Bài làm:

Ta có: \(E=\frac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)

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

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

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

Thay \(x=2\left(\sqrt{3}+1\right)\) vào thì giá trị của E là:

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

\(E=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{2\sqrt{3}}}{\sqrt{12+4+8\sqrt{3}-4}+4+2\sqrt{3}}\)

\(E=\frac{\sqrt{3}+1+\sqrt{2\sqrt{3}}}{2\sqrt{3+2\sqrt{3}}+4+2\sqrt{3}}\)

Bài làm:

Ta có:

\(D=\frac{1}{\sqrt{h+2\sqrt{h-1}}}+\frac{1}{\sqrt{h-2\sqrt{h-1}}}\)

\(D=\frac{1}{\sqrt{\left(h-1\right)+2\sqrt{h-1}+1}}+\frac{1}{\sqrt{\left(h-1\right)-2\sqrt{h-1}+1}}\)

\(D=\frac{1}{\sqrt{\left(\sqrt{h-1}+1\right)^2}}+\frac{1}{\sqrt{\left(\sqrt{h-1}-1\right)^2}}\)

\(D=\frac{1}{\left|\sqrt{h-1}+1\right|}+\frac{1}{\left|\sqrt{h-1}-1\right|}\)

Tại h = 3 thì giá trị của D là:

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

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

Bn tự vẽ hình nha:)

Mà đề cho BH,CH rồi thì tính toán rì nữa-.-?

1) Ta có: \(\hept{\begin{cases}AH^2=BH.CH=2.8=16\\BC=BH+CH=2+8\end{cases}}\Rightarrow\hept{\begin{cases}AH=4\left(cm\right)\\BC=10\left(cm\right)\end{cases}}\)

2) Ta có: \(\hept{\begin{cases}AH^2=BH.CH=2.2=4\\BC=BH+CH=2+2\end{cases}}\Rightarrow\hept{\begin{cases}AH=2\left(cm\right)\\BC=4\left(cm\right)\end{cases}}\)

3) Ta có: \(\hept{\begin{cases}AH^2=BH.CH=\sqrt{5}.4\sqrt{5}=\left(2\sqrt{5}\right)^2\\BC=BH+CH=\sqrt{5}+4\sqrt{5}\end{cases}}\Rightarrow\hept{\begin{cases}AH=2\sqrt{5}\left(cm\right)\\BC=5\sqrt{5}\left(cm\right)\end{cases}}\)

Nếu đề bắt tính cả AB,AC thì ib mk làm lại cho:)

lớn hơn hay = thế ạ

Ta có :

\(a^2b+b^2c+c^2a\ge\frac{9a^2b^2c^2}{1+2a^2b^2c^2}\)

\(\Leftrightarrow\left(a^2b+b^2c+c^2a\right)\left(1+2a^2b^2c^2\right)\ge9a^2b^2c^2\)

\(\Leftrightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^{3v}+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)(*)

Áp dụng BĐT AM-GM ta có:

\(a^2b+a^4b^3c^2+a^3b^2c^4\ge3\sqrt[3]{a^9b^6c^6}=3a^3b^2c^2\)

\(b^2c+a^2b^4c^3+a^4b^3c^2\ge3a^2b^3c^2\)

\(c^2a+a^3b^2c^4+a^2b^4c^4\ge3a^2b^2c^3\)

Cộng theo vế

\(\Rightarrow a^2b+b^2c+c^2a+2a^4b^3c^2+2a^2b^4c^3+2a^3b^2c^4\ge3a^2b^2c^2\left(a+b+c\right)\)

Vậy $(*)$ đúng

Do đó ta có đpcm

#Cừu

Đặt \(\hept{\begin{cases}S_{OAB}=x\\S_{OBC}=y\\S_{OCA}=z\end{cases}}\)

Có: \(\frac{OA}{OD}=\frac{S_{AOB}}{S_{ODB}}=\frac{S_{AOC}}{S_{ODC}}=\frac{x+z}{y}\)

\(\Rightarrow\frac{R}{OD}=\frac{x+z}{y}\)

\(\Rightarrow OD=R.\frac{y}{x+z}\)

Tương tự, có: \(OD+OE+OF=R\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)

\(\ge\frac{3}{2}R.\)(BĐT Nesbitt)

Bài làm:

đk: \(x>0;x\ne9;x\ne25\)

Ta có: \(\left(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}+\frac{36}{9-x}\right)\div\left(\frac{\sqrt{x}-5}{3\sqrt{x}-x}\right)\)

\(=\left[\frac{\left(3+\sqrt{x}\right)^2-\left(3-\sqrt{x}\right)^2+36}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\div\left[\frac{\sqrt{x}-5}{\left(3-\sqrt{x}\right)\sqrt{x}}\right]\)

\(=\frac{12\sqrt{x}+36}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\frac{\left(3-\sqrt{x}\right)\sqrt{x}}{\sqrt{x}-5}\)

\(=\frac{12\left(3+\sqrt{x}\right)}{3+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{12\sqrt{x}}{\sqrt{x}-5}\)

đk: \(a\ge0;a\ne1\)

Ta có:

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

\(B=\frac{1}{2\left(1+\sqrt{a}\right)}+\frac{1}{2\left(1-\sqrt{a}\right)}-\frac{a^2+2}{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)

\(B=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)+\left(1+\sqrt{a}\right)\left(a+\sqrt{a}+1\right)-2\left(a^2+2\right)\left(1+\sqrt{a}\right)}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)

\(B=\frac{2a+2\sqrt{a}+2-2a^2\sqrt{a}-2a^2-4-4\sqrt{a}}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)

\(B=\frac{-2a^2\sqrt{a}-2a^2+2a-2\sqrt{a}-2}{2\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)

\(B=\frac{-a^2\sqrt{a}-a^2+a-\sqrt{a}-1}{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}\)

Tại \(a=\sqrt{2}\) thì giá trị của B là:

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

\(B\approx3,45267\)

\(ĐKXĐ:x>1\)

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

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

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

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

\(=\frac{-\left(a^2+a+1\right)}{\left(a-1\right)\left(a^2+a+1\right)}+\frac{a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-a^2-a-1+a^2+2}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)

\(=\frac{-\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}=\frac{-1}{a^2+a+1}\)

Với \(a=\sqrt{2}\)( thỏa mãn ĐKXĐ ), ta có:

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

:(

\(A=\frac{3+x^2}{y+z}+\frac{3+y^2}{z+x}+\frac{3+z^2}{x+y}\)

\(=3\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)+\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)\)

\(\ge3\cdot\frac{9}{2\left(x+y+z\right)}+\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}\)

\(=\frac{27}{2\cdot3}+\frac{3}{2}=6\)

Đẳng thức xảy ra tại x=y=z=1