A B C x

có \(\sin x=\frac{AB}{BC}\)  và \(\cos x=\frac{AC}{BC}\)

\(\Rightarrow\hept{\begin{cases}\sin^2x=\frac{AB^2}{BC^2}\\\cos^2x=\frac{AC^2}{BC^2}\end{cases}}\)

\(\Rightarrow\sin^2x+\cos^2x=\frac{AB^2+AC^2}{BC^2}=\frac{BC^2}{BC^2}=1\left(pytago\right)\)

\(7\sqrt{x}=42\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)

\(\sqrt{x}>5\Leftrightarrow x>25\)

\(\sqrt{x}< 3=x< 9\)

\(3\sqrt{x}>25\Leftrightarrow\sqrt{x}>\frac{25}{3}\Leftrightarrow x>\frac{625}{9}\)

\(M=\frac{a^3-8a+\left(a^2-16\right)\sqrt{a^2-9}-5a^2+48}{a^3-8a+\left(a^2-16\right)\sqrt{a^2-9}+5a^2-48}\)

\(=\frac{\left(a-4\right)^2\left(a+3\right)+\left(a-4\right)\left(a+4\right)\sqrt{\left(a-3\right)\left(a+3\right)}}{\left(a+4\right)^2\left(a-3\right)+\left(a-4\right)\left(a+4\right)\sqrt{\left(a-3\right)\left(a+3\right)}}\)

\(=\frac{\sqrt{a+3}\left(a-4\right)\left[\left(a-4\right)\sqrt{a+3}+\left(a+4\right)\sqrt{a-3}\right]}{\sqrt{a-3}\left(a+4\right)\left[\left(a+4\right)\sqrt{a-3}+\left(a-4\right)\sqrt{a+3}\right]}\)

\(=\frac{\sqrt{a+3}\left(a-4\right)}{\sqrt{a-3}\left(a+4\right)}\)

O A B C D H M

a, xét tam giác CHA và tg CHO có : CH chung

AH = HO do H là trđ của AO (gt)

^CHA = ^CHO = 90

=> tg CHA = tg CHO (2cgv)

=> CH = CO

có AB _|_ CD => A là điểm chính giữa của cung CD => AC = AD mà OC  = OD 

=> AC = CO = OD = DA

=> ACOD là hình thoi

b, C thuộc đường tròn đường kính AB => ^ACB = 90 => AC _|_ CB

có AC // DO do ACOD là hình thoi 

=> DO _|_ CB  

M là trung điểm của dây BC (Gt) => OM _|_ BC (định lí)

=> D;O;M thẳng hàng

c, xét tg ACB có ^ACB = 90 và CH _|_ AB

=> AH.HB = CH^2

=> 4AH.HB = 4CH^2

=> 4AH.HB = (2CH)^2

mà 2CH = CD

=> CD^2 = 4AH.HB

AD BĐT AM-GM ta có

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

\(\Rightarrow c^3+\frac{a^3+b^3}{2}\ge c^3+\frac{a+b}{c}\ge2c\sqrt{a+b}\)

Tượng tự \(\Rightarrow b^3+\frac{a^3+c^3}{2}\ge b^3+\frac{a+c}{b}\ge2b\sqrt{a+c}\); \(a^3+\frac{b^3+c^3}{2}\ge a^3+\frac{b+c}{a}\ge2a\sqrt{b+c}\)

Cộng vế theo vế các bđt trên ta được: \(a^3+b^3+c^3\ge a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}\)(đpcm)

Đẳng thức xảy ra <=> a=b=c=\(\sqrt[3]{2}\)

Bài 2 : 

a, \(3\sqrt{x-7}-4=11\Leftrightarrow3\sqrt{x-7}=15\)ĐK : x >= 7 

\(\Leftrightarrow\sqrt{x-7}=5\Leftrightarrow x-7=25\Leftrightarrow x=32\)

b, \(\sqrt{\frac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)ĐK : \(x\le2\)

\(\Leftrightarrow\frac{5}{2}\sqrt{2-x}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)

\(\Leftrightarrow\left(\frac{5}{2}-8+3\right)\sqrt{2-x}=-10\Leftrightarrow\frac{-5}{2}\sqrt{2-x}=-10\)

\(\Leftrightarrow\sqrt{2-x}=4\Leftrightarrow2-x=16\Leftrightarrow x=-14\)

c, \(\frac{\sqrt{x}}{\sqrt{x}-2}=2\)ĐK : \(x\ge0;x\ne4\)

\(\Rightarrow\sqrt{x}=2\sqrt{x}-4\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)

d, \(5\sqrt{2x-3}+8\sqrt{8x-12}-\frac{1}{3}\sqrt{18x-27}=15\)ĐK : \(x\ge\frac{3}{2}\)

\(\Leftrightarrow5\sqrt{2x-3}+16\sqrt{2x-3}-\sqrt{2x-3}=15\)

\(\Leftrightarrow20\sqrt{2x-3}=15\Leftrightarrow\sqrt{2x-3}=\frac{3}{4}\Leftrightarrow2x-3=\frac{9}{16}\Leftrightarrow x=\frac{57}{32}\)

\(\frac{A}{B}=\frac{\frac{2\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}}{3-\sqrt{x}}-\frac{3x+3}{x-9}}{\frac{\sqrt{x}+1}{\sqrt{x}-3}}=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

Trả lời:

\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}-\frac{\sqrt{x}}{3-\sqrt{x}}-\frac{3x+3}{x-9}\) \(\left(ĐKXĐ:x\ge0;x\ne9\right)\)

\(=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3x+3}{x-9}\)

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

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

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

Ta có: \(P=\frac{A}{B}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{-3}{\sqrt{x}+3}\)

=> ĐPCM

1. Ta có : \(3\sqrt{x}+2\ge2\forall x\ge0\)(*)

 \(A=\frac{6}{3\sqrt{x}+2}\)nguyên <=> \(3\sqrt{x}+2\in\left\{3;6\right\}\)( do (*) )

<=> \(x\in\left\{\frac{1}{9};\frac{16}{9}\right\}\). Mà x\(\in\)Z

=> Không có giá trị x nguyên nào để A nguyên

2. Với x\(\in\)R thì \(x\in\left\{\frac{1}{9};\frac{16}{9}\right\}\)