\(\dfrac{1}{2}x=\dfrac{1}{2}+\dfrac{3}{5}\\ \)

\(\dfrac{1}{2}x=\dfrac{5}{10}+\dfrac{6}{10}\)

\(\dfrac{1}{2}x=\dfrac{11}{10}\\ x=\dfrac{22}{10}\)

\(x=\dfrac{11}{5}\)

\(\dfrac{1}{2}x-\dfrac{3}{5}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{11}{10}\)

\(\Rightarrow x=\dfrac{11}{5}\)

=> 0,5x = 1,1

=> x=2,2

d
c

Chào bạn, bạn có thể cho mình biết chổ bạn cần giúp đc hông? Mình sẽ làm thêm.

5. D

6. C

II

1. B

2. A

III

1. B => before

2. A => take

3. D => station

4. B => more serious

IV

1. use

2. improving

3. were talking

4. would go

5. will have

6. collect

7. going

8. disappointed

PTHH : 2Al     +     6HCl  --> 2AlCl3   +    3H2 ↑   (1)

n_AlCl3 = \(\dfrac{m}{M}=\dfrac{13,35}{27+35,5.3}=0.1\left(mol\right)\) 

Từ (1) => n_HCl   =   2n_H2  = 0.2 (mol)

=> m_HCl = n.M  =  0.2 x  36.5 = 7.3 (g)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)

Lời giải:

a. ĐKXĐ: $x>0; x\neq 4$

b.

\(M=\sqrt{x}.\left[\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right].\frac{x-4}{2\sqrt{x}}\)

\(=\frac{2\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.\frac{x-4}{2}=\frac{2\sqrt{x}}{x-4}.\frac{x-4}{2}=\sqrt{x}\)

c. Để $M>3\Leftrightarrow \sqrt{x}>3\Leftrightarrow x>9$

Kết hợp đkxđ suy ra $x>9$ thì $M>3$

\(d_{\dfrac{A}{H_2}}=32\\ M_A=32.2=64\left(\dfrac{g}{mol}\right)\)

\(m_S=\dfrac{50.64}{100}=32g\\ m_O=64-32=32g\\ n_S=\dfrac{32}{32}=1mol\\ n_O=\dfrac{32}{16}=2mol\\ CTHH:SO_2\\ \Rightarrow B\)

D

D

Câu 4: ĐKXĐ: x>=1/2

Ta có: \(2\left(x-\sqrt{2x^2+5x-3}\right)=1+x\left(\sqrt{2x-1}-2\sqrt{x+3}\right)\)

=>\(2x-2\sqrt{2x^2+5x-3}=1+x\cdot\sqrt{2x-1}-2x\cdot\sqrt{x+3}\)

=>\(2x-1-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}-x\cdot\sqrt{2x-1}+2x\cdot\sqrt{x+3}=0\)

=>\(2x-1-x\cdot\sqrt{2x-1}-2\cdot\sqrt{\left(2x-1\right)\left(x+3\right)}+2x\cdot\sqrt{x+3}=0\)

=>\(\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\cdot\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)

=>\(\left(\sqrt{2x-1}-\sqrt{4x+12}\right)\left(\sqrt{2x-1}-x\right)=0\)

TH1: \(\sqrt{2x-1}-\sqrt{4x+12}=0\)

=>\(\sqrt{2x-1}=\sqrt{4x+12}\)

=>4x+12=2x-1

=>2x=-13

=>\(x=-\frac{13}{2}\) (loại)

TH2: \(\sqrt{2x-1}-x=0\)

=>\(\sqrt{2x-1}=x\)

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

=>x-1=0

=>x=1(nhận)

5:

Th1: m=0

=>6x-27=0

=>x=27/6(loại)

TH2: m<>0

Δ=(6m-6)^2-4m(9m-27)

=36m^2-72m+36-36m^2+108m=36m+36

Để phương trình có hai nghiệm pb thì 36m+36>0

=>m>-1

x1+x2=x1x2

=>6(m-1)=9(m-3)

=>9m-27=6m-6

=>3m=21

=>m=7