Đặt \(log_2x=t\Rightarrow t\ge4\)

Phương trình trở thành: \(\sqrt{t^2-2t-3}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{\left(t+1\right)\left(t-3\right)}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{t+1}=m\sqrt{t-3}\)

\(\Leftrightarrow m=\sqrt{\dfrac{t+1}{t-3}}\)

Hàm \(f\left(t\right)=\sqrt{\dfrac{t+1}{t-3}}\) nghịch biến khi \(t\ge4\)

\(\lim\limits_{t\rightarrow+\infty}\sqrt{\dfrac{t+1}{t-3}}=1\) ; \(f\left(4\right)=\sqrt{5}\)

\(\Rightarrow1< f\left(t\right)\le\sqrt{5}\Rightarrow1< m\le\sqrt{5}\)

Đáp án D

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

PTHH: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

PTHH: K₂O + 2HCl → 2KCl + H₂O

Lần sau em đăng câu hỏi đúng box để mọi người hỗ trợ nha

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

PTHH: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

PTHH: K₂O + 2HCl → 2KCl + H₂O

1 knowing the answer, they told us

2 the unique handicraft, we didn't like it

3 We looked into these machines so that we could find out the method 

We looked into these machines so as to find out the method 

4 Jack's poverty, everyone looks up to him

5 In order to choose the best chair, Linda looked through this catalouge 

6 is not as dirty as this one

7 I didn't play as successfully as Linda

8 A dog can't run as fast as a horse

A dog runs more slowly than a horse

9 No rivers in the world is as long as the Nile

10 No one in my family talks as politely as Linda

Bài 1: 

a: \(\sqrt{0.49a^2}=-0.7a\)

b: \(\sqrt{25\left(a-7\right)^2}=5a-35\)

c: \(\sqrt{a^4\left(a-2\right)^2}=a^2\cdot\left(a-2\right)\)

d: \(\dfrac{1}{a-3b}\cdot\sqrt{a^6\left(a-3b\right)^2}\)

\(=\dfrac{1}{a-3b}\cdot a^3\cdot\left(a-3b\right)=a^3\)

Bài 2: 

a: \(2\left(x+y\right)\cdot\sqrt{\dfrac{1}{x^2+2xy+y^2}}\)

\(=2\left(x+y\right)\cdot\dfrac{1}{x+y}\)

=2

b: \(\dfrac{3x}{7y}\cdot\sqrt{\dfrac{49y^2}{9x^2}}\)

\(=\dfrac{3x}{7y}\cdot\dfrac{-7y}{3x}\)

=-1

\(1,\\ a,\sqrt{0,49a^2}\left(a< 0\right)=-0,7a\\ b,\sqrt{25\left(7-a\right)^2}\left(a\ge7\Leftrightarrow7-a\le0\right)=5\left(a-7\right)\\ c,\sqrt{a^4\left(a-2\right)^2}\left(a>0\right)=\left[{}\begin{matrix}a^2\left(a-2\right),\forall a\ge2\\a^2\left(2-a\right),\forall0< a< 2\end{matrix}\right.\\ d,\dfrac{1}{a-3b}\sqrt{a^6\left(a-3b\right)^2}\left(a>3b>0\right)=\dfrac{1}{a-3b}\cdot a^3\left(a-3b\right)=a^3\)

1 was arrested before the garden by the boy yesterday

2 his cats taken care of carefully by Jack everyday

3 Has the country been protected from Covid-19 successfully

4 is reported that the man is running out of money

is reported to be running out of money\

5 has been said that the thief got out of the prison

has been said to have got out of the prison

6 was thought that Mary had turned down the job

Mary was thought to have  turned down the job

\(1,\\ a,=\dfrac{\left(3+2\sqrt{3}\right)\sqrt{3}}{3}+\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{1}\\ =\dfrac{3\sqrt{3}+6}{3}+\sqrt{2}=\sqrt{3}+1+\sqrt{2}\\ b,=\left(\dfrac{\sqrt{5}+\sqrt{2}}{3}-\dfrac{\sqrt{5}-\sqrt{2}}{3}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}+3}{3}\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\\ =\dfrac{2\sqrt{2}+3}{3\left(3+2\sqrt{2}\right)}=\dfrac{1}{3}\)

\(2,\\ A=2x+\sqrt{\left(x-3\right)^2}=2x+\left|x-3\right|\\ =2\left(-5\right)+\left|-5-3\right|=-10+8=-2\\ B=\dfrac{\sqrt{\left(2x+1\right)^2}}{\left(x-4\right)\left(x+4\right)}\left(x-4\right)^2=\dfrac{\left|2x+1\right|\left(x-4\right)}{x+4}\\ B=\dfrac{17\cdot4}{12}=\dfrac{17}{3}\)