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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)
Câu 2:
\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)
\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(D=4\sqrt{\dfrac{1}{3}}+5\sqrt{12}-6\sqrt{27}\)
\(=\dfrac{4}{9}\sqrt{3}+5.2\sqrt{3}-6.3\sqrt{3}\)
\(=\dfrac{4}{9}\sqrt{3}+10\sqrt{3}-18\sqrt{3}\)
\(=-\dfrac{68}{9}\sqrt{3}\)
b) \(E=\dfrac{2}{\sqrt{3}-1}-\sqrt{4-2\sqrt{3}}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}\)
\(=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
c) \(F=\dfrac{\sqrt{15}-\sqrt{10}}{\sqrt{3}-\sqrt{2}}+\dfrac{3}{2-\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}+\dfrac{3\left(2+\sqrt{5}\right)}{-1}\)
\(=\sqrt{5}-6-3\sqrt{5}=-2\sqrt{5}-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi DH là khoảng cách thấp nhất từ máy bay đến mặt đất, khi đó AC có độ dài lớn nhất là 2,2m. Dựng hình chữ nhật DHEK => DH = EK
Do BA = BE = BC = 1,5m cố định nên tam giác ACE vuông tại A
Xét tam giác ACE vuông tại A có cos\(\widehat{ECA}\) = \(\dfrac{CA}{CE}=\dfrac{2,2}{3}\) => \(\widehat{ECA}\) \(\approx\) 42o50'
BA = BC => tam giác ABC cân tại B => \(\widehat{BAC}=\widehat{BCA}\) = \(\widehat{ECA}\) \(\approx\) 42o50'
=> \(\widehat{DBK}\) = \(\widehat{BAC}+\widehat{BCA}\) = 2.\(\widehat{BCA}\) = 85o40'
Xét tam giác DBK vuông tại D có: BK = BD. cos\(\widehat{DBK}\)
= 4.cos85o40' \(\approx\) 0,3022
=> DH = KE \(\approx\) 1,5 - 0,3022 \(\approx\)1,2 (m)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=-2\left(x^2-\dfrac{1}{2}x\right)=-2\left(x^2-2.x.\dfrac{1}{4}+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=-2\left(x^2-2x.\dfrac{1}{4}+\dfrac{1}{16}\right)+\dfrac{1}{8}=-2\left(x-\dfrac{1}{4}\right)^2+\dfrac{1}{8}\le\dfrac{1}{8}\)
\(\Rightarrow A_{max}=\dfrac{1}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a, \(\dfrac{2}{3}\) + \(\dfrac{1}{5}\). \(\dfrac{10}{7}\)
= \(\dfrac{2}{3}\) + \(\dfrac{2}{7}\)
= \(\dfrac{20}{21}\)
b, \(\dfrac{7}{12}\) - \(\dfrac{27}{7}\). \(\dfrac{1}{18}\)
= \(\dfrac{7}{12}\) - \(\dfrac{3}{14}\)
= \(\dfrac{31}{84}\)
c, \(\dfrac{3}{10}\). \(\dfrac{-5}{6}\) - \(\dfrac{1}{8}\)
= - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\)
= - \(\dfrac{3}{8}\)
d, - \(\dfrac{4}{9}\): \(\dfrac{8}{3}\) + \(\dfrac{1}{18}\)
= - \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\)
= - \(\dfrac{1}{9}\)
e, {[(\(\dfrac{1}{2}\) - \(\dfrac{2}{3}\))2 : 2 ] - 1}. \(\dfrac{4}{5}\)
= {[ (-\(\dfrac{1}{6}\))2 : 2] - 1}. \(\dfrac{4}{5}\)
= { [\(\dfrac{1}{36}\) : 2] - 1}. \(\dfrac{4}{5}\)
= { \(\dfrac{1}{72}\) - 1}. \(\dfrac{4}{5}\)
=- \(\dfrac{71}{72}\).\(\dfrac{4}{5}\)
= -\(\dfrac{71}{90}\)
![](https://rs.olm.vn/images/avt/0.png?1311)