-5-\dfrac{5}{3}=-\dfrac{20}{3}\)

Vậy \(\left\{{}\begin{matrix}a=\dfrac{5}{3}\\b=-\dfrac{20}{3}\\c=3\end{matrix}\right.\) thỏa mãn đề bài

1) Khi \(x=9\Rightarrow B=\dfrac{\sqrt{9}-5}{\sqrt{9}+2}=\dfrac{3-5}{3+2}=-\dfrac{2}{5}\)

2) \(...\Rightarrow A=\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{x+4\sqrt{x}+4-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{\left(\sqrt{x}+2\right)^2-9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-5}\)

\(\Rightarrow P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-5}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

3) \(P>\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}>\dfrac{1}{3}\)

\(\Leftrightarrow3\sqrt{x}-3>\sqrt{x}+2\)

\(\Leftrightarrow2\sqrt{x}>5\)

\(\Leftrightarrow\sqrt{x}>\dfrac{5}{2}\)

\(\Leftrightarrow x>\dfrac{25}{4}\)

\(...\Rightarrow u_n=\dfrac{1}{2}.\dfrac{2n^2-3+\dfrac{5}{2}}{2n^2-3}=\dfrac{1}{2}+\dfrac{5}{2\left(2n^2-3\right)}\)

Ta thấy \(\dfrac{1}{2}+\dfrac{5}{2.\left(-3\right)}\le u_n\le\dfrac{1}{2}+0\left(n\ge0\right)\)

\(\Rightarrow-\dfrac{1}{3}\le u_n\le\dfrac{1}{2}\)

\(\Rightarrow u_n\) bị chặn trên bởi \(\dfrac{1}{2}\) và bị chặn dưới bởi \(-\dfrac{1}{3}\)

a) \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{JA}+2\overrightarrow{JB}+3\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow2\left(\overrightarrow{JA}+\overrightarrow{JB}\right)+3\left(\overrightarrow{JB}+\overrightarrow{JC}\right)=\overrightarrow{0}\)

\(\Leftrightarrow4\overrightarrow{JM}+6\overrightarrow{JN}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{JM}=-\dfrac{3}{2}\overrightarrow{JN}\)

\(\Rightarrow M;N:J\) thẳng hàng \(\Rightarrow\) Đúng

b) Sai (Chứng minh trên)

c) \(2\overrightarrow{IA}+3\overrightarrow{IC}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{IJ}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow5\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{JC}=\overrightarrow{0}\)

mà \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Rightarrow5\overrightarrow{IJ}-5\overrightarrow{JB}=\overrightarrow{0}\) (trừ vế với vế)

\(\Leftrightarrow\overrightarrow{IJ}=\overrightarrow{JB}\)

\(\Rightarrow J\) là trung điểm \(BI\Rightarrow\) Đúng

d) \(\overrightarrow{AE}=\dfrac{5}{7}\overrightarrow{AB}\)

\(\Leftrightarrow\overrightarrow{AJ}+\overrightarrow{JE}=\dfrac{5}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}\)

\(\Leftrightarrow\overrightarrow{JE}=-\dfrac{2}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}=\dfrac{2}{7}\overrightarrow{JA}+\dfrac{5}{7}\overrightarrow{JB}\left(1\right)\)

Ta lại có  \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{JC}=-\dfrac{2}{3}\overrightarrow{JA}-\dfrac{5}{3}\overrightarrow{JB}\left(2\right)\)

\(\left(1\right);\left(2\right)\) Ta thấy \(\dfrac{\dfrac{2}{7}}{-\dfrac{2}{3}}=\dfrac{\dfrac{5}{7}}{-\dfrac{5}{3}}=-\dfrac{3}{7}\)

\(\Rightarrow\overrightarrow{JE}=-\dfrac{3}{7}\overrightarrow{JC}\)

\(\Rightarrow C,E,J\) thẳng hàng \(\Rightarrow\) Đúng

\(\left|2x+5\right|=\left|3x+10\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+5=3x+10\\2x+5=-3x-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\5x=-15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-3\end{matrix}\right.\)

Vậy nghiệm của phương trình cho là \(x\in\left\{-3;-5\right\}\)

\(f\left(1\right)=\dfrac{3.1^2-5}{1+2}=-\dfrac{2}{3}\)

\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{3x^2-5}{x+2}=\dfrac{3.1^2-5}{1+2}=-\dfrac{2}{3}\)

\(\Rightarrow f\left(1\right)=\lim\limits_{x\rightarrow1}f\left(x\right)=-\dfrac{2}{3}\)

\(\Rightarrow\) Hàm số \(f\left(x\right)\) liên tục tại \(x=1\)