\(\lim\limits\left[\left(1-n\right)\left(\sqrt{n^2-6n}-\sqrt[3]{n^3-27n^2}\right)\right]\)
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a.
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BC\\BC\perp AB\left(gt\right)\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp CD\\AD\perp CD\left(gt\right)\end{matrix}\right.\) \(\Rightarrow CD\perp\left(SAD\right)\)
b.
\(\left\{{}\begin{matrix}SA\perp\left(ABCD\right)\Rightarrow SA\perp BD\\BD\perp AC\left(\text{ hai đường chéo hình vuông}\right)\end{matrix}\right.\)
\(\Rightarrow BD\perp\left(SAC\right)\)
Mà (SAC) đi qua trung điểm O của BD
\(\Rightarrow\left(SAC\right)\) là mp trung trực của BD
c.
Theo câu a ta có: \(\left\{{}\begin{matrix}BC\perp\left(SAB\right)\\AH\in\left(SAB\right)\end{matrix}\right.\) \(\Rightarrow BC\perp AH\)
Mà \(AH\perp SB\left(gt\right)\)
\(\Rightarrow AH\perp\left(SBC\right)\Rightarrow AH\perp SC\)
Lại có \(AI\perp SC\left(gt\right)\)
\(\Rightarrow SC\perp\left(AIH\right)\) (1)
Tương tự, ta chứng minh được \(AK\perp\left(SCD\right)\Rightarrow AK\perp SC\)
\(\Rightarrow SC\perp\left(AIK\right)\) (2)
(1);(2) \(\Rightarrow\left(AIH\right)\) trùng \(\left(AIK\right)\) hay 3 đường AH, AI, AK cùng nằm trong 1 mp
a.
\(2x^2+3x-5>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -\dfrac{5}{2}\end{matrix}\right.\)
b.
\(3x-7>0\Rightarrow x>\dfrac{7}{3}\)
c.
\(\dfrac{2x+1}{x+3}>0\Rightarrow\left[{}\begin{matrix}x>-\dfrac{1}{2}\\x< -3\end{matrix}\right.\)
\(4u_n=\dfrac{4\left(-9n^2+7n-2024\right)}{2n+1}=-18n+23+\dfrac{8119}{2n+1}\)
\(8119=23.353\) có 4 ước số dương nên dãy có 4 số hạng nguyên
20.
\(F=\dfrac{a^{\dfrac{2}{3}}\left(a^{-\dfrac{2}{3}}-a^{\dfrac{1}{3}}\right)}{a^{\dfrac{1}{8}}\left(a^{\dfrac{3}{8}}-a^{-\dfrac{1}{8}}\right)}=\dfrac{a^{\dfrac{2}{3}-\dfrac{2}{3}}-a^{\dfrac{2}{3}+\dfrac{1}{3}}}{a^{\dfrac{1}{8}+\dfrac{3}{8}}-a^{\dfrac{1}{8}-\dfrac{1}{8}}}=\dfrac{a^0-a}{a^{\dfrac{1}{2}}-a^0}=\dfrac{1-a}{\sqrt{a}-1}=-1-\sqrt{a}\)
\(\Rightarrow F=-1-2017^{1009}\)
21.
\(K=\sqrt[3]{\dfrac{2}{3}\sqrt[3]{\dfrac{2}{3}.\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}}}=\sqrt[3]{\dfrac{2}{3}\sqrt[3]{\dfrac{2}{3}^{\dfrac{3}{2}}}}=\sqrt[3]{\dfrac{2}{3}.\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}}\)
\(=\sqrt[3]{\left(\dfrac{2}{3}\right)^{\dfrac{3}{2}}}=\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}\)
22.
