Ta có \(M=\dfrac{\sqrt{a}+2}{\sqrt{a}-2}=1+\dfrac{4}{\sqrt{a}-2}\)

Để \(M\) nhận giá trị nguyên thì \(4⋮\left(\sqrt{a}-2\right)\Rightarrow\left(\sqrt{a}-2\right)\inƯ\left(4\right)\)

\(\Rightarrow\sqrt{a}-2\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{3;1;4;0;6\right\}\)

\(\Rightarrow a\in\left\{\sqrt{3};1;2;0;\sqrt{6}\right\}\)

mà a là số hữu tỉ nên \(a\in\left\{1;2;0\right\}\)

report nha 

a. \(\left(x-1\right)^3=27\Leftrightarrow x-1=3\Leftrightarrow x=4\)

b. \(x^2+x=0\Leftrightarrow x\left(x+1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

c. \(\left(2x+1\right)^2=25\) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

d. \(\left(2x-3\right)^2=36\) \(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

e. \(5^{x+2}=625\) \(\Leftrightarrow5^{x+2}=5^4\) \(\Rightarrow x+2=4\Rightarrow x=2\)

\(E=\sqrt{\left(2-\sqrt{3}\right)^2}-\dfrac{1}{2}\sqrt{12}\)

\(=2-\sqrt{3}-\sqrt{3}=2-2\sqrt{3}\)

Câu 1:

f. \(-\dfrac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)

\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)

\(=6x^2yz^2-5x^2yz\)

Câu 3: có bạn làm abc r nên mình làm d nha

d. \(2x\left(3x-1\right)+\left(4-2x\right)3x=7\)

\(\Leftrightarrow6x^2-2x+12x-6x^2=7\)

\(\Leftrightarrow10x=7\)

\(\Leftrightarrow x=\dfrac{7}{10}\)

c. \(\left|\dfrac{8}{4}-\left|x-\dfrac{1}{4}\right|\right|-\dfrac{1}{2}=\dfrac{3}{4}\)

\(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{8}{4}-x+\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{8}{4}+x-\dfrac{1}{4}\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left|\dfrac{9}{4}-x\right|-\dfrac{1}{2}=\dfrac{3}{4}\\\left|\dfrac{7}{4}+x\right|-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}\dfrac{9}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\\x=\dfrac{9}{4}-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\\\left[{}\begin{matrix}\dfrac{7}{4}+x-\dfrac{1}{2}=\dfrac{3}{4}\\-\dfrac{7}{4}-x-\dfrac{1}{2}=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\end{matrix}\right.\\\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=-3\end{matrix}\right.\) 

Ở nơi x=9/4-1/2 là x-9/4-1/2 nha

1. to look → looking

2. dislikes → likes

3. to not → not to

4. Right

5. to cook → cooking

6. take → taking

7. Right

8. to socialise → socialising

9. to watch → watching

10. Right