\(4sin^2x-2\left(\sqrt{3}+\sqrt{2}\right)sinx+\sqrt{6}=0\)

\(\Leftrightarrow4sin^2x-2\sqrt{3}sinx-2\sqrt{2}sinx+\sqrt{6}=0\)

\(\Leftrightarrow2sinx\left(2sinx-\sqrt{3}\right)-\sqrt{2}\left(2sinx-\sqrt{3}\right)=0\)

\(\Leftrightarrow\left(2sinx-\sqrt{3}\right)\left(2sinx-\sqrt{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=\dfrac{\sqrt{3}}{2}\\sinx=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\dfrac{2\pi}{3}+k2\pi\\x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

a: Ta có: \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

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

b: Ta có: \(\left(\dfrac{1}{2}\right)^x=\dfrac{1}{32}\)

nên x=5

1 was sung

2 was bitten

3 was stolen

4 Was - given

5 weren't made

6 was cooked

7 weren't sent

8 were told

9 is drunk

10 was destroyed

1, was sung

2,was bitten

3,was stolen

4,was -Given

5, Weren't made

6,was cooked

7,weren't sent

8, were told

9,was drunk

10, was destroyed

Xét ΔDBC có 

BA là đường trung tuyến

BA=DC/2

Do đó: ΔDCB vuông tại B

- Vậy bạn có hiểu bài của anh Nguyễn Lê Phước Thịnh không? Vì đây là kiến thức lớp 8.

Câu 4: 

a: Ta có: R(x)=0

\(\Leftrightarrow2x+3=0\)

hay \(x=-\dfrac{3}{2}\)

b: Ta có: H(x)=0

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)