ĐK: \(x\ge-1;y\ge3;z\ge1\)

\(\sqrt{x+1}+\sqrt{y-3}+\sqrt{z-1}\le\frac{x+1+1+y-3+1+z-1+1}{2}=\frac{x+y+z}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x=0\\y=4\\z=2\end{cases}\left(tm\right)}\)

Đặt: \(P=\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)

=> \(2P=2\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)

\(2P=\left(\sqrt{2}.\sqrt{2+\sqrt{3}}-\sqrt{2}.\sqrt{3+\sqrt{5}}\right)^2\)

\(2P=\left(\sqrt{4+2\sqrt{3}}-\sqrt{6+2\sqrt{5}}\right)^2\)

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

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

\(2P=\left(\sqrt{3}-\sqrt{5}\right)^2=3+5-2\sqrt{15}=8-2\sqrt{15}\)

=> \(P=4-\sqrt{15}\)

AB.AC = BC.AH ( hệ thức trong tam giác vuông )
<=> AB²AC² = BC²AH²
<=> AH² = AB²AC² / BC²
<=> AH² = AB²AC² / AB²+AC² ( Tính chất Pytago )
<=> 1/AH² = AB²+AC² / AB²AC²
<=> 1/AH² = 1/AB² + 1/AC²

=> đpcm

\(ĐKXĐ:x\ge1\)

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

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

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

\(\Leftrightarrow2x-2\sqrt{x^2-1}=4\)

\(\Leftrightarrow2\sqrt{x^2-1}=2x-4\)

\(\Leftrightarrow4.\left(x^2-1\right)=\left(2x-4\right)^2\)

\(\Leftrightarrow4x^2-4=4x^2-16x+16\)

\(\Leftrightarrow4x^2-4x^2+16x=16+4\)

\(\Leftrightarrow16x=20\)

\(\Leftrightarrow x=\frac{5}{4}\)(t/m ĐKXĐ)

<=> x - 1 - x - 1 + 2.\(\sqrt{\left(x-1\right)\left(x+1\right)}\) = 4

<=> 2.\(\sqrt{x^2-1}\)= 6

<=> \(\sqrt{x^2-1}=3\)

<=> \(\sqrt{x^2-1}=\sqrt{9}\)

<=> x² - 1 = 9

<=> x² = 10

<=> x = \(\pm\sqrt{10}\)

a, Năng lượng giao động:

Ta có : \(E=\frac{1}{2}kA^2=500mJ\)

b,Vận tốc:

Vận tốc lớn nhất của vật đc tính như sau :

\((E_đ)_{max}=E=\frac{1}{2}mv^2_{max}\Rightarrow v_{max}=\sqrt{\frac{2E}{m}}=1,00m.s^{-1}\)

Khi vận tốc lớn nhất, thế năng nhỏ nhất . Ta có :

              \(E_t=0\Leftrightarrow x=0\): vị trí cân bằng

c, Vị trí vật tại đó : E_d = E_t

Ta có :         2E_t = E           => \(kx^2=\frac{1}{2}kA^2\)

\(\Rightarrow\)                 \(x=\pm\frac{A}{\sqrt{2}}=\pm0,5.\sqrt{2}\approx\pm7,0\left(cm\right)\)

\(a,\)\(A=\left(\frac{x-2\sqrt{3x}+3}{x-3}\right)\left(\sqrt{4x}+\sqrt{12}\right).\)

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

\(=\frac{\left(\sqrt{x}-\sqrt{3}\right)^22\left(\sqrt{x}+\sqrt{3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

\(b,x=4-2\sqrt{3}\)\(=3-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)

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

\(\Rightarrow A=2\left(\sqrt{x}-\sqrt{3}\right)=2\left(\sqrt{3}-1-\sqrt{3}\right)=2.\left(-1\right)=-2\)