Viết đề thiếu giả thiết rồi, thoi mình cứ giả sử tam giác ABC vuông tại A, đường cao AH

=>\(\hept{\begin{cases}AB^2=BH.BC\\AC^2=CH.BC\end{cases}}\Rightarrow\frac{AB^2}{AC^2}=\frac{BH.BC}{CH.BC}=\frac{BH}{CH}\)

Tìm đkxđ mà ?

a) đk: \(\frac{x}{3}\ge0\Rightarrow x\ge0\)

b) đk: \(4-x\ge0\Rightarrow x\le4\)

c) Vì \(1+x^2\ge1>0\left(\forall x\right)\)

=> Xác định với mọi x thực

d) Vì \(x^2+6\ge6>0\left(\forall x\right)\)

=> \(-\frac{5}{x^2+6}< 0\left(\forall x\right)\) => không tồn tại x để căn thức xác định

e) đk: \(\frac{2}{x^2}>0\Rightarrow x^2>0\Rightarrow x>0\)

f) đk: \(\frac{1}{-1+x}>0\)

=> \(-1+x>0\Rightarrow x>1\)

g) đk: \(\frac{4}{x+3}>0\Leftrightarrow x+3>0\Rightarrow x>-3\)

h) Vì: \(x^2-2x+1=\left(x-1\right)^2\ge0\left(\forall x\right)\)

=> căn thức xác định với mọi giá trị của x

đk: \(x\ge0\)

\(x\sqrt{x}+4\sqrt{x}+12=7x\)

\(\Leftrightarrow\left(x+4\right)\sqrt{x}=7x-12\)

\(\Leftrightarrow\left(x+4\right)^2\cdot x=\left(7x-12\right)^2\)

\(\Leftrightarrow x^3+8x^2+16x=49x^2-168x+144\)

\(\Leftrightarrow x^3-41x^2+184x-144=0\)

\(\Leftrightarrow\left(x^3-x^2\right)-\left(40x^2-40x\right)+\left(144x-144\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-40x+144\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)\left(x-36\right)=0\)

=> \(x\in\left\{1;4;36\right\}\)

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

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

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

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

\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}\right)^2+2\cdot\sqrt{3}\cdot1+1^2}\) 

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

\(=\left(\sqrt{3}-1\right)|\sqrt{3}+1|\)    

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

\(=\left(\sqrt{3}\right)^2-1^2\)  

\(=3-1\)   

\(=2\)

Ta có: AB/AC=3/4 => AB/3=AC/4
=>. Đặt AB/3=AC/4=k
=> AB=3k ; AC=4k
Vì tg ABC vuông tại A
Áp dụng định lý Py-ta-go vào tg vuông ABC ta có:
=> AB^2 + AC^2 = BC^2
=> (3k)^2 + (4k)^2 = 15^2
=> 9k^2 + 16k^2 = 225
=> 25k^2 = 225
=> k^2=9 => k=3
=> AB=3k=3.3=9 cm
AC=4k=4.3=12 cm

Bài 1:

a)    \(=5.|2a|-5a^2\)

b)    \(=7\left(a-1\right)+5a=12a-7\)

c)    \(|a-2|-5\sqrt{a+2}\)

Bài 2:

a)    \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)

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

\(=2\sqrt{2}\)

c)    \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)

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

a) \(5\sqrt{4a^2}-5a^2\)

\(=5.|2a|-5a^2\)

b) \(7\sqrt{\left(a-1\right)^2}+5a\)

\(=7\left(a-1\right)+5a\)

\(=12a-7\)

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

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

bài 2:

a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)

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

\(=8-2\sqrt{2}\)

b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=2\sqrt{2}\)

c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)

\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)

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

Ta có: \(A=\left(\frac{1}{\sqrt{x-1}}+\frac{1}{\sqrt{x+1}}\right)^2\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)    \(\left(ĐK:x\ge1\right)\)

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

   \(\Leftrightarrow A=\frac{x+1+x-1+2\sqrt{x^2-1}}{x^2-1}\times\frac{x^2-1}{2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=\frac{2x+2\sqrt{x^2-1}}{2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=x+\sqrt{1-x^2}-\sqrt{1-x^2}\)

   \(\Leftrightarrow A=x\)

Học tốt

ĐKXĐ : ...............

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

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

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

\(A=\frac{x+1+2\sqrt{x^2-1}+x-1}{2}-\sqrt{1-x^2}\)

\(A=\frac{2x+2\sqrt{x^2-1}-2\sqrt{1-x^2}}{2}\)

\(A=\frac{2x+2\sqrt{x^2-1}+2\sqrt{x^2-1}}{2}\)

\(A=\frac{2x+4\sqrt{x^2-1}}{2}\)

\(A=x+2\sqrt{x^2-1}\)

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

<=> \(xy+\sqrt{x^2+1}\sqrt{y^2+1}-1=-x\sqrt{x^2+1}-y\sqrt{y^2+1}\)--->Bình phương 2 vế:

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

                                                                                                     \(x^2\left(x^2+1\right)+y^2\left(y^2+1\right)+2xy\sqrt{x^2+1}\sqrt{y^2+1}\)

<=>\(2\left(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\right)=\left(x^2-y^2\right)^2\ge0\)=>\(1-xy-\sqrt{x^2+1}\sqrt{y^2+1}\ge0\)

<=>\(1-xy\ge\sqrt{x^2+1}\sqrt{y^2+1}>0\)---> Bình phương 2 vế:

\(1+x^2y^2-2xy\ge\left(x^2+1\right)\left(y^2+1\right)\)<=>\(0\ge\left(x+y\right)^2\ge0\)<=>\(x+y=0\Leftrightarrow x=-y\Rightarrow x^2=y^2\)

--> Thay vào A---> \(A=\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=\left(x+\sqrt{y^2+1}\right)\left(y+\sqrt{x^2+1}\right)=1\)