\(\sqrt[]{x\sqrt[]{x\sqrt[]{x.x^{\dfrac{1}{2}}}}}=\sqrt[]{x\sqrt[]{x\sqrt[]{x^{\dfrac{3}{2}}}}}=\sqrt[]{x\sqrt[]{x.x^{\dfrac{3}{4}}}}=\sqrt[]{x\sqrt[]{x^{\dfrac{7}{4}}}}=\sqrt[]{x.x^{\dfrac{7}{8}}}\)
\(=\sqrt[]{x^{\dfrac{15}{8}}}=x^{\dfrac{15}{16}}\)
\(x^{\dfrac{15}{16}}:x^{\dfrac{11}{16}}=x^{\dfrac{4}{16}}=x^{\dfrac{1}{4}}=\sqrt[4]{x}\)
23.
A đúng, do \(a^{\sqrt[]{3}}< a^{\sqrt[]{5}}\Rightarrow\dfrac{1}{a^{\sqrt[]{3}}}>\dfrac{1}{a^{\sqrt[]{5}}}\)
24.
A đúng
\(\left\{{}\begin{matrix}a+c=2b\\ab=c^2\end{matrix}\right.\) \(\Rightarrow2ab=2c^2\Rightarrow a\left(a+c\right)=2c^2\)
\(\Rightarrow2c^2-ac-a^2=0\Rightarrow2c^2-2ac+ac-a^2=0\)
\(\Rightarrow2c\left(c-a\right)+a\left(c-a\right)=0\)
\(\Rightarrow\left(c-a\right)\left(2c+a\right)=0\Rightarrow\left[{}\begin{matrix}c=a\left(loại\right)\\2c+a=0\end{matrix}\right.\)
\(\Rightarrow c=-\dfrac{a}{2}\) \(\Rightarrow a;c\) trái dấu, mà \(a< c\Rightarrow a< 0< c\)
\(b=\dfrac{c^2}{a}=\dfrac{c^2}{2c}=\dfrac{c}{2}\Rightarrow b\) cùng dấu c \(\Rightarrow b>0\)
Mà b nguyên nên b nhỏ nhất bằng 1, khi đó \(c=2b=2\)
Bài này biểu diễn ngược hơi mệt xíu, cộng trừ mấy lần mới ra:
Gọi O là tâm đáy thì \(\overrightarrow{SO}=\dfrac{1}{2}\overrightarrow{SA}+\dfrac{1}{2}\overrightarrow{SC}=\dfrac{1}{2}\overrightarrow{SB}+\dfrac{1}{2}\overrightarrow{SD}\) (1)
\(\Rightarrow\overrightarrow{SA}+\overrightarrow{SC}=\overrightarrow{SB}+\overrightarrow{SD}\) (2)
Bây giờ tìm cách đưa \(\overrightarrow{SA};\overrightarrow{SB};\overrightarrow{SC};\overrightarrow{SD}\) biểu diễn qua \(\overrightarrow{SM};\overrightarrow{SN};\overrightarrow{SG}\) là được
Với \(\overrightarrow{SB};\overrightarrow{SD}\) đơn giản: \(\overrightarrow{SB}+\overrightarrow{SD}=2\overrightarrow{SO}=3\overrightarrow{SG}\)
\(\overrightarrow{SA}=\overrightarrow{SM}+\overrightarrow{MA}=\overrightarrow{SM}+\overrightarrow{ON}=\overrightarrow{SM}+\overrightarrow{OS}+\overrightarrow{SN}=\overrightarrow{SM}-\dfrac{3}{2}\overrightarrow{SG}+\overrightarrow{SN}\)
Đặt \(\overrightarrow{SC}=x.\overrightarrow{SH}\)
Thế vào (2):
\(\Rightarrow\overrightarrow{SM}-\dfrac{3}{2}\overrightarrow{SG}+\overrightarrow{SN}+x.\overrightarrow{SH}=3\overrightarrow{SG}\)
\(\Rightarrow\overrightarrow{SM}=\dfrac{9}{2}\overrightarrow{SG}-\overrightarrow{SN}-x.\overrightarrow{SH}\)
Câu 22:
