a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=8^2+15^2=64+225=289=17^2\)

=>BC=17

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH=\frac{8\cdot15}{17}=\frac{120}{17}\) ≃7,1

Xét ΔABC vuông tại A có AH là đường cao

nên \(BH\cdot BC=BA^2\)

=>\(BH=\frac{8^2}{17}=\frac{64}{17}\) ≃3,8

b: Xét ΔABC vuông tại A có sin B\(=\frac{AC}{BC}=\frac{15}{17}\)

nên \(\hat{B}\) ≃62 độ

c: Xét ΔBAC có BD là phân giác

nên \(\frac{AD}{AB}=\frac{CD}{BC}\)

=>\(\frac{AD}{8}=\frac{CD}{17}\)

mà AD+CD=AC=15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{AD}{8}=\frac{CD}{17}=\frac{AD+CD}{8+17}=\frac{15}{25}=0,6\)

=>\(AD=0,6\cdot8=4,8\)

ΔABD vuông tại A

=>\(S_{ABD}=\frac12\cdot AB\cdot AD=\frac12\cdot4.8\cdot8=4\cdot4,8=19,2\)

Xét ΔBAD vuông tại A và ΔBED vuông tại E có

BD chung

\(\hat{ABD}=\hat{EBD}\)

Do đó: ΔBAD=ΔBED

=>\(S_{BED}=S_{BAD}=19,2\)

\(S_{ABED}=S_{BAD}+S_{BED}=19,2+19,2=38,4\)

1 Mark Twan was born in Missouri in 1835

2 The lesson is so difficult that nobody can understand it

3 A man who is called a tax collector collects taxes

1. She was so lazy that she was punished by her teacher.

2. Alice started playing tennis 2 years ago.

3. My uncle is a famous stamp collector.

4. Can I meet you at between 8 am and 9 am?

5. Can you wait for me for a while?

1. She was so lazy that she was punished by her teacher.

2. Alice started playing tennis 2 years ago.

3. My uncle is a famous stamp collector.

4. Can I meet you at between 8 am and 9 am?

5. Can you wait for me for a while?

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn nhé. Viết thế này khó đọc quá.

Lời giải:
$B=\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$=\frac{\sqrt{x}(\sqrt{x}-1)+(\sqrt{x}-4)}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=\frac{x-4}{(\sqrt{x}-1)(\sqrt{x}+1)}$

Do đó:

$A:B=\frac{\sqrt{x}-2}{\sqrt{x}+1}: \frac{x-4}{(\sqrt{x}-1)(\sqrt{x}+1)}$

$=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+2)}$

$=\frac{\sqrt{x}-1}{\sqrt{x}+2}$

Để $A:B< \frac{1}{2}\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+2}< \frac{1}{2}$

$\Leftrightarrow \frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{1}{2}<0$

$\Leftrightarrow \frac{\sqrt{x}-4}{2(\sqrt{x}+2)}<0$

$\Leftrightarrow \sqrt{x}-4<0$

$\Leftrightarrow 0\leq x< 16$

Kết hợp đkxđ suy ra $0\leq x< 16; x\neq 1$

Khi  $x=6-2\sqrt{5}=(\sqrt{5}-1)^2$ thì $\sqrt{x}=\sqrt{5}-1$
$A=\frac{\sqrt{5}-1-2}{\sqrt{5}-1+1}=\frac{5-3\sqrt{5}}{5}$

1 She could play table tennis if my friend asked 

2 Tam doesn't take part in the football match training

3 I must help my parents to work on the farm in the afternoon

4 Students should design and stick to a timetable

1 She could play table tennis if my friend asked 

2 Tam doesn't take part in the football match training

3 I must help my parents to work on the farm in the afternoon

4 Students should design and stick to a timetable

I used to keep a diary to write down special events