4 C

5 D

6 D

7 C

8 C

9 A

10 B

11 B

12 A

13 D

14 A

15 A

16 A

17 B

18 A

19 B

20 C

21 C

22 B

23 B

1 A

2 B

3 B

4 B

5 D

6 D

Câu a chắc em ghi đề nhầm, H,B,E thẳng hàng nên làm sao nó là tam giác được.

HFB và HEC đồng dạng thì đúng

a.

Do BE, CF là các đường cao \(\Rightarrow\widehat{BFH}=\widehat{CEH}=90^0\)

Xét hai tam giác HFB và HEC có:

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^0\\\widehat{BHF}=\widehat{CHE}\left(\text{đối đỉnh}\right)\end{matrix}\right.\) 

\(\Rightarrow\Delta HFB\sim\Delta HEC\left(g.g\right)\)

b.

Do AD là đường cao ứng với BC và H thuộc AD \(\Rightarrow\widehat{BDH}=90^0\)

Xét hai tam giác BDH và BEC có:

\(\left\{{}\begin{matrix}\widehat{BDH}=\widehat{BEC}=90^0\\\widehat{DBH}-chung\end{matrix}\right.\)

\(\Rightarrow\Delta BDH\sim\Delta BEC\left(g.g\right)\)

\(\Rightarrow\dfrac{BH}{BC}=\dfrac{BD}{BE}\Rightarrow BH.BE-BD.BC\)

a.

\(P=\left(\dfrac{x-1}{x+1}-\dfrac{x}{x-1}-\dfrac{3x+1}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)

\(=\left(\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\left(\dfrac{x^2-2x+1-\left(x^2+x\right)+3x+1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\)

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

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

b.

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

Do Q nguyên tố nên Q cũng là số tự nhiên \(\Rightarrow\dfrac{10}{2x+1}\) là số tự nhiên

\(\Rightarrow2x+1=Ư\left(10\right)\)

Mà 2x+1 luôn lẻ nên ta chỉ cần xét các ước tự nhiên lẻ của 10

\(\Rightarrow2x+1=\left\{1;5\right\}\)

\(\Rightarrow x=\left\{0;2\right\}\)

Với \(x=0\Rightarrow Q=10\) ko phải SNT (loại)

Với \(x=2\Rightarrow Q=2\) là SNT (thỏa mãn)

Vậy \(x=2\) thì Q là SNT

a.

Do BE, CF là các đường cao \(\Rightarrow\widehat{BFH}=\widehat{CEH}=90^0\)

Xét hai tam giác BHF và CHE có:

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^0\\\widehat{BHF}=\widehat{CHE}\left(\text{đối đỉnh}\right)\end{matrix}\right.\) 

\(\Rightarrow\Delta BHF\sim\Delta CHE\left(g.g\right)\)

\(\Rightarrow\dfrac{HB}{HC}=\dfrac{HF}{HE}\Rightarrow HE.HB=HC.HF\)

b.

Xét hai tam giác BAE và CAF có:

\(\left\{{}\begin{matrix}\widehat{A}-chung\\\widehat{BEA}=\widehat{CFA}=90^0\end{matrix}\right.\)

\(\Rightarrow\Delta BAE\sim\Delta CAF\left(g.g\right)\)

\(\Rightarrow\dfrac{AE}{AF}=\dfrac{AB}{AC}\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét hai tam giác AEF và ABC có:

\(\left\{{}\begin{matrix}\widehat{A}-chung\\\dfrac{AE}{AB}=\dfrac{AF}{AC}\left(cmt\right)\end{matrix}\right.\)

\(\Rightarrow\Delta AEF\sim\Delta ABC\left(c.g.c\right)\)

a) Xét \(\Delta BHF\) và \(\Delta CHE\) có:

\(\left\{{}\begin{matrix}\widehat{BFH}=\widehat{CEH}=90^{\circ}\left(CF\bot AB;BE\bot AC;BE\cap CF=\left\{H\right\}\right)\\\widehat{BHF}=\widehat{CHE}\left(\text{hai góc đối đỉnh}\right)\end{matrix}\right.\)

\(\Rightarrow \Delta BHF\backsim \Delta CHE\) \(\left(g.g\right)\Rightarrow\dfrac{HB}{HC}=\dfrac{HF}{HE}\) (các cạnh tương ứng)

\(\Rightarrow HE\cdot HB=HC\cdot HF\) (đpcm)

b) Xét \(\Delta ABE\) và \(\Delta ACF\) có: \(\left\{{}\begin{matrix}\widehat{AEB}=\widehat{AFC}=90^{\circ}\left(BE\bot AC;CF\bot AB\right)\\\widehat{BAC}\text{ chung}\end{matrix}\right.\)

\(\Rightarrow \Delta ABE \backsim \Delta ACF\) \(\left(g.g\right)\Rightarrow\dfrac{AB}{AC}=\dfrac{AE}{AF}\) (các cạnh tương ứng)

\(\Rightarrow\dfrac{AE}{AB}=\dfrac{AF}{AC}\)

Xét \(\Delta AEF\) và \(\Delta ABC\) có: \(\left\{{}\begin{matrix}\dfrac{AE}{AB}=\dfrac{AF}{AC}\left(cmt\right)\\\widehat{BAC}\text{ chung}\end{matrix}\right.\)

\(\Rightarrow \Delta AEF \backsim \Delta ABC\) \(\left(c.g.c\right)\) (đpcm)

\(\text{#}Toru\)

\(\dfrac{x+1}{99}+\dfrac{x+4}{96}+\dfrac{x+7}{93}+\dfrac{x+9}{91}+4=0\) (sửa đề)

\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+7}{93}+1\right)+\left(\dfrac{x+9}{91}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{96}+\dfrac{x+100}{93}+\dfrac{x+100}{91}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{96}+\dfrac{1}{93}+\dfrac{1}{91}\right)=0\)

\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{96}+\dfrac{1}{93}+\dfrac{1}{91}>0\))

\(\Leftrightarrow x=-100\)

A

I

1 B

2 D

3 A

4 C

5 D

6 B

7 A

8 C

II

1 C

2 C

3 A

4 B

5 D

6 A

7 B

8 D

B

I

1 B

2 C

3 C

4 B

5 D

Thiếu đề b ơi

đề là gì ạ

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1 She no longer writes to me as often as she used to

2 I wrote this software program on my own

3 The exam had high standard. Consequently, many students fail the exams

4 Crops in the area can't grow well as there is little rain during the dry season

5 Not everyone was aware of the severity of environmental issues

6 Air pollution leads to many respiratory illnesses

7 The polluted environment causes birds to leave their habitats

8 Aquatic life suffers or dies due to thermal pollution