ta có \(\frac{1}{AO^2}=\frac{1}{AB^2}+\frac{1}{AD^2}\Rightarrow AD=\sqrt{117}cm\)

ta có \(\widehat{ADB}=\widehat{ACD}\text{ (do cùng phụ với góc }\widehat{CDB}\text{)}\) nên \(\Delta ADB~\Delta DAC\left(g.g\right)\Rightarrow\frac{DA}{DC}=\frac{AB}{AD}\Rightarrow DC=\frac{DA^2}{AB}=\frac{9\sqrt{13}}{2}\)

ta có \(S_{ABCD}=\frac{1}{2}AD\left(AB+CD\right)=\frac{507}{4}cm^2\)

Xét tam giác ABC vuông tại A, đường cao AH

Ta có : \(\frac{HB}{HC}=\frac{1}{4}\Rightarrow HB=\frac{1}{4}HC\)

* Áp dụng hệ thức : \(AH^2=BH.HC=\left(\frac{1}{4}HC\right)HC=\frac{1}{4}HC^2\)

\(\Rightarrow196=\frac{1}{4}HC^2\Leftrightarrow HC^2=784\Leftrightarrow HC=28\)cm 

=> HB = 28/4 = 7 cm 

=> BC = HB + HC = 28 + 7 = 35 cm 

Áp dụng định lí Pytago tam giác AHB vuông tại H 

\(AB^2=BH^2+AH^2=49+196=245\Rightarrow AB=7\sqrt{5}\)cm 

* Áp dụng hệ thức : \(AH.BC=AB.AC\Rightarrow AC=\frac{AH.BC}{AB}=14\sqrt{5}\)cm

Chu vi tam giác ABC là : \(P_{ABC}=AB+AC+BC=35+21\sqrt{5}\)cm 

Xét tam giác ABC vuông tại A, đường cao AH

Áp dụng định lí Pytago tam giác ABC vuông tại A

\(BC^2=AB^2+AC^2=144+256=400\Rightarrow BC=20\)cm 

* Áp dụng hệ thức : \(AB^2=BH.BC\Rightarrow BH=\frac{AB^2}{BC}=\frac{144}{20}=\frac{36}{5}\)cm 

* Áp dụng hệ thức : \(AC^2=CH.BC\Rightarrow CH=\frac{AC^2}{BC}=\frac{256}{20}=12,8\)cm 

Vì AD là đường pg nên \(\frac{AB}{AC}=\frac{BD}{DC}\Rightarrow\frac{DC}{AC}=\frac{BD}{AB}\)

Áp dụng tunhs chất dãy tỉ số bằng nhau 

\(\frac{DC}{AC}=\frac{BD}{AB}=\frac{BC}{AB+AC}=\frac{20}{28}=\frac{5}{7}\)

\(\Rightarrow BD=\frac{5}{7}.AB=\frac{5}{7}.12=\frac{60}{7}\)cm 

=> \(HD=BD-BH=\frac{60}{7}-\frac{36}{5}=\frac{48}{35}\)cm 

\(A=\left(\frac{1}{\sqrt{a}+2}+\frac{1}{\sqrt{a}-2}\right):\frac{\sqrt{a}}{a-4}\)\(ĐKXĐ:x\ge0;x\ne4\)

\(A=\frac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}+2\right).\left(\sqrt{a}-2\right)}.\frac{a-4}{\sqrt{a}}\)

\(A=\frac{2\sqrt{a}}{a-4}.\frac{a-4}{\sqrt{a}}\)

\(A=2\)

\(B=\left(\frac{4x}{\sqrt{x}-1}-\frac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\frac{\sqrt{x}+1}{x^2}\) \(ĐKXĐ:x\ne1,x>0\)

\(B=\left(\frac{4x.\left(\sqrt{x}-2\right)-\sqrt{x}+2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}-2\right)}\right).\frac{\sqrt{x}-1}{x^2}\)

\(B=\frac{\left(\sqrt{x}-2\right).\left(4x-1\right)}{\left(\sqrt{x}-2\right)}.\frac{1}{x^2}\)

\(B=\frac{4x-1}{1}.\frac{1}{x^2}\)

\(B=\frac{4x-1}{x^2}\)

Vậy.....

