Xét tam giác vuông AHB và CHA có :

      góc AHB = góc CHA = 90độ 

      góc ABH = góc CAH ( cùng phụ với góc C )

Vậy tam giác AHB đồng dạng tam giác CHA ( g.g )

Suy ra : \(\frac{AH}{HC}=\frac{AB}{CA}\)    ( 1 )

Theo đề bài \(\frac{AB}{AC}=\frac{3}{4}\) và AH = 12cm  ( 2 )

Từ ( 1 ) và ( 2 ) suy ra : \(\frac{12}{HC}=\frac{3}{4}\Rightarrow HC=\frac{12.4}{3}=16\) ( cm )

Theo hệ thức liên hệ giữa đường cao và hình chiếu , ta có :

\(AH^2=HB.HC\Rightarrow HB=\frac{AH^2}{HC}=\frac{12^2}{16}=9\) ( cm )

Vậy BH = 9cm , HC = 16cm

Học tốt

\(ĐKXĐ:x\ge4\)

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

\(=\sqrt{\left(\sqrt{x-4}\right)^2+2.2\sqrt{x-4}+2^2}-\sqrt{x-4}\)

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

\(=\sqrt{x-4}+2-\sqrt{x-4}=2\)( vì \(x\ge4\)nên \(\sqrt{x-4}\ge0\))

Trả lời:

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)

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

\(A=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)

\(A=\frac{\sqrt{2}.\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{4+2\sqrt{3}}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=\frac{\sqrt{2}.\left(\sqrt{3}+1\right)}{\sqrt{2}.\left(\sqrt{3}+1\right)}\)

\(A=1\)

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

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

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

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

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

mình nghĩ đề như này mới đúng chứ :))

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

<=> 2x - 5 = 5

<=> 2x = 5 + 5

<=> 2x = 10

<=> x = 5

\(\sqrt{2x}-5=\sqrt{5}\left(dk:x\ge0\right)\)

\(< =>\sqrt{2x}=\sqrt{5}\left(1+\sqrt{5}\right)\)

\(< =>\sqrt{x}=\frac{\sqrt{5}\left(1+\sqrt{5}\right)}{\sqrt{2}}\)

\(< =>x=\frac{5\left(1+\sqrt{5}\right)^2}{2}\)

Từ 2a + b + c = 0 <=> a + a + b + c = 0 <=> a + c = -(a + b)

Ta có: VT = 2a³ + b³ + c³ = (a³  + b³) + (a³ + c³)

= (a + b)(a² - ab + b²) + (a + c)(a² - ac + c²)

= (a + b)(a² + 2ab + b²) - 3ab(a + b) + (a + c)(a² + 2ac + c²) - 3ac(a + c)

= (a + b)³ - 3ab(a + b) + (a + c)³ - 3ac(a + c)

= (a + b)³ - (a + b)³ - 3ab(a + b) + 3ac(a + b)

= -3a(a + b)(b - c) = 3a(a + b)(c - b) = VP

=> VT = VP => đpcm

Bất đẳng thức cần chứng minh tương đương: \(\frac{1}{a^4\left(b+1\right)\left(c+1\right)}+\frac{1}{b^4\left(c+1\right)\left(a+1\right)}+\frac{1}{c^4\left(a+1\right)\left(b+1\right)}\ge\frac{3}{4}\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=1\end{cases}}\)và ta đưa BĐT cần chứng minh về dạng \(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}\ge\frac{3}{4}\)

Áp dụng BĐT AM - GM, ta được:\(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y+1}{8}+\frac{z+1}{8}\ge\frac{3}{4}x\)

Tương tự: \(\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z+1}{8}+\frac{x+1}{8}\ge\frac{3}{4}y\); \(\frac{z^3}{\left(x+1\right)\left(y+1\right)}+\frac{x+1}{8}+\frac{y+1}{8}\ge\frac{3}{4}z\)

Cộng theo vế của 3 BĐT trên, ta được: \(\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}+\)\(\frac{x+y+z+3}{4}\ge\frac{3}{4}\left(x+y+z\right)\)

\(\Rightarrow\frac{x^3}{\left(y+1\right)\left(z+1\right)}+\frac{y^3}{\left(z+1\right)\left(x+1\right)}+\frac{z^3}{\left(x+1\right)\left(y+1\right)}\)\(\ge\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{1}{2}.3\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c = 1