\(n_{KOH}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\)(mol)

PTHH : 2KOH + H₂SO₄ ---> K₂SO₄ + 2H₂O

                2           : 1          :     1        :   2

         0.2mol           0.1mol

\(m_{H_2SO_4}=n.M=0,1.98=9,8\left(g\right)\)

=> \(m_{ddH_2SO_4}=\dfrac{m_{H_2SO_4}.100\%}{C\%}=\dfrac{9,8.100\%}{35\%}=28\left(g\right)\)

ĐKXĐ : x;y \(\ne0\)

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{2}{y}=-2\\\dfrac{3}{x}+\dfrac{2}{y}=7\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\\dfrac{1}{x}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9+\dfrac{1}{y}=-1\\x=\dfrac{1}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{10}\\x=\dfrac{1}{9}\end{matrix}\right.\)

a) \(\left\{{}\begin{matrix}x+y=3\\-2x+y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\3x=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2+y=3\\x=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

Vậy HPT 1 nghiệm (1;2)

b) \(\left\{{}\begin{matrix}2x+6y=11\\4x-9y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+12y=22\\4x-9y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-9y=1\\21y=21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-9.1=1\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2,5\\y=1\end{matrix}\right.\)

Vậy HPT 1 nghiệm (2,5 ; 1)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

Vậy HPT 1 nghiệm (2 ; 1)

11 He asked his daughter what kind of sport were you interested in  

12 I asked the neighboor what time your husband left for work 

13) You hate watching violent films , don't you ?

14) They have just repaired the street ìn front of my house,haven't they ?

15) The first TV picture wasn't produced by Baird in 1926,did he ?

16) This kind of shoes was bought last year,didn't she 

My mom asked me how your course took

The man asked me how many students in the class were there 

Giả sử d cắt trục tung tại A ; trục hoành tại B 

=> Tọa độ A(0;2) ; Tọa độ B(\(\left(\dfrac{-2}{m+3};0\right)\)

S_AOB = 4 

=> \(\dfrac{AO.OB}{2}=4\)

=> AO.OB = 8

<=> \(2.\dfrac{-2}{m+3}=8\)

<=> \(m=-\dfrac{7}{2}\)

1) He has bought his computer for one year

2) He last ate octopus 20 years ago

3) They have played tennis for 3 mohths

4) I have met ny husband since 1990

5) I've become head of the Stock Department  for 4 years

6) My mother started studying English 15 days ago 

c) Ta có \(\widehat{ECF}=\widehat{ECA}+\widehat{ACB}+\widehat{FCB}=2\widehat{ACB}=180^o\)

(Vì \(\widehat{ECA}=\widehat{ACH};\widehat{HCB}=\widehat{FCB}\))

=> E;C;F thẳng hàng 

mà EC = CF 

=> C trung điểm EF

mà I trung điểm AB

=> CI đường trung bình hình thang EABF

=> EA//CI//FB

=> \(\widehat{ECI}=90^{\text{o}}\)

=> EF tiếp tuyến (I) 

Hình tự vẽ

a) BF ; AE tiếp tuyến 

=> \(\widehat{BFE}=\widehat{EFB}=90^{\text{o}}\)

Ta có \(\widehat{BFE}+\widehat{EFB}=180^{\text{o}}\)

=> FB//AE 

b) Xét tam giác vuông ACE ; ACH 

AC² = AE² + CE² = AH² + HC² 

=> AE = AH (CE = HC)

Tương tự ta có FB = HB

lại có \(\widehat{ACB}=90^{\text{o}}\left(\text{thuộc (I) ; đường kính AB}\right)\)

Xét tam giác vuông ABC vuông tại C ; đường cao AH có

AH.AB = CH² = AE.FB 

Nam said that he had to finish his homework before going to out