a)

\(B=\frac{x+3}{x-9}+\frac{2}{\sqrt{x}-3}-\frac{1}{3-\sqrt{x}}\)

\(\Leftrightarrow\frac{x+3}{x-9}+\frac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\frac{x+3+2\sqrt{x}+6+\sqrt{x}+3}{x-9}\)

\(\Leftrightarrow\frac{3\sqrt{x}+x+14}{x-9}\)

cái ngôn ngữ j zậy ta ? =_=

\(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\)

\(< \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+...+\frac{1}{\sqrt{n^2}}\)(\(n\)số hạng) 

\(=\frac{n}{\sqrt{n^2}}=1\)

Ta có đpcm. 

a/

\(S_{ABC}=S_{ABD}+S_{BCD}\)

\(\Rightarrow AB.BC.\sin\widehat{ABC}=AB.BD.\sin\widehat{ABD}+BC.BD.\sin\widehat{CBD}\)

\(\Rightarrow6.12.\sin120^o=6.BD.\sin60^o+12.BD.\sin60^o=18.BD.\sin60^o\)

\(\Rightarrow BD=\frac{72.\sin120^o}{18.\sin60^o}=\frac{4.2.\sin60^o.\cos60^o}{\sin60^o}=8.\cos60^o=4cm\)

b/

Ta có \(BM=CM=\frac{BC}{2}=\frac{12}{2}=6cm=AB\)

=>Ta, giác ABM cân tại B mà BD là phân giác của \(\widehat{B}\Rightarrow BD\perp AM\) (trong tg cân đường phân giác của góc ở đỉnh đồng thời là đường cao)

c/ Theo tính chất đường phân giác trong tam giác

Ta có \(\frac{AD}{AB}=\frac{CD}{BC}\Rightarrow\frac{AD}{CD}=\frac{AB}{BC}=\frac{6}{12}=\frac{1}{2}\) 

\(\Rightarrow AD=\frac{AC}{3}\)

Gọi E là giao của BD và AM

Xét tg vuông ABE có \(\widehat{BAE}=90^o-\widehat{ABD}=90^o-60^o=30^o\Rightarrow BE=\frac{AB}{2}=\frac{6}{2}=3cm\)

\(\Rightarrow DE=BD-BE=4-3=1cm\)

Ta có \(AE^2=AB^2-BE^2=6^2-3^2=27\)

Xét tg vuông ADE có

\(AD=\sqrt{AE^2+DE^2}=\sqrt{28}=2\sqrt{7}cm\)

Mà \(AD=\frac{AC}{3}\Rightarrow AC=3AD=3.2\sqrt{7}=6\sqrt{7}cm\)

x, y > 0

\(=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\cdot\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]\cdot\frac{\sqrt{xy}\left(x+y\right)}{\left(\sqrt{x^3}+\sqrt{y^3}\right)+\left(x\sqrt{y}+y\sqrt{x}\right)}\)

\(=\frac{2\sqrt{xy}+x+y}{xy}\cdot\frac{\sqrt{xy}\left(x+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{xy}}\cdot\frac{\left(x+y\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(x+y\right)}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

Ta có : \(A=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{16}}=\frac{\sqrt{x}+\sqrt{y}}{4}=\frac{\sqrt{x}}{4}+\frac{\sqrt{y}}{4}\ge2\sqrt{\frac{\sqrt{xy}}{4\cdot4}}=2\sqrt{\frac{\sqrt{16}}{16}}=1\)( AM-GM )

Dấu "=" xảy ra <=> x = y = 4 . Vậy MinA = 1

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

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

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

b, \(A=7\Leftrightarrow\left(\sqrt{x}\right)^3-1=7\left(\sqrt{x}-1\right)^2\)

\(\Leftrightarrow\sqrt{x}^3-7x+14\sqrt{x}-8=0\)

\(\Leftrightarrow\left(\sqrt{x}^3-4x\right)-\left(3x-12\sqrt{x}\right)+\left(2\sqrt{x}-8\right)=0\)

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

=> Tìm x

d, \(A< 1\Leftrightarrow\left(\sqrt{x}\right)^3-1< \left(\sqrt{x}-1\right)^2\)

\(\Leftrightarrow\sqrt{x}^3-1-\left(\sqrt{x}-1\right)^2< 0\)

\(\Leftrightarrow\sqrt{x}^3-x+2\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}.\left(x-\sqrt{x}+2\right)< 0\)

Mà \(x-\sqrt{x}+2=\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow\sqrt{x}< 0\)vô lí

=> Không tìm được x

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

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

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

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

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

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

b, Đặt \(\sqrt{x}=a,\left(a\ge0\right)\)\(\Rightarrow P=\frac{1+a^3}{a^2-a}\), để chứng minh P > 1

thì ta chứng minh \(1+a^3>a^2-a\)

\(\Leftrightarrow a^3-a^2+a+1>0\Leftrightarrow\left(a-1\right)^3+2\left(a^2-a+1\right)>0\)

mà \(a^2-a+1=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall a\)

\(\Rightarrow2\left(a^2-a+1\right)\ge\frac{3}{2},a\ge0\)nên \(\left(a-1\right)^3\ge1\Rightarrow a^3-a^2+a+1\ge\frac{1}{2}\)hay \(P>1\)