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đ r

ĐR

i: ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}+1\\x\le-\sqrt{5}+1\end{matrix}\right.\)

l:ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le-5\end{matrix}\right.\)

m: ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le3\end{matrix}\right.\)

n: ĐKXĐ: \(x\in R\)

11:

i: \(=\dfrac{2x\left(3x-1\right)+1}{3x-1}=2x+\dfrac{1}{3x-1}\)

h: \(=\dfrac{1}{2}\cdot\dfrac{54x^3+2x^2-2x+2}{-2x+1}\)

\(=\dfrac{1}{2}\cdot\dfrac{54x^3-27x^2+29x^2-14.5x+12.5x-6.25+8.25}{-2x+1}\)

\(=\dfrac{-1}{2}\left(27x^2+14.5x+6.25\right)+\dfrac{-1}{2}\cdot\dfrac{8.25}{2x-1}\)

I.4. My family sometimes plays games together

5. Did you play badminton when you was small?

~HT~

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Câu g,h,i,k,l,m của bài 1 hay bài 2 ấy bạn?

g) \(\sqrt{7-4\sqrt{3}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}=\left|2-\sqrt{3}\right|+\left|\sqrt{3}-1\right|=2-\sqrt{3}+\sqrt{3}-1=1\)

h) \(\sqrt{11+6\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}=3-\sqrt{2}-\sqrt{2}+1=4\)

i) \(\sqrt{15+6\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(3+\sqrt{6}\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}=3+\sqrt{6}+3-\sqrt{6}=6\)

k) \(\sqrt{33+8\sqrt{2}}-\sqrt{33-8\sqrt{2}}=\sqrt{\left(4\sqrt{2}+1\right)^2}-\sqrt{\left(4\sqrt{2}-1\right)^2}=4\sqrt{2}+1-4\sqrt{2}+1=2\)

m: \(\Leftrightarrow\left\{{}\begin{matrix}-3x< =-4\\2x< 5\end{matrix}\right.\Leftrightarrow\dfrac{4}{3}< x< \dfrac{5}{2}\)