Câu 9:

a) Ta có: \(9x^2-16=0\)

\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)

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

b) Ta có: \(4x^2=13\)

\(\Leftrightarrow x^2=\dfrac{13}{4}\)
\(\Leftrightarrow x\in\left\{\dfrac{\sqrt{13}}{2};-\dfrac{\sqrt{13}}{2}\right\}\)

c) Ta có: \(2x^2+9=0\)

\(\Leftrightarrow2x^2=-9\)(Vô lý)

d) Ta có: \(-x^2+324=0\)

\(\Leftrightarrow x^2=324\)

\(\Leftrightarrow\left[{}\begin{matrix}x=18\\x=-18\end{matrix}\right.\)

Bài 11:

a: Ta có: \(P=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}\)

\(=\sqrt{x}\cdot\left(\sqrt{x}-1\right)\)

\(=x-\sqrt{x}\)

b: Để P=2 thì \(x-\sqrt{x}-2=0\)

hay x=4

Bài 10:

a: Ta có: \(A=\left(1+\dfrac{\sqrt{x}}{x+1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right)\)

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

\(=\dfrac{x+\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}-1}\)

b: Để A<0 thì \(\sqrt{x}-1< 0\)

hay x<1

Kết hợp ĐKXĐ, ta được: \(0\le x< 1\)

Để A=-1 thì \(x+\sqrt{x}+1=-\sqrt{x}+1\)

\(\Leftrightarrow x=0\)

c: Thay x=4 vào A, ta được:

\(A=\dfrac{4+2+1}{2-1}=7\)

Câu 10:

$\sin ^2x=0\Leftrightarrow \sin x=0$

$\Rightarrow x=k\pi$ với $k$ nguyên.

Trong các khoảng đã cho chỉ có khoảng ở đáp án A chứa $k\pi$ với $k$ nguyên.

Câu 11:

PT\(\Leftrightarrow 2\sin x\cos x-\sin x-2+4\cos x=0\)

\(\Leftrightarrow 2\cos x(\sin x+2)-(\sin x+2)=0\)

\(\Leftrightarrow (2\cos x-1)(\sin x+2)=0\)

Vì $\sin x\geq -1$ nên $\sin x+2\geq 1>0$

$\Rightarrow 2\cos x-1=0$

$\Leftrightarrow \cos x=\frac{1}{2}=\cos \frac{\pi}{3}$

$\Rightarrow x=\frac{\pi}{3}+2k\pi$ hoặc $x=-\frac{\pi}{3} +2k\pi$ với $k$ nguyên.

Đáp án B.

1. If you use compact light bulbs, you will save a lot of energy

2. English is spoken in the world by many people

1 If you use compact light bulbs, you will save a lot of energy

2 English is spoken in the world

a) Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

Chất oxh: Cl₂, chất khử: Cl₂

b) \(n_{Cl_2}=\dfrac{17,92}{22,4}=0,8\left(mol\right);n_{NaOH}=0,5.4=2\left(mol\right)\)

PTHH: Cl₂ + 2NaOH --> NaClO + NaCl + H₂O

_____0,8---->1,6--------->0,8---->0,8

=> \(\left\{{}\begin{matrix}C_{M\left(NaCl\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaClO\right)}=\dfrac{0,8}{0,5}=1,6M\\C_{M\left(NaOH\right)}=\dfrac{2-1,6}{0,5}=0,8M\end{matrix}\right.\)