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

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

\(\frac{B}{A}< \frac{-1}{3}\Leftrightarrow\frac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}+\frac{1}{3}< 0\Leftrightarrow\frac{-9}{3\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{3\left(\sqrt{x}-3\right)}< 0\)

\(\Leftrightarrow\frac{\sqrt{x}-12}{3\left(\sqrt{x}-3\right)}< 0\). Xét hai trường hợp :

1.\(\hept{\begin{cases}\sqrt{x}-12>0\\3\left(\sqrt{x}-3\right)< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>144\\x< 9\end{cases}}\left(loai\right)\)

2. \(\hept{\begin{cases}\sqrt{x}-12< 0\\3\left(\sqrt{x}-3\right)>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 144\\x>9\end{cases}}\Rightarrow9< x< 144\)

Vậy ... 

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

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

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

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

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

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

\(\frac{B}{A}=\frac{-3}{\sqrt{x}+3}\) mà \(\frac{B}{A}< -\frac{1}{3}\)

\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}< -\frac{1}{3}\)

\(\Leftrightarrow\frac{-3}{\sqrt{x}+3}+\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{-9-\sqrt{x}-3}{3\left(\sqrt{x}+3\right)}< 0\) mà \(3\left(\sqrt{x}+3\right)>0\)

\(\Leftrightarrow-12-\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}>-12\left(luondung\right)\)

H A B C I K

a, xét tam giác AHB có : ^AHB = 90 và HI _|_ AB => AI.AB = AH^2

xét tam giác AHC có : ^AHC = 90 và HK _|_ AC => AK.AC = AH^2

=> AI.AB = AK.AC

b, xét tam giác AHC có ^AHC = 90 \(\Rightarrow\sin\widehat{C}=\frac{AH}{AC}\Leftrightarrow\sin^2\widehat{C}=\frac{AH^2}{AC^2}\)

\(\Rightarrow\sin^2\widehat{C}\cdot AC=\frac{AH^2}{AC}\)    mà \(AH^2=AK\cdot AC\left(câua\right)\)

\(\Rightarrow\sin^2\widehat{C}\cdot AC=AK\)

a.Xét tam giác vuông AHC có đường cao HK ta có : \(AK.AC=AH^2\)

Xét tam giác vuông AHB có đường cao HI ta có : \(AI.AB=AH^2\) vậy \(AI.AB=AK.AC\)

b. ta có \(AK=\frac{AH^2}{AC}=\frac{AH^2}{AC^2}.AC=AC.sin^2C\)

c. ta có :

\(\frac{1}{4}=\frac{S_{AKI}}{S_{ABC}}=\frac{AK.AI}{AB.AC}=\frac{AK}{AB.AC}.\frac{AK.AC}{AB}=\frac{AK^2}{AB^2}\) nên \(AK=\frac{1}{2}AB\) tương tự \(AI=\frac{1}{2}AC\)

\(\Rightarrow KI=\frac{1}{2}CB\Rightarrow AH=\frac{1}{2}CB\Rightarrow\text{AH là đường trung tuyến của tam giác vuong}\)

AH vừa là đường trung tuyến vừa là đường cao nên ABC vuông cân

ĐÁP ÁN

hÌNH

HT

b8 :

\(ab+bc+ca=5\Rightarrow a^2+5=a^2+ab+bc+ca\)

\(\Rightarrow a^2+5=a\left(a+b\right)+c\left(a+b\right)=\left(a+c\right)\left(a+b\right)\)

tương tự có \(b^2+5=\left(b+c\right)\left(b+a\right)\) và \(c^2+5=\left(c+a\right)\left(c+b\right)\) 

thay vào A ta đc :

\(A=a\sqrt{\frac{\left(b+c\right)\left(b+a\right)\left(c+a\right)\left(c+b\right)}{\left(a+b\right)\left(a+c\right)}}+b\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(c+a\right)\left(c+b\right)}{\left(b+a\right)\left(b+c\right)}}+\)

\(c\sqrt{\frac{\left(a+b\right)\left(a+c\right)\left(b+a\right)\left(b+c\right)}{\left(c+a\right)\left(c+b\right)}}\)

\(A=a\sqrt{\left(b+c\right)^2}+b\sqrt{\left(c+a\right)^2}+c\sqrt{\left(b+a\right)^2}\)

\(A=a\left|b+c\right|+b\left|c+a\right|+c\left|b+a\right|\)

\(A=ab+ac+bc+ba+cb+ca\) do a;b;c dương

\(A=2\left(ab+bc+ca\right)=2\cdot5=10\)

b9: 

\(x_o=\sqrt[3]{a+\sqrt{a^2+b^3}}-\sqrt[3]{\sqrt{a^2+b^3}-a}\)

\(\Leftrightarrow x_o^3=a+\sqrt{a^2+b^3}-\sqrt{a^2+b^3}+a-3^3\sqrt{\left(a+\sqrt{a^2+b^3}\right)\cdot\left(\sqrt{a^2+b^3}-a\right)}\cdot x_o\)

\(\Leftrightarrow x_o^3=2a-3\sqrt[3]{a^2+b^3-a^2}\cdot x_o\)

\(\Leftrightarrow x_o^3=2a-3bx_O\)

\(\Leftrightarrow x_o^3+3bx_o-2a=0\left(đpcm\right)\)

bài 9 dấu <=> thứ nhất do dài quá bị xuống dòng, như này này b 

\(a+\sqrt{a^2+b^3}-\sqrt{a^2+b^3}+a-3\sqrt[3]{\left(\sqrt{a^2+b^3}+a\right)\left(\sqrt{a^2+b^3-a}\right)}\cdot x_o\)

A B C 60 O 3m

tam giác ABC vuông tại A có \(\tan\widehat{C}=\frac{AB}{AC}\) nên \(\tan60^o=\frac{AB}{3}\)

\(\Rightarrow AB=3\sqrt{3}\)

1m 30 o A B C

tam giác ABC có \(\cos\widehat{B}=\frac{AB}{BC}\) nên \(\cos\widehat{60}=\frac{1}{BC}\)

\(\Rightarrow\frac{1}{2}=\frac{1}{BC}\Leftrightarrow BC=2\)

vậy miếng tôn dài 2m

A B C 3 2

tam giác ABC vuông tại A \(\Rightarrow\tan\widehat{C}=\frac{AB}{AC}=\frac{3}{2}\)

bấm máy tính shift -> tan -> 3/2 -> ^o'" 

\(\Rightarrow\widehat{C}\simeq56^o\)

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

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

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

\(P=2\sqrt{x}+1\)

b, x = 9/4 (thỏa mãn) => \(P=2\sqrt{\frac{9}{4}}+1=2\cdot\frac{3}{2}+1=4\)

c, \(P\left(\sqrt{x}-2\right)=\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)< 0\)

mà \(2\sqrt{x}+1>0\) \(\Rightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow x< 4\) kết hợp đk : x >= 0 và x khác 1

\(\Rightarrow0\le x\le4\) và x khác 1