1+1= Mấy? Khó quá mọi người giải thích giúp ạ!!!
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Lời giải:
\(\lim\limits_{x\to 2-}y=\lim\limits_{x\to 2-}\frac{\sqrt{4-x^2}}{(x-2)(x-3)}=\lim\limits_{x\to 2-}\frac{\sqrt{2+x}}{\sqrt{2-x}(x-3)}=-\infty \) nên $x=2$ là TCĐ
Vì \(x\in [-2;2)\) nên không tồn tại \(\lim\limits_{x\to +\infty }y\) nên đths không có TCN
Còn $x=3$ không thể là TCĐ vì tại $x=3$ thì $\sqrt{4-x^2}$ không tồn tại .
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a: =>x-4=0 hoặc x+5=0
=>x=4 hoặc x=-5
b: =>39/7:x=13
hay x=3/7
c: \(\Leftrightarrow\left(4.5-2x\right)=\dfrac{11}{4}:\dfrac{4}{9}=\dfrac{99}{16}\)
\(\Leftrightarrow2x=-\dfrac{27}{16}\)
hay x=-27/32
d: \(\Leftrightarrow x\cdot\dfrac{19}{15}=684\)
hay x=540
a. \(\left[{}\begin{matrix}x-4=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
b.\(\Leftrightarrow\dfrac{39}{7}:x=13\)
\(\Leftrightarrow x=13.\dfrac{39}{7}\)
\(\Leftrightarrow x=\dfrac{507}{7}\)
c.\(\Leftrightarrow4,5-2x=\dfrac{99}{16}\)
\(\Leftrightarrow-2x=\dfrac{27}{16}\)
\(\Leftrightarrow x=-\dfrac{27}{32}\)
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15.
\(\Delta'=m^2+m-2>0\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -2\end{matrix}\right.\)
Đáp án B
16.
\(\dfrac{\pi}{2}< a< \pi\Rightarrow\dfrac{\pi}{4}< \dfrac{a}{2}< \dfrac{\pi}{2}\Rightarrow\dfrac{\sqrt{2}}{2}< sin\dfrac{a}{2}< 1\Rightarrow\dfrac{1}{2}< sin^2\dfrac{a}{2}< 1\)
\(sina=\dfrac{3}{5}\Leftrightarrow sin^2a=\dfrac{9}{25}\Leftrightarrow4sin^2\dfrac{a}{2}.cos^2\dfrac{a}{2}=\dfrac{9}{25}\)
\(\Leftrightarrow sin^2\dfrac{a}{2}\left(1-sin^2\dfrac{a}{2}\right)=\dfrac{9}{100}\Leftrightarrow sin^4\dfrac{a}{2}-sin^2\dfrac{a}{2}+\dfrac{9}{100}=0\)
\(\Rightarrow\left[{}\begin{matrix}sin^2\dfrac{a}{2}=\dfrac{1}{10}< \dfrac{1}{2}\left(loại\right)\\sin^2\dfrac{a}{2}=\dfrac{9}{10}\end{matrix}\right.\)
\(\Rightarrow sin\dfrac{a}{2}=\dfrac{3\sqrt{10}}{10}\)
17.
Áp dụng công thức trung tuyến:
\(AM=\dfrac{\sqrt{2\left(AB^2+AC^2\right)-BC^2}}{2}=\dfrac{\sqrt{201}}{2}\)
18.
\(\Leftrightarrow x^2+2x+4>m^2+2m\) ; \(\forall x\in\left[-2;1\right]\)
\(\Leftrightarrow m^2+2m< \min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)\)
Xét \(f\left(x\right)=x^2+2x+4\) trên \(\left[-2;1\right]\)
\(-\dfrac{b}{2a}=-1\in\left[-2;1\right]\) ; \(f\left(-2\right)=4\) ; \(f\left(-1\right)=3\) ; \(f\left(1\right)=7\)
\(\Rightarrow\min\limits_{\left[-2;1\right]}\left(x^2+2x+4\right)=f\left(1\right)=3\)
\(\Rightarrow m^2+2m< 3\Leftrightarrow m^2+2m-3< 0\)
\(\Rightarrow-3< m< 1\Rightarrow m=\left\{-2;-1;0\right\}\)
Đáp án C
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45 If there were eggs in the fried, I'd make a cake for you
46 She's as beautiful as her friend
47 If your test score is high, your father will give you a reward
48 My son is taller than may daughter
49 If Nam had a camera, he'd take some pictures of his trip
50 Phong doens't have enough money, so he can't travel
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Bài `13`
\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =\sqrt{9\cdot3}+\sqrt{16\cdot3}-\sqrt{36\cdot3}-\sqrt{4\cdot3}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =\left(3+4-6-2\right)\sqrt{3}\\ =-\sqrt{3}\\ b,\left(\sqrt{28}+\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\\ =\left(\sqrt{4\cdot7}+\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{4\cdot21}\\ =\left(2\sqrt{7}+2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =2\cdot7+2\sqrt{21}-7+2\sqrt{21}\\ =14+2\sqrt{21}-7+2\sqrt{21}\\ =7+4\sqrt{21}\)
Bằng 2 nhé tớ search mạng rồi
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