1 How long do you live in HN

2 How often did you go to the movies

3 What did Mrs Robinson buy

4 When was your father in Nghe An

5 How do they go to school

6 Why didn't she go to class

7 Who left home at 7o'clock yesterday

8 What subject does he study in the high school

9 What is your favorite hobby

10 How far is it

cảm mơn nhìuu ạ:33

Trích mẫu thử

Sục khí $CO_2$ tới dư vào mẫu thử : 

- mẫu thử tạo kết tủa trắng là $Ba(OH)_2$
$Ba(OH)_2 + CO_2 \to BaCO_3 + H_2O$

- mẫu thử không hiện tượng gì là $NaCl$

a: \(\sqrt{9x^2-12x+4}=7\)

=>\(\sqrt{\left(3x-2\right)^2}=7\)

=>|3x-2|=7

=>\(\left[\begin{array}{l}3x-2=7\\ 3x-2=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=9\\ 3x=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac53\end{array}\right.\)

b: \(\sqrt{4x^2-4x+1}=4x+1\)

=>\(\sqrt{\left(2x-1\right)^2}=4x+1\)

=>|2x-1|=4x+1

=>\(\begin{cases}4x+1\ge0\\ \left.\left(4x+1\right)^2=\left(2x-1\right)^2\right.\end{cases}\Rightarrow\begin{cases}x\ge-\frac14\\ \left(4x+1+2x-1\right)\left(4x+1-2x+1\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac14\\ 6x\left(2x+2\right)=0\end{cases}\Rightarrow x=0\)

c: \(\sqrt{x^2-10x+25}=2x+11\)

=>\(\sqrt{\left(x-5\right)^2}=2x+11\)

=>|x-5|=2x+11

=>\(\begin{cases}2x+11\ge0\\ \left(2x+11\right)^2=\left(x-5\right)^2\end{cases}\Rightarrow\begin{cases}2x\ge-11\\ \left(2x+11+x-5\right)\left(2x+11-x+5\right)=0\end{cases}\)

=>\(\begin{cases}x\ge-\frac{11}{2}\\ \left(3x+6\right)\left(x+16\right)=0\end{cases}\Rightarrow x=-2\)

d: ĐKXĐ: x>=2 hoặc x<=-2

\(\sqrt{x^2-4}=3\)

=>\(x^2-4=3^2=9\)

=>\(x^2=13\)

=>\(\left[\begin{array}{l}x=\sqrt{13}\left(nhận\right)\\ x=-\sqrt{13}\left(nhận\right)\end{array}\right.\)

e: \(\sqrt{x^2+5}=x+2\)

=>\(\begin{cases}x+2\ge0\\ x^2+5=\left(x+2\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-2\\ x^2+4x+4=x^2+5\end{cases}\)

=>\(\begin{cases}x\ge-2\\ 4x=1\end{cases}\Rightarrow x=\frac14\)

f:ĐKXĐ:x>=3 hoặc x<=0

Ta có: \(\sqrt{x^2-3x}=x+1\)

=>\(\begin{cases}x+1\ge0\\ x^2-3x=\left(x+1\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-1\\ x^2-3x=x^2+2x+1\end{cases}\)

=>\(\begin{cases}x\ge-1\\ -5x=1\end{cases}\Rightarrow x=-\frac15\) (nhận)

g: ĐKXĐ: x>=-3

\(5\cdot\sqrt{4x+12}-2\cdot\sqrt{9x+27}=12\)

=>\(5\cdot2\sqrt{x+3}-2\cdot3\sqrt{x+3}=12\)

=>\(4\cdot\sqrt{x+3}=12\)

=>\(\sqrt{x+3}=3\)

=>x+3=9

=>x=6(nhận)

h: ĐKXĐ: x>=5

Ta có: \(27\cdot\sqrt{\frac{x-5}{9}}+3\cdot\sqrt{4x-20}=30\)

=>\(9\sqrt{x-5}+3\cdot2\cdot\sqrt{x-5}=30\)

=>\(15\cdot\sqrt{x-5}=30\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

i: ĐKXĐ: x>=2

\(12\cdot\sqrt{\frac{x-2}{4}}-2\cdot\sqrt{4x-8}=18\)

=>\(12\cdot\frac{\sqrt{x-2}}{2}-2\cdot2\sqrt{x-2}=18\)

=>\(2\cdot\sqrt{x-2}=18\)

=>\(\sqrt{x-2}=9\)

=>x-2=81

=>x=83(nhận)

k: ĐKXĐ: x>=-1/2

Ta có: \(5\cdot\sqrt{8x+4}-3\cdot\sqrt{\frac{2x+1}{9}}=15+2\cdot\sqrt{18x+9}\)

=>\(5\cdot2\cdot\sqrt{2x+1}-3\cdot\frac{\sqrt{2x+1}}{3}-2\cdot3\sqrt{2x+1}=15\)

=>\(3\cdot\sqrt{2x+1}=15\)

=>\(\sqrt{2x+1}=5\)

=>2x+1=25

=>2x=24

=>x=12(nhận)

Bài 13:

$6-2\sqrt{5}=5-2\sqrt{5}.\sqrt{1}+1$

$=(\sqrt{5}-1)^2$

Tương tự: $6+2\sqrt{5}=(\sqrt{5}+1)^2$
Do đó:
$M=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}$

$=|\sqrt{5}+1|-|\sqrt{5}-1|=(\sqrt{5}+1)-(\sqrt{5}-1)$

$=2$

Bài 14:

a.

$M=\sqrt{4+2\sqrt{4}.\sqrt{5}+5}-\sqrt{4-2\sqrt{4}.\sqrt{5}+5}$

$=\sqrt{(\sqrt{4}+\sqrt{5})^2}-\sqrt{(\sqrt{4}-\sqrt{5})^2}$

$=|\sqrt{4}+\sqrt{5}|-|\sqrt{4}-\sqrt{5}|$

$=2+\sqrt{5}-(\sqrt{5}-2)=4$

b.

$N=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}$

$=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$

$=|\sqrt{7}-1|-|\sqrt{7}+1|$

$=(\sqrt{7}-1)-(\sqrt{7}+1)=-2$