Giúp mk với ạ tối trước 10h mk phải nộp r

a: Xét tứ giác OACD có \(\widehat{OAC}+\widehat{ODC}=90^0+90^0=180^0\)

nên OACD là tứ giác nội tiếp

c: Xét (O) có

CA,CD là các tiếp tuyến

Do đó: CA=CD

=>C nằm trên đường trung trực của AD(1)

Ta có: OA=OD

=>O nằm trên đường trung trực của AD(2)

Từ (1),(2) suy ra CO là đường trung trực của AD

=>CO\(\perp\)AD tại H

Xét (O) có

ΔKAB nội tiếp

AB là đường kính

Do đó: ΔKAB vuông tại K

=>AK\(\perp\)CB tại K

Xét ΔCAB vuông tại A có AK là đường cao

nên \(CK\cdot CB=CA^2\left(3\right)\)

Xét ΔCAO vuông tại A có AH là đường cao

nên \(CH\cdot CO=CA^2\left(4\right)\)

Từ (3),(4) suy ra \(CK\cdot CB=CH\cdot CO\)

=>\(\dfrac{CK}{CO}=\dfrac{CH}{CB}\)

Xét ΔCKH và ΔCOB có

\(\dfrac{CK}{CO}=\dfrac{CH}{CB}\)

\(\widehat{KCH}\) chung

Do đó: ΔCKH~ΔCOB

=>\(\widehat{CKH}=\widehat{COB}=180^0-\widehat{IOA}\)

Xét ΔOAI có OA=OI

nên ΔOAI cân tại O

=>\(\widehat{IAO}=\dfrac{180^0-\widehat{IOA}}{2}\)

=>\(\widehat{IAO}=\dfrac{\widehat{CKH}}{2}\)

=>\(\widehat{CKH}=2\cdot\widehat{IAO}\)

bk : bán kính

đk: đường kính

nt:nội tiếp

t^2 : tiếp tuyến

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\)

|P|>P

=>P<0

=>\(\dfrac{3\sqrt{x}}{\sqrt{x}-2}< 0\)

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

mà x nguyên

nên \(x\in\left\{0;1;2;3\right\}\)

1 She had her car repaired yesterday
2 You've got to see the manager tomorrow morning
3 Every time we rang, there was never any answer
4 Don't call him that insulting name

1 She had her car repaired yesterday

2 You've got to see the manager tomorrow morning

3 Every time we rang, there were no answers

4 Don't call him that insulting name

Tks so much 👍<3

b) \(B=\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-1}\left(x\ge0,x\ne1\right)\\ =\dfrac{1}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-1+\sqrt{x}\left(\sqrt{x}+1\right)+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ \)

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

c)

 \(AB\le8\Leftrightarrow\dfrac{4\sqrt{x}}{x-1}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\le8\\ \Leftrightarrow\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\le8\\ \Leftrightarrow\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\le8\\ \Leftrightarrow4\sqrt{x}\le8\left(x-2\sqrt{x}+1\right)\\ \) ( Nhân cả 2 vế BPT cho \(\left(\sqrt{x}-1\right)^2>0\) )

\(\Leftrightarrow8x-16\sqrt{x}+8\ge4\sqrt{x}\\ \Leftrightarrow8x-20\sqrt{x}+8\ge0\\ \Leftrightarrow2x-5\sqrt{x}+2\ge0\\ \)

\(\Leftrightarrow\left(2x-4\sqrt{x}\right)-\left(\sqrt{x}-2\right)\ge0\\ \Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)\ge0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)\ge0\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2\ge0\\2\sqrt{x}-1\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2\le0\\2\sqrt{x}-1\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}\ge2\\\sqrt{x}\ge\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}\le2\\\sqrt{x}\le\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\\ \)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge4\\x\ge\dfrac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le4\\x\le\dfrac{1}{4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge4\\x\le\dfrac{1}{4}\end{matrix}\right.\)

Kết hợp ĐK: \(x\ge0,x\ne1\)

Kết luận: \(x\ge4\) hoặc \(0\le x\le\dfrac{1}{4}\) thì \(AB\le8\)

Lời giải:

ĐK: $x\geq 0; x\neq 1$

$AB=\frac{4\sqrt{x}}{(\sqrt{x}-1)^2}\leq 8$

$\Rightarrow 4\sqrt{x}\leq 8(\sqrt{x}-1)^2$
$\Leftrightarrow \sqrt{x}\leq 2(\sqrt{x}-1)^2$
$\Leftrightarrow \sqrt{x}\leq 2(x-2\sqrt{x}+1)$

$\Leftrightarrow 2x-5\sqrt{x}+2\geq 0$

$\Leftrightarrow (\sqrt{x}-2)(2\sqrt{x}-1)\geq 0$

$\Leftrightarrow \sqrt{x}\geq 2$ hoặc $\sqrt{x}\leq \frac{1}{2}$

$\Leftrightarrow x\geq 4$ hoặc $0\leq x\leq \frac{1}{4}$

Kết hợp đkxđ suy ra $x\geq 4$ hoặc $0\leq x\leq \frac{1}{4}$

\(B=\dfrac{2\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\)

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

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

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

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

\(B=\dfrac{2x-3}{\sqrt{x}-1}+\dfrac{3-\sqrt{x}}{x-1}\)

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

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

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

\(=\dfrac{2x\sqrt{x}+2x-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+1}=\dfrac{2x+4\sqrt{x}}{\sqrt{x}+1}\)

Mình sửa đề nhé;-; Đề trước lỗi á