Xét tam giác \(ABC\)vuông tại \(A\):

\(BC^2=AB^2+AC^2\)(định lí Pythagore) 

\(=12^2+16^2=400\)

\(\Leftrightarrow BC=20\left(cm\right)\)

\(AB^2=BD.BC\Leftrightarrow BD=\frac{AB^2}{BC}=\frac{12^2}{20}=7,2\left(cm\right)\)

Xét tam giác \(ABC\)phân giác \(AD\): 

\(\frac{AB}{BD}=\frac{AC}{CD}\)(tính chất đường phân giác) 

\(=\frac{AB+AC}{BD+CD}=\frac{12+16}{20}=1,4\)

\(\Leftrightarrow BD=\frac{AB}{1,4}=\frac{12}{1,4}=\frac{60}{7}\left(cm\right)\)

\(HD=\left|BD-BH\right|=\frac{48}{35}\left(cm\right)\)

\(A=2+2\sqrt{28n^2+1}\)là số tự nhiên mà \(n\)là số tự nhiên nên \(\sqrt{28n^2+1}\)là số tự nhiên. 

Suy ra \(28n^2+1=k^2\)(với \(k\inℕ\))

\(\Leftrightarrow k^2-1=28n^2\)

Suy ra \(k\)lẻ nên \(k=2m+1\).

\(\left(2m+1\right)^2-1=28n^2\)

\(\Leftrightarrow m^2+m=7n^2\)

\(\Rightarrow\orbr{\begin{cases}m⋮7\\m+1⋮7\end{cases}}\)

- \(m=7p\)

\(p\left(7p+1\right)=n^2\)

mà \(\left(p,7p+1\right)=1\)nên \(\hept{\begin{cases}p=a^2\\7p+1=b^2\end{cases}}\)

\(A=2+2\sqrt{28n^2+1}=2+2k=2+4m+2=4+28p\)

\(=4\left(1+7p\right)=4b^2\)là một số chính phương. 

- \(m+1=7p\)

\(p\left(7p-1\right)=n^2\)

mà \(\left(p,7p-1\right)=1\)nên \(\hept{\begin{cases}p=a^2\\7p-1=b^2\end{cases}}\)

\(b^2+1=7p\Rightarrow b^2\equiv6\left(mod7\right)\)

Không có giá trị nào thỏa mãn. 

Do đó ta có đpcm. 

Ta có: \(A=2+2\sqrt{28n^2+1}\) là số chính phương

\(\Leftrightarrow2+2\sqrt{28n^2+1}⋮2\)

\(\Rightarrow2+2\sqrt{28n^2+1}=4\)

\(\Rightarrow\sqrt{28n^2+1}=1\)

\(\Rightarrow28n^2+1=1^2\)

\(\Rightarrow28n^2=0\Rightarrow n=0\)

Vậy A là SCP với n=0

đúng rồi đấy giỏi quá

a) ĐKXĐ: \(x\ge0\); \(1-4x\ne\)0; \(2\sqrt{x}-1\ne0\); \(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\ne\)0

<=> \(x\ge0\); x \(\ne\)1/4

Ta có:  \(A=\left(\frac{\sqrt{x}-4x}{1-4x}-1\right):\left(\frac{1+2x}{1-4x}-\frac{2\sqrt{x}}{2\sqrt{x}-1}-1\right)\)

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

\(A=\frac{\sqrt{x}-1}{1-4x}\cdot\frac{1-4x}{6x+4x+2\sqrt{x}}\)

\(A=\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\)

b)Với x \(\ge\)0 và x \(\ne\)1/4

Ta có: A > A² <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\left(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)^2\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\left(1-\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\right)>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10x+2\sqrt{x}-\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}\cdot\frac{10+\sqrt{x}+1}{10x+2\sqrt{x}}>0\)

<=> \(\sqrt{x}-1>0\) <=> \(x>1\)

c) Với x\(\ge\)0 và x \(\ne\)1/4 (1)

Ta có: \(\left|A\right|>\frac{1}{4}\) <=> \(\orbr{\begin{cases}A>\frac{1}{4}\\A< -\frac{1}{4}\end{cases}}\)

TH1: \(A>\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}>\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)>10x+2\sqrt{x}\)

<=> \(4\sqrt{x}-4>10x+2\sqrt{x}\)

<=> \(10x-2\sqrt{x}+4< 0\)(vô liia  vì \(10x-2\sqrt{x}+4>0\))

TH2: \(A< -\frac{1}{4}\) <=> \(\frac{\sqrt{x}-1}{10x+2\sqrt{x}}< -\frac{1}{4}\)

<=> \(4\left(\sqrt{x}-1\right)< -10x-2\sqrt{x}\)

<=> \(4\sqrt{x}-4+10x+2\sqrt{x}< 0\)