\(\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x\cdot\sqrt{x\cdot\sqrt{x\cdot x^{\dfrac{1}{2}}}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x\cdot\sqrt{x\cdot\sqrt{x^{\dfrac{3}{2}}}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x\cdot\sqrt{x\cdot x^{\dfrac{3}{4}}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x\cdot\sqrt{x^{\dfrac{5}{4}}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x\cdot x^{\dfrac{5}{8}}}:x^{\dfrac{11}{16}}\)
\(=\sqrt{x^{\dfrac{13}{8}}}:x^{\dfrac{11}{16}}=x^{\dfrac{13}{16}}:x^{\dfrac{11}{16}}=x^{\dfrac{1}{8}}\)
=>Chọn C
Câu 23:
\(\dfrac{1}{3}< \dfrac{1}{2}\)
=>\(a^{\dfrac{1}{3}}< a^{\dfrac{1}{2}}=\sqrt{a}\)(Vì a>1)
=>Loại C
\(a^{2018}< a^{2019}\)(do a>1)
=>\(\dfrac{1}{a^{2018}}>\dfrac{1}{a^{2019}}\)
=>Loại D
\(\dfrac{\sqrt[3]{a^2}}{a}=a^{\dfrac{2}{3}}:a=a^{-\dfrac{1}{3}}< 1\)
=>Loại B
=>Chọn A
Câu 21:
\(K=\sqrt[3]{\dfrac{2}{3}\cdot\sqrt[3]{\dfrac{2}{3}\cdot\sqrt{\dfrac{2}{3}}}}\)
\(=\sqrt[3]{\dfrac{2}{3}\cdot\sqrt[3]{\dfrac{2}{3}\cdot\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}}}\)
\(=\sqrt[3]{\dfrac{2}{3}\cdot\sqrt[3]{\left(\dfrac{2}{3}\right)^{\dfrac{3}{2}}}}\)
\(=\sqrt[3]{\dfrac{2}{3}\cdot\left(\dfrac{2}{3}\right)^{\dfrac{3}{2}\cdot\dfrac{1}{3}}}=\sqrt[3]{\dfrac{2}{3}\cdot\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}}\)
\(=\sqrt[3]{\left(\dfrac{2}{3}\right)^{\dfrac{3}{2}}}=\left(\dfrac{2}{3}\right)^{\dfrac{3}{2}\cdot\dfrac{1}{3}}=\left(\dfrac{2}{3}\right)^{\dfrac{1}{2}}=\sqrt{\dfrac{2}{3}}\)
=>Chọn B
Câu 20:
\(F=\dfrac{a^{\dfrac{2}{3}}\left(\sqrt[3]{a^{-2}}-\sqrt[3]{a}\right)}{a^{\dfrac{1}{8}}\cdot\left(\sqrt[8]{a^3}-\sqrt[8]{a^{-1}}\right)}\)
\(=\dfrac{a^{\dfrac{2}{3}}\cdot\left(a^{-\dfrac{2}{3}}-a^{\dfrac{1}{3}}\right)}{a^{\dfrac{1}{8}}\cdot\left(a^{\dfrac{3}{8}}-a^{-\dfrac{1}{8}}\right)}\)
\(=\dfrac{a^{\dfrac{2}{3}}\cdot a^{-\dfrac{2}{3}}-a^{\dfrac{2}{3}}\cdot a^{\dfrac{1}{3}}}{a^{\dfrac{1}{8}}\cdot a^{\dfrac{3}{8}}-a^{\dfrac{1}{8}}\cdot a^{-\dfrac{1}{8}}}\)
\(=\dfrac{1-a}{a^{\dfrac{1}{2}}-1}=\dfrac{1-a}{\sqrt{a}-1}=-\sqrt{a}-1\)
Thay \(a=2017^{2018}\) vào F, ta được:
\(F=-\sqrt{2017^{2018}}-1=-1-2017^{1009}\)
\(\left(1-n\right)\left(\dfrac{-6n}{\sqrt[2]{n^2-6n}+n}+\dfrac{27n^2}{n^2+n\sqrt[3]{n^3-27n^2}+\sqrt[3]{\left(n^3-27n^2\right)^2}}\right)\)
Ngoặc sau giới hạn hữu hạn tới \(\dfrac{27}{3}-\dfrac{6}{2}=6>0\), ngoặc trước tới âm vô cùng, nên giới hạn bằng âm vô cùng