Đặt căn thức là A

Ta có: \(\sqrt{2}A=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}+2\sqrt{12}\)

\(\Rightarrow\sqrt{2}A=\sqrt{\left(\sqrt{12-1}\right)^2}+\sqrt{\left(\sqrt{12+1}\right)^2}+2\sqrt{12}\)

\(\Rightarrow\sqrt{2}A=\sqrt{12}-1+\sqrt{12}+1+2\sqrt{12}\)

\(\Rightarrow\sqrt{2}A=4\sqrt{2}\)

\(\Rightarrow A=4\sqrt{6}\)

\(A=\sqrt{\left(\frac{\sqrt{x}}{2}\right)^2-2.\frac{\sqrt{x}}{2}.\frac{1}{6}+\frac{1}{36}+\frac{35}{36}}\)

\(=\sqrt{\left(\frac{\sqrt{x}}{2}-\frac{1}{6}\right)^2+\frac{35}{36}\ge}\sqrt{\frac{35}{36}}\)

Dấu "=" khi x = 1/9

Ta có: \(1=abc\le\left(\frac{a+b+c}{3}\right)^3\)

\(\Rightarrow a+b+c\ge3\).

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{3}{2}\)

Dấu \(=\)khi \(a=b=c=1\).

Bài 7. 

a) \(\sqrt{x+9}=3\)(ĐK: \(x\ge-9\))

\(\Leftrightarrow x+9=3^2\)

\(\Leftrightarrow x=0\)(thỏa mãn) 

b) \(\sqrt{2x^2+2}=3x-1\)

\(\Rightarrow2x^2+2=\left(3x-1\right)^2\)

\(\Leftrightarrow7x^2-6x-1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}\)

Thử lại chỉ có \(x=1\)thỏa mãn.

c) \(\sqrt{x^2-2x+1}=19x-1\)

\(\Rightarrow x^2-2x+1=\left(19x-1\right)^2\)

\(\Leftrightarrow360x^2-36x=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{10}\end{cases}}\)

Thử lại chỉ có \(x=\frac{1}{10}\)thỏa mãn. 

d) \(\sqrt{x^2-x-6}=\sqrt{x-3}\)(ĐK: \(x\ge3\)) 

\(\Rightarrow x^2-x-6=x-3\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\left(l\right)\\x=3\left(tm\right)\end{cases}}\)

e) \(\sqrt{4x^2+4x+1}=\sqrt{x^2+12x+36}\)

\(\Leftrightarrow\left|2x+1\right|=\left|x+6\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=x+6\\2x+1=-x-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-\frac{7}{3}\end{cases}}\)

g) \(\sqrt{x+4\sqrt{x-4}}=2\)(ĐK: \(x\ge4\))

\(\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}-2\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-4}-2=2\\\sqrt{x-4}-2=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=20\\x=4\end{cases}}\)(thỏa mãn) 

Bài 10. 

b) \(\sqrt{x^2-2x+4}=x-1\)

\(\Rightarrow x^2-2x+4=\left(x-1\right)^2\)

\(\Leftrightarrow0x=3\)(vô nghiệm) 

c) \(\sqrt{x^2-6x+9}=4-x\)

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

\(\Rightarrow\orbr{\begin{cases}x-3=4-x\\x-3=x-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=7\\0x=1\end{cases}}\)

\(\Leftrightarrow x=\frac{7}{2}\)(thử lại thỏa mãn) 

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

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

Với \(x\ge2\): 

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

\(\Leftrightarrow x=3\)(thỏa mãn) 

Với \(1\le x< 2\):

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

\(\Leftrightarrow0x=2\)(vô nghiệm) 

Với \(x< 1\):

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

\(\Leftrightarrow x=0\)(thỏa mãn) 

\(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)(ĐK: \(x\ge0,x\ne1\)) 

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

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

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

\(A=\frac{5}{\sqrt{x}}\)

\(\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}=\frac{5}{\sqrt{x}}\)

\(\Rightarrow x=5\left(x+\sqrt{x}+1\right)\)

\(\Leftrightarrow4x+5\sqrt{x}+1=0\)(vô nghiệm do \(x\ge0\)) 

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

\(=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)(vì \(x\ne1\))

Do đó \(A< \frac{1}{3}\).