<=> \(10x+6\sqrt{x}-4< 0\)

<=> \(5x+3\sqrt{x}-2< 0\)

<=> \(\left(5\sqrt{x}-2\right)\left(\sqrt{x}+1\right)< 0\)

<=> \(x< \frac{4}{25}\) (2)

Từ (1) và (2) => \(0\le x< \frac{4}{25}\)

\(\frac{\sqrt{x}+3}{\sqrt{x}+1}-\frac{5}{1-\sqrt{x}}+\frac{4}{x-1}=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+7\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+6\sqrt{x}+\sqrt{x}+6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)

a) Xét tứ giác \(AKHI\)có: \(\widehat{KAI}=\widehat{AKH}=\widehat{HIA}=90^o\)

nên tứ giác \(AKHI\)có ba góc vuông nên \(AKHI\)là hình chữ nhật. 

b) \(\Delta AKH=\Delta KAI\left(c.g.c\right)\)

\(\Rightarrow\widehat{AHK}=\widehat{KIA}\)(hai góc tương ứng) 

mà \(\widehat{AHK}=\widehat{ACB}\)(vì cùng phụ với \(\widehat{HAC}\)) 

nên \(\widehat{KIA}=\widehat{ACB}\)

Xét tam giác \(AIK\)và tam giác \(ACB\)có: 

\(\widehat{IAK}=\widehat{CAB}\)(góc chung) 

\(\widehat{KIA}=\widehat{BCA}\)(cmt) 

\(\Rightarrow\Delta AIK~\Delta ACB\left(g.g\right)\)

\(\Rightarrow\frac{AI}{AC}=\frac{AK}{AB}\)(hai cặp cạnh tương ứng) 

\(\Rightarrow AI.AB=AK.AC\).

c) \(AI.AB=AK.AC\Leftrightarrow\frac{AB}{AC}=\frac{AK}{AI}\)

Xét tam giác \(ABK\)và tam giác \(ACI\):

\(\widehat{A}\)chung

\(\frac{AB}{AC}=\frac{AK}{AI}\)(cmt)

\(\Rightarrow\Delta ABK~\Delta ACI\left(c.g.c\right)\)

\(\Rightarrow\widehat{ABK}=\widehat{ACI}\)(hai góc tương ứng)

ĐK : x ≥ -1/2

\(\Leftrightarrow\sqrt{2x+1}-\frac{3}{2}\sqrt{4\left(2x+1\right)}+\sqrt{25\left(2x+1\right)}=0\)

\(\Leftrightarrow\sqrt{2x+1}-3\sqrt{2x+1}+5\sqrt{2x+1}=0\)

\(\Leftrightarrow3\sqrt{2x+1}=0\Leftrightarrow x=-\frac{1}{2}\left(tm\right)\)

11. \(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)

\(A=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(A=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(A=\frac{x-1}{\sqrt{x}}\)

12. \(M=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b}{a-b}-\frac{b}{b-\sqrt{ab}}+\frac{a}{\sqrt{ab}+a}\right)-\frac{\sqrt{\left(\sqrt{a}-\sqrt{b}\right)^2}}{2}\)

\(M=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\frac{a+b+\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)+\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)-\frac{\sqrt{b}-\sqrt{a}}{2}\)

\(M=\frac{a+b}{\sqrt{a}+\sqrt{b}}:\frac{a+b+\sqrt{ab}+b+a-\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}-\frac{\sqrt{b}-\sqrt{a}}{2}\)

\(M=\frac{a+b}{\sqrt{a}+\sqrt{b}}\cdot\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{2\left(a+b\right)}-\frac{\sqrt{b}-\sqrt{a}}{2}\)

\(M=\frac{\sqrt{a}-\sqrt{b}}{2}-\frac{\sqrt{b}-\sqrt{a}}{2}=\frac{\sqrt{a}-\sqrt{b}-\sqrt{b}+\sqrt{a}}{2}=\sqrt{a}-\sqrt{b}\)

13) \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

\(P=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(P=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

15) \(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}-1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\)

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

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

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

Ta có: \(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}=\frac{2}{\sqrt{a}+\sqrt{c}}\) 

\(\Leftrightarrow\frac{\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)}=\frac{2}{\sqrt{a}+\sqrt{c}}\) 

\(\Leftrightarrow\left(\sqrt{a}+\sqrt{c}\right)\left(\sqrt{a}+\sqrt{c}+2\sqrt{b}\right)=2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{b}+\sqrt{c}\right)\) 

\(\Leftrightarrow a+\sqrt{ac}+2\sqrt{ab}+\sqrt{ac}+c+2\sqrt{bc}=2\left(\sqrt{ab}+\sqrt{ac}+b+\sqrt{bc}\right)\)

\(\Leftrightarrow a+c=2b\) (luôn đúng)