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)

a) Xét tam giác \(ADC\)vuông tại \(D\): 

\(tan\widehat{ACD}=\frac{AD}{DC}=\frac{1}{2}\Rightarrow\widehat{ACD}=arctan\frac{1}{2}\)

b) Xét tam giác \(ADC\)vuông tại \(D\): 

\(AC^2=AD^2+DC^2=AD^2+4AD^2=5AD^2\)

\(\Leftrightarrow AD=\sqrt{\frac{AC^2}{5}}=\sqrt{\frac{25^2}{5}}=5\sqrt{5}\left(cm\right)\)

\(AB=AD=5\sqrt{5}\left(cm\right),CD=2AD=10\sqrt{5}\left(cm\right)\).

c) Xét tam giác \(ADC\)vuông tại \(D\): 

\(DH=\frac{AD.DC}{AC}=\frac{10\sqrt{5}.5\sqrt{5}}{25}=10\left(cm\right)\)

\(AH=\frac{AD^2}{AC}=\frac{AB^2}{AC}\Leftrightarrow\frac{AB}{AC}=\frac{AH}{AB}\)

Xét tam giác \(ABH\)và tam giác \(ACB\):

\(\widehat{A}\)chung

\(\frac{AB}{AC}=\frac{AH}{AB}\)

suy ra \(\Delta ABH~\Delta ACB\left(c.g.c\right)\)

\(\Rightarrow\widehat{ABH}=\widehat{ACB}\)

\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)

\(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)

\(\Leftrightarrow\left[\left(x-1\right)-2\sqrt{x-1}+1\right]+\left[\left(y-2\right)-4\sqrt{y-2}+4\right]+\left[\left(z-3\right)-6\sqrt{z-3}+9\right]=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)

ĐK: \(x\ge1,y\ge2,z\ge3\).

\(x+y+z+8=2\sqrt{x-1}+4\sqrt{y-2}+6\sqrt{z-3}\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+1+y-2-4\sqrt{y-2}+4+z-3-6\sqrt{z-3}+9=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-1\right)^2+\left(\sqrt{y-2}-2\right)^2+\left(\sqrt{z-3}-3\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}-1=0\\\sqrt{y-2}-2=0\\\sqrt{z-3}-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=6\\z=12\end{cases}}\)(thỏa mãn)

\(a,\sqrt{x^2-6x+9}=4-x\)

\(\sqrt{\left(x-3\right)^2}=4-x\)

\(\left|x-3\right|=4-x\)

\(\orbr{\begin{cases}x-3=4-x\\x-3=x-4\end{cases}\orbr{\begin{cases}x=\frac{7}{2}\left(TM\right)\\0x=-1\left(KTM\right)\end{cases}}}\)

\(b,\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

\(\left|x-1\right|+\left|x-2\right|=3\)

\(TH1:1\le x\)

\(1-x+2-x=3\)

\(x=0\left(TM\right)\)

\(TH2:1< x\le2\)

\(x-1+2-x=3\)

\(0x=2\left(KTM\right)\)

\(TH3:2< x\)

\(x-1+x-2=3\)

\(x=3\left(TM\right)\)

\(c,\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=5\)

\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5\)

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

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

\(\sqrt{2x-3}+1>0;\sqrt{2x-3}+4>0\)

\(\sqrt{2x-3}+1+\sqrt{2x-3}+4=5\)

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

\(x=\frac{3}{2}\left(TM\right)\)

\(d,\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(ĐKXĐ:x\ge3\)

\(\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\sqrt{\left(x-3\right)\left(x+3\right)}+x-3=0\)

\(\left(x-3\right)\left(x+3\right)=\left(3-x\right)^2\)

\(\left(x-3\right)\left(x+3\right)=9-6x+x^2\)

\(x^2-9=9-6x+x^2\)

\(18-6x=0\)

\(x=3\left(TM\right)\)