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)

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

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

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

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

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

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

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

\(TH1:1\le x\)

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

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

\(TH2:1< x\le2\)

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

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

\(TH3:2< x\)

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

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

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

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

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

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

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

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

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

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

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

\(ĐKXĐ:x\ge3\)

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

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

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

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

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

\(18-6x=0\)

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

\(B=\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\)

\(=\sqrt{a-1+2\sqrt{a-1}+1}+\sqrt{a-1-2\sqrt{a-1}+1}\)

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

\(=\left|\sqrt{a-1}+1\right|+\left|\sqrt{a-1}-1\right|\)

\(=\sqrt{a-1}+1+1-\sqrt{a-1}\)(vì \(1\le a\le2\))

\(=2\).

\(x+y+z\le2xyz+\sqrt{2}\)

\(\Leftrightarrow x+y+z-2xyz\le\sqrt{2}\)

\(\Leftrightarrow x\left(1-2yz\right)+y+z\le\sqrt{2}\)

Ta có: \(x\left(1-2yz\right)+y+z\le\sqrt{\left[x^2+\left(y+z\right)^2\right]\left[\left(1-2yz\right)^2+1\right]}\)( bđt bunhiacopxki )

Ta cần chứng minh \(\sqrt{\left[x^2+\left(y+z\right)^2\right]\left[\left(1-2yz\right)^2+1\right]}\le\sqrt{2}\)

\(\Leftrightarrow\left[x^2+y^2+z^2+2yz\right]\left[1-4yz+4y^2z^2+1\right]\le2\)

\(\Leftrightarrow\left(1+2yz\right)\left(2-4yz+4y^2z^2\right)\le2\)

\(\Leftrightarrow2-4yz+4y^2z^2+4yz-8y^2z^2+8y^3z^3\le2\)

\(\Leftrightarrow8y^3z^3-4y^2z^2\le0\)

\(\Leftrightarrow4y^2z^2\left(2yz-1\right)\le0\)(1) 

Ta có: \(1=x^2+y^2+z^2\ge y^2+z^2\ge2yz\)

\(\Rightarrow yz\le\frac{1}{2}\)

\(\Rightarrow\left(1\right)\)đúng

Vậy đẳng thức đc chứng minh

ok bạn

a)\(-5\sqrt{80}+4\sqrt{45}-2\sqrt{245}\)

\(=-20\sqrt{5}+12\sqrt{5}-14\sqrt{5}\)

\(=\left(-20+12-14\right)\sqrt{5}=-22\sqrt{5}\)

b)\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{9-6\sqrt{3}+3}-\sqrt{12-6\sqrt{12}+9}\)

\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)

\(=\left|3-\sqrt{3}\right|-\left|\sqrt{12}-3\right|\)

\(=3-\sqrt{3}-\sqrt{12}+3\)(do \(3>\sqrt{3};\sqrt{12}>3\))

\(=6-\sqrt{12}-\sqrt{3}\)

\(=6-2\sqrt{3}-\sqrt{3}=6-3\sqrt{3}\)

c)\(\sqrt{7-\sqrt{40}}-\sqrt{7+\sqrt{40}}\)

\(=\sqrt{5-2\sqrt{5}.\sqrt{2}+2}-\sqrt{5+2\sqrt{5}.\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)(do \(\sqrt{5}>\sqrt{2}\))

\(=-2\sqrt{2